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10 tháng 9 2017
a/ \(\dfrac{1}{3}-\dfrac{2}{5}+3x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-1}{15}+3x=\dfrac{3}{4}\)
\(\Leftrightarrow3x=\dfrac{49}{60}\)
\(\Leftrightarrow x=\dfrac{49}{180}\)
Vậy....
b/ \(\dfrac{3}{2}-1+4x=\dfrac{2}{3}-7x\)
\(\Leftrightarrow\dfrac{1}{2}+4x=\dfrac{2}{3}-7x\)
\(\Leftrightarrow4x+7x=\dfrac{2}{3}-\dfrac{1}{2}\)
\(\Leftrightarrow11x=\dfrac{1}{6}\)
\(\Leftrightarrow x=\dfrac{1}{66}\)
Vậy....
c/ \(2\left(\dfrac{3}{4}-5x\right)=\dfrac{4}{5}-3x\)
\(\Leftrightarrow\dfrac{3}{2}-10x=\dfrac{4}{5}-3x\)
\(\Leftrightarrow-10x+3x=\dfrac{4}{5}-\dfrac{3}{2}\)
\(\Leftrightarrow-7x=-\dfrac{7}{10}\)
\(\Leftrightarrow x=-\dfrac{1}{10}\)
Vậy .....
10 tháng 9 2017
d/ \(4\left(\dfrac{1}{2}-x\right)-5\left(x-\dfrac{3}{10}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow2-4x-5x-\dfrac{3}{2}=\dfrac{7}{4}\)
\(\Leftrightarrow2+\left(-4x\right)+\left(-5x\right)+\left(\dfrac{-3}{2}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow-9x+\dfrac{1}{2}=\dfrac{7}{4}\)
\(\Leftrightarrow-9x=\dfrac{5}{4}\)
\(\Leftrightarrow x=-\dfrac{5}{36}\)
NT
2.(x-3)+3x+0.5=\(\dfrac{3}{4}\)
4x+2+4x=272
(1,2-5x).(2\(\dfrac{1}{8}\) +1/2 x)=0
GIÚP MÌNH VỚI !!!!
2
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
30 tháng 9 2023
\(2\left(x-3\right)+3x+0,5=\dfrac{3}{4}\\ \Leftrightarrow2x-6+3x+\dfrac{1}{2}=\dfrac{3}{4}\\ \Leftrightarrow x\left(2+3\right)=\dfrac{3}{4}-\dfrac{1}{2}+6\\ \Leftrightarrow5x=\dfrac{25}{4}\\ \Leftrightarrow x=\dfrac{25}{4}:5=\dfrac{5}{4}\\ ---\\ 4^{x+2}+4^x=272\\ \Leftrightarrow4^x\left(4^2+1\right)=272\\ \Leftrightarrow4^x.17=272\\ \Leftrightarrow4^x=\dfrac{272}{17}=16=4^2\\ Vậy:x=2\\ ----\\ \left(1,2-5x\right)\left(2\dfrac{1}{8}+\dfrac{1}{2}x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}1,2-5x=0\\2,125+0,5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=1,2\\0,5x=-2,125\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1,2}{5}=0,24\\x=\dfrac{-2,125}{0,5}=-4,25\end{matrix}\right.\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 10 2023
a) \(2\left(x-3\right)+3x+0,5=\dfrac{3}{4}\)
\(\Rightarrow2x-6+3x+\dfrac{1}{2}=\dfrac{3}{4}\)
\(\Rightarrow5x-6=\dfrac{3}{4}-\dfrac{1}{2}\)
\(\Rightarrow5x-6=\dfrac{1}{4}\)
\(\Rightarrow5x=\dfrac{1}{4}+6\)
\(\Rightarrow5x=\dfrac{25}{4}\)
\(\Rightarrow x=\dfrac{25}{4}:5\)
\(\Rightarrow x=\dfrac{5}{4}\)
b) \(4^{x+2}+4^x=272\)
\(\Rightarrow4^x\cdot4^2+4^x\cdot1=272\)
\(\Rightarrow4^x\cdot\left(16+1\right)=272\)
\(\Rightarrow4^x\cdot17=272\)
\(\Rightarrow4^x=16\)
\(\Rightarrow4^x=4^2\)
\(\Rightarrow x=2\)
c) \(\left(1,2-5x\right)\left(2\dfrac{1}{8}+\dfrac{1}{2}x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}1,2-5x=0\\\dfrac{15}{8}+\dfrac{1}{2}x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}5x=1,2\\\dfrac{1}{2}x=-\dfrac{15}{8}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1,2}{5}\\x=-\dfrac{15}{8}:\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{25}\\x=-\dfrac{15}{4}\end{matrix}\right.\)
8 tháng 3 2023
a: =1/2x^3*x^2-1/2x^3*6x-1/2x^3*10
=1/2x^5-3x^4-5x^3
b: =-3x^2*5x^3+3x^2*4x^2-3x^2*3x+3x^2*3x
=-15x^5+12x^4-9x^3+9x^2
c: \(=3x\cdot5x^2-3x\cdot2x-3x=15x^3-6x^2-3x\)
d: \(=\dfrac{1}{2}x^2y\cdot2x^3-\dfrac{1}{2}x^2y\cdot\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)
TD
30 tháng 1 2022
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
=>\(\dfrac{4\left(3x-2y\right)}{4.4}=\dfrac{3\left(2z-4x\right)}{3.3}=\dfrac{2\left(4y-3z\right)}{2.2}\)
=>\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có
\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\)
=>\(\dfrac{12x-8y}{16}=0\)
=>12x-8y=0
=>12x=8y
=>\(\dfrac{12x}{24}=\dfrac{8y}{24}\)
=>\(\dfrac{x}{2}=\dfrac{y}{3}\)(1)
Lại có \(\dfrac{8y-6z}{4}=0\)
=>8y-6z=0
=>8y=6z
=>\(\dfrac{8y}{24}=\dfrac{6z}{24}\)
=>\(\dfrac{y}{3}=\dfrac{z}{4}\)(2)
từ (1) và (2)=>\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)\)
3 tháng 3 2022
\(a.ĐK:x\ne3;1\)
\(\Rightarrow\dfrac{1}{2\left(x-3\right)}+\dfrac{3x-10}{\left(x-1\right)\left(x-3\right)}=\dfrac{7}{2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)+2\left(3x-10\right)}{2\left(x-1\right)\left(x-3\right)}=\dfrac{7\left(x-1\right)\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow x-1+2\left(3x-10\right)=7\left(x-1\right)\left(x-3\right)\)
\(\Leftrightarrow x-1+6x-20=7\left(x^2-4x+3\right)\)
\(\Leftrightarrow7x-21=7x^2-28x+21\)
\(\Leftrightarrow7x^2-35x+42=0\)
\(\Leftrightarrow7\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)
b.\(ĐK:x\ne2;4\)
\(\Rightarrow\dfrac{x-1}{x-2}-\dfrac{x+3}{4-x}=\dfrac{2}{\left(x-2\right)\left(4-x\right)}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(4-x\right)-\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(4-x\right)}=\dfrac{2}{\left(x-2\right)\left(4-x\right)}\)
\(\Leftrightarrow\left(x-1\right)\left(4-x\right)-\left(x+3\right)\left(x-2\right)=2\)
\(\Leftrightarrow4x-x^2-4+x-x^2+2x-3x+6-2=0\)
\(\Leftrightarrow-2x^2+4x=0\)
\(\Leftrightarrow-2x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)
3 tháng 3 2022
a: \(\Leftrightarrow\dfrac{1}{2\left(x-3\right)}+\dfrac{3x-10}{\left(x-1\right)\left(x-3\right)}=\dfrac{7}{2}\)
\(\Leftrightarrow x-1+2\left(3x-10\right)=7\left(x-1\right)\left(x-3\right)\)
\(\Leftrightarrow7\left(x^2-4x+3\right)=x-1+6x-20=7x-21\)
\(\Leftrightarrow\left(x-3\right)\left(7x-7\right)-7\left(x-3\right)=0\)
=>(x-3)(7x-14)=0
=>x=3(loại) hoặc x=2(nhận)
b: \(\Leftrightarrow\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)=-2\)
\(\Leftrightarrow x^2-5x+4+x^2+x-6=-2\)
\(\Leftrightarrow2x^2-4x=0\)
=>2x(x-2)=0
=>x=0(nhận) hoặc x=2(loại)
20 tháng 9 2017
a/ \(\left(4x-5\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-5=0\\3x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy ............
b/ \(\dfrac{x+1}{2016}+\dfrac{x+2}{2015}=\dfrac{x+3}{2014}+\dfrac{x+4}{2013}\)
\(\Leftrightarrow\left(\dfrac{x+1}{2016}+1\right)+\left(\dfrac{x+2}{2015}+1\right)=\left(\dfrac{x+3}{2014}+1\right)+\left(\dfrac{x+4}{2013}+1\right)\)
\(\Leftrightarrow\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}=\dfrac{x+2017}{2014}+\dfrac{x+2017}{2013}\)
\(\Leftrightarrow\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}-\dfrac{x+2017}{2014}-\dfrac{x+2017}{2013}=0\)
\(\Leftrightarrow x+2017\left(\dfrac{1}{2016}+\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\right)=0\)
Mà \(\dfrac{1}{2016}+\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\ne0\)
\(\Leftrightarrow x+2017=0\)
\(\Leftrightarrow x=-2017\)
Vậy ..
LN
20 tháng 9 2017
\(\left(4x-5\right)\left(3x+2\right)=0\)
\(\)\(\Rightarrow\left[{}\begin{matrix}4x-5=0\\3x+2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
\(\dfrac{x+1}{2016}+\dfrac{x+2}{2015}=\dfrac{x+3}{2014}+\dfrac{x+4}{2013}\)
\(\Rightarrow\dfrac{x+1}{2016}+1+\dfrac{x+2}{2015}+1=\dfrac{x+3}{2014}+1+\dfrac{x+4}{2013}+1\)
\(\Rightarrow\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}=\dfrac{x+2017}{2014}+\dfrac{x+2017}{2013}\)
\(\Rightarrow\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}-\dfrac{x+2017}{2014}-\dfrac{x+2017}{2013}=0\)
\(\Rightarrow\left(x+2017\right)\left(\dfrac{1}{2016}+\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\right)=0\)
Vì \(\dfrac{1}{2016}+\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\ne0\)
Nên:
\(x+2017=0\Rightarrow x=-2017\)
ND
Nguyễn Đức Trí
VIP
21 tháng 8 2023
a) \(P=\dfrac{2x+5}{x+3}\inℤ\left(x\inℤ;x\ne-3\right)\)
\(\Rightarrow2x+5⋮x+3\)
\(\Rightarrow2x+5-2\left(x+3\right)⋮x+3\)
\(\Rightarrow2x+5-2x-6⋮x+3\)
\(\Rightarrow-1⋮x+3\)
\(\Rightarrow x+3\in\left\{-1;1\right\}\)
\(\Rightarrow x\in\left\{-4;-2\right\}\)
b) \(P=\dfrac{3x+4}{x+1}\inℤ\left(x\inℤ;x\ne-1\right)\)
\(\Rightarrow3x+4⋮x+1\)
\(\Rightarrow3x+4-3\left(x+1\right)⋮x+1\)
\(\Rightarrow3x+4-3x-3⋮x+1\)
\(\Rightarrow1⋮x+1\)
\(\Rightarrow x+1\in\left\{-1;1\right\}\)
\(\Rightarrow x\in\left\{-2;0\right\}\)
c) \(P=\dfrac{4x-1}{2x+3}\inℤ\left(x\inℤ;x\ne-\dfrac{3}{2}\right)\)
\(\Rightarrow4x-1⋮2x+3\)
\(\Rightarrow4x-1-2\left(2x+3\right)⋮2x+3\)
\(\Rightarrow4x-1-4x-6⋮2x+3\)
\(\Rightarrow-7⋮2x+3\)
\(\Rightarrow2x+3\in\left\{-1;1;-7;7\right\}\)
\(\Rightarrow x\in\left\{-2;-1;-5;2\right\}\)
21 tháng 8 2023
a) P=\(\dfrac{2x+5}{x+3}=\dfrac{2\left(x+3\right)-2}{x+3}=\dfrac{2\left(x+3\right)}{x+3}-\dfrac{2}{x+3}=2-\dfrac{2}{x+3}\)
để \(P\inℤ\) thì \(\dfrac{2}{x+3}\inℤ\) hay 2 ⋮ (x-3) ⇒x+3 ϵ Ư2= (2,-2,1,-1)
ta có bảng sau:
| x+3 | 2 | -2 | 1 | -1 |
| x | -1 | -5 | -2 | -4 |
Vậy x \(\in-1,-2,-5,-4\)
DT
Đỗ Thanh Hải
CTVVIP
8 tháng 8 2017
a) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\)
\(20x^2-16x-34=10x^2+3x-34\)
\(10x^2-19x=0\)
\(x\left(10x-19\right)=0\)
\(\Leftrightarrow x=0\)
hoặc \(10x-19=0\)
\(\Leftrightarrow x=\dfrac{19}{10}\)
Vạy ..............
b) \(\dfrac{x-1}{x+5}=\dfrac{6}{7}\)
\(\Leftrightarrow1-\dfrac{x-1}{x+5}=1-\dfrac{6}{7}\)
\(\Leftrightarrow\dfrac{x+5}{x+5}-\dfrac{x-1}{x+5}=\dfrac{7}{7}-\dfrac{6}{7}\)
\(\Leftrightarrow\dfrac{\left(x+5\right)-\left(x-1\right)}{x+5}=\dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{x+5-x+1}{x+5}=\dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{\left(x-x\right)+\left(5+1\right)}{x+5}=\dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{6}{x+5}=\dfrac{1}{7}\)
\(\Leftrightarrow x+5=42\)
\(\Leftrightarrow x=37\)
24 tháng 9 2022
1: \(\Leftrightarrow3x+4=2\)
=>3x=-2
=>x=-2/3
2: \(\Leftrightarrow7x-7=6x-30\)
=>x=-23
3: =>\(5x-5=3x+9\)
=>2x=14
=>x=7
4: =>9x+15=14x+7
=>-5x=-8
=>x=8/5
2.(\(\dfrac{1}{4}\) - 3\(x\)) = \(\dfrac{1}{5}\) - 4\(x\)
\(\dfrac{1}{2}\) - 6\(x\) = \(\dfrac{1}{5}\) - 4\(x\)
- 4\(x\) + 6\(x\) =\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\)
2\(x\) = \(\dfrac{3}{10}\)
\(x\) = \(\dfrac{3}{10}\): 2
\(x=\dfrac{3}{20}\)
Vậy \(x=\dfrac{3}{20}\)