tim x:
a. \(3\dfrac{1}{3}x+16=13,25\)
b. \(x-43=\left(57-x\right)-50\)
K
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23 tháng 1 2021
a) Ta có: \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\)
\(\Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90}{15}-\dfrac{5\left(1-2x\right)}{15}\)
\(\Leftrightarrow3x-9=90-5+10x\)
\(\Leftrightarrow3x-9=10x+85\)
\(\Leftrightarrow3x-10x=85+9\)
\(\Leftrightarrow-7x=94\)
hay \(x=-\dfrac{94}{7}\)
Vậy: \(S=\left\{-\dfrac{94}{7}\right\}\)
b) Ta có: \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)
\(\Leftrightarrow\dfrac{2\left(3x-2\right)}{12}-\dfrac{60}{12}=\dfrac{3\left(3-2x-14\right)}{12}\)
\(\Leftrightarrow6x-4-60=9-6x-42\)
\(\Leftrightarrow6x-64=-6x-33\)
\(\Leftrightarrow6x+6x=-33+64\)
\(\Leftrightarrow12x=31\)
hay \(x=\dfrac{31}{12}\)
Vậy: \(S=\left\{\dfrac{31}{12}\right\}\)
c) Ta có: \(3\left(x-1\right)+3=5x\)
\(\Leftrightarrow3x-3+3=5x\)
\(\Leftrightarrow3x-5x=0\)
\(\Leftrightarrow-2x=0\)
hay x=0
Vậy: S={0}
d) Ta có: \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)
\(\Leftrightarrow\dfrac{x+1}{100}+1+\dfrac{x+2}{99}+1=\dfrac{x+3}{98}+1+\dfrac{x+4}{97}+1\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}=\dfrac{x+101}{98}+\dfrac{x+101}{97}\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\)
\(\Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\)
mà \(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\ne0\)
nên x+101=0
hay x=-101
Vậy: S={-101}
23 tháng 1 2021
a) \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\\ \Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90-5\left(1-2x\right)}{15}\\ \Leftrightarrow3x-9=90-5+10x\\ \Leftrightarrow3x-10x=90-5+9\\ \Leftrightarrow-7x=94\\ \Leftrightarrow x=\dfrac{-94}{7}\)
Vậy \(x=\dfrac{-94}{7}\) là nghiệm của pt
b) \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\\ \Leftrightarrow\dfrac{2\left(3x-2\right)-60}{12}=\dfrac{9-6\left(x+7\right)}{12}\\ \Leftrightarrow6x-4-60=9-6x-42\\ \Leftrightarrow6x+6x=9-42+4+60\\ \Leftrightarrow12x=31\\ \Leftrightarrow x=\dfrac{31}{12}\)
Vậy \(x=\dfrac{31}{12}\) là nghiệm của pt
c) \(3\left(x-1\right)+3=5x\\ \Leftrightarrow3x+3+3=5x\\ \Leftrightarrow5x-3x=3+3\\ \Leftrightarrow2x=6\\ \Leftrightarrow x=3\)
Vậy x = 3 là nghiệm của pt
d) \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\\ \Leftrightarrow\left(\dfrac{x+1}{100}+1\right)+\left(\dfrac{x+2}{99}+1\right)=\left(\dfrac{x+3}{98}+1\right)+\left(\dfrac{x+4}{97}+1\right)\\ \Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\\ \Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\\ \Leftrightarrow x+101=0\\ \Leftrightarrow x=-101\)
Vậy x = -101 là nghiệm của pt
e) \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}=-4\\ \Leftrightarrow\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{53-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)=0\\ \Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}=0\\ \Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}\right)=0\\ \Leftrightarrow100-x=0\\ \Leftrightarrow x=100\)
Vậy x = 100 là nghiệm của pt
f) \(\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\\ \Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)=0\\ \Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\\ \Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\\ \Leftrightarrow x-100=0\\ \Leftrightarrow x=100\)
Vậy x = 100 là nghiệm của pt
28 tháng 2 2022
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
8 tháng 3 2017
b)\(\dfrac{x+14}{86}+\dfrac{x+15}{85}+\dfrac{x+16}{84}+\dfrac{x+17}{83}+\dfrac{x+116}{4}=0\)
\(\Leftrightarrow\dfrac{x+14}{86}+1+\dfrac{x+15}{85}+1+\dfrac{x+16}{84}+1+\dfrac{x+17}{83}+1+\dfrac{x+116}{4}-4=0\)
\(\Leftrightarrow\dfrac{x+100}{86}+\dfrac{x+100}{85}+\dfrac{x+100}{84}+\dfrac{x+100}{83}+\dfrac{x+100}{4}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{86}+\dfrac{1}{85}+\dfrac{1}{84}+\dfrac{1}{83}+\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x+100=0\).Do \(\dfrac{1}{86}+\dfrac{1}{85}+\dfrac{1}{84}+\dfrac{1}{83}+\dfrac{1}{4}\ne0\)
\(\Leftrightarrow x=-100\)
c)\(\dfrac{1}{\left(x^2+5\right)\left(x^2+4\right)}+\dfrac{1}{\left(x^2+4\right)\left(x^2+3\right)}+\dfrac{1}{\left(x^2+3\right)\left(x^2+2\right)}+\dfrac{1}{\left(x^2+2\right)\left(x^2+1\right)}=-1\)
\(\Leftrightarrow\dfrac{1}{\left(x^2+1\right)\left(x^2+2\right)}+\dfrac{1}{\left(x^2+2\right)\left(x^2+3\right)}+...+\dfrac{1}{\left(x^2+4\right)\left(x^2+5\right)}=-1\)
\(\Leftrightarrow\dfrac{1}{x^2+1}-\dfrac{1}{x^2+2}+\dfrac{1}{x^2+2}-\dfrac{1}{x^2+3}+...+\dfrac{1}{x^2+4}-\dfrac{1}{x^2+5}=-1\)
\(\Leftrightarrow\dfrac{1}{x^2+1}-\dfrac{1}{x^2+5}=-1\)\(\Leftrightarrow\dfrac{4}{x^4+6x^2+5}=-1\)
\(\Leftrightarrow\dfrac{x^4+6x^2+9}{x^4+6x^2+5}=0\Leftrightarrow x^4+6x^2+9=0\)
\(\Leftrightarrow\left(x^2+3\right)^2>0\forall x\) (vô nghiệm)
K
14 tháng 1 2019
a, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\Leftrightarrow\left(\dfrac{59-x}{49}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)
\(\Leftrightarrow\dfrac{100-x}{45}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\)
\(\Leftrightarrow\left(100-x\right).\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\)
Mà \(\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)\ne0\)
\(\Rightarrow100-x=0\)
\(\Rightarrow x=100\)
Vậy \(S=\left\{100\right\}\)
K
14 tháng 1 2019
b, \(6x^2-5x+3=2x-3x\left(3-2x\right)\)
\(\Leftrightarrow6x^2-5x+3=2x-9x+6x^2\)
\(\Leftrightarrow6x^2-5x+3=-7x+6x^2\)
\(\Leftrightarrow6x^2-5x+3+7x-6x^2=0\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow2x=-3\)
\(\Leftrightarrow x=\dfrac{-3}{2}\)
Vậy \(S=\left\{\dfrac{-3}{2}\right\}\)
DC
6 tháng 3 2022
a) \(x-\dfrac{5}{6}=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}+\dfrac{5}{6}\)
\(x=\dfrac{4}{3}\)
vậy x = ....
b) \(3\dfrac{1}{3}x+16\dfrac{3}{4}=-13,25\)
\(\dfrac{10}{3}\)\(x+\dfrac{67}{4}=-13,25\)
\(\dfrac{10}{3}x=\left(-13,25\right)-\dfrac{67}{4}\)
\(\dfrac{10}{3}x=-30\)
\(x=\left(-30\right):\dfrac{10}{3}\)
\(x=-9\)
vậy x =...
sai mog bn thông cảm!!!
AH
Akai Haruma
Giáo viên
11 tháng 12 2017
Bài 1:
Đặt \(\left\{\begin{matrix} 5x+3=a\\ 2x+4=b\end{matrix}\right.\) \(\Rightarrow 3x-1=a-b\)
PT trở thành:
\(a^3-b^3=(a-b)^3\)
\(\Leftrightarrow (a-b)(a^2+ab+b^2)=(a-b)^3\)
\(\Leftrightarrow (a-b)[a^2+ab+b^2-(a^2-2ab+b^2)]=0\)
\(\Leftrightarrow 3ab(a-b)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=0\\b=0\\a=b\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{5}\\x=-2\\5x+3=2x+4\Leftrightarrow x=\dfrac{1}{3}\end{matrix}\right.\)
Thử lại thấy đều thỏa mãn
Vậy \(x\in\left\{\frac{-3}{5};-2;\frac{1}{3}\right\}\)
AH
Akai Haruma
Giáo viên
11 tháng 12 2017
Bài 2:
\(\frac{x-1}{2013}+\frac{x-2}{2012}-\frac{x-3}{2011}=\frac{x-4}{2010}\)
\(\Leftrightarrow \frac{x-1}{2013}-1+\frac{x-2}{2012}-1-\left(\frac{x-3}{2011}-1\right)=\frac{x-4}{2010}-1\)
\(\Leftrightarrow \frac{x-2014}{2013}+\frac{x-2014}{2012}-\frac{x-2014}{2011}=\frac{x-2014}{2010}\)
\(\Leftrightarrow (x-2014)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\) (1)
Thấy rằng \(2013> 2011; 2012> 2010\Rightarrow \frac{1}{2013}< \frac{1}{2011}; \frac{1}{2012}< \frac{1}{2010}\)
\(\Rightarrow \frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}< 0\) (2)
Từ (1),(2) suy ra \(x-2014=0\Leftrightarrow x=2014\)
Bài 3:
Đặt \(\left\{\begin{matrix} 2x-5=a\\ x-2=b\end{matrix}\right.\Rightarrow x-3=a-b\)
PT trở thành: \(a^3-b^3=(a-b)^3\)
\(\Leftrightarrow (a-b)(a^2+ab+b^2)-(a-b)(a^2-2ab+b^2)=0\)
\(\Leftrightarrow 3ab(a-b)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=0\\b=0\\a-b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=2\\x-3=0\Leftrightarrow x=3\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{5}{2}; 2; 3\right\}\)
23 tháng 1
a: \(\left\{{}\begin{matrix}\dfrac{x}{35}-y=2\\y-\dfrac{x}{50}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x-35y}{35}=2\\\dfrac{50y-x}{50}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-35y=70\\-x+50y=50\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}15y=120\\x-35y=70\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=8\\x=70+35y=70+35\cdot8=350\end{matrix}\right.\)
b: ĐKXĐ: x<>0 và y<>0
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{16}\\\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{4}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{3}{y}=\dfrac{3}{16}\\\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{y}=\dfrac{3}{16}-\dfrac{1}{4}=\dfrac{-1}{16}\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{16}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=48\\\dfrac{1}{x}=\dfrac{1}{16}-\dfrac{1}{48}=\dfrac{2}{48}=\dfrac{1}{24}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=24\\y=48\end{matrix}\right.\left(nhận\right)\)
5 tháng 1 2021
a) Ta có: \(\left(x^2-16\right)\left(\dfrac{x}{4}-\dfrac{4x+5}{3}\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\left(\dfrac{3x-16x-20}{12}\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\cdot\left(-13x-20\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\\-13x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\-13x=20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=\dfrac{-20}{13}\end{matrix}\right.\)
Vậy: \(x\in\left\{4;-4;\dfrac{-20}{13}\right\}\)
b) Ta có: \(\left(4x-1\right)\left(x+5\right)=x^2-25\)
\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(4x-1-x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-5;\dfrac{-4}{3}\right\}\)
c) Ta có: \(x\left(x+3\right)^3-\dfrac{x}{4}\cdot\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\cdot\left[x\left(x+3\right)^2-\dfrac{1}{4}x\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left[x\left(x^2+6x+9\right)-\dfrac{1}{4}x\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^3+6x^2+9x-\dfrac{1}{4}x\right)=0\)
\(\Leftrightarrow\left(x+3\right)\cdot x\cdot\left(x^2+6x+\dfrac{35}{4}\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x^2+6x+9-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left[\left(x+3\right)^2-\dfrac{1}{4}\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x+3-\dfrac{1}{2}\right)\left(x+3+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x+\dfrac{5}{2}\right)\left(x+\dfrac{7}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+\dfrac{5}{2}=0\\x+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-3;-\dfrac{5}{2};-\dfrac{7}{2}\right\}\)
a. \(3\dfrac{1}{3}x+16=13,25\)
=> x + 16 = 13,25
=> x = 13,25 - 16
=> x = \(-\dfrac{11}{4}\)
b. x - 43 = (57 - x) - 50
=> 2x = 57 - 50 + 43
=> 2x = 7 + 43
=> 2x = 50
=> x = 50 : 2
=> x = 25
a, \(3\dfrac{1}{3}x\) + 16 = 13,25
\(\dfrac{10}{3}x\) + 16 = 13,25
\(\dfrac{10}{3}x\) = -2,75
x = -2,75 : \(\dfrac{10}{3}\)
x = -\(\dfrac{33}{40}\)
b, x - 43 = ( 57 - x ) - 50
x - 43 = 57 - x - 50
x + x = 57 - 50 + 43
2x = 50
x = 25