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17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)
LP
2
MT
2 tháng 7 2015
a/ (x-5)^2-49=0
<=>(x-5)2-72
<=>(x-5-7)(x-5+7)=0
<=>(x-12)(x+2)=0
<=>x-12=0 hoặc x+2=0
<=>x=12 hoặc x=-2
vậy x=12 hoặc x=-2
b/ (x+11)^2=121
<=>(x+11)2-121=0
<=>(x+11)2-112=0
<=>(x+11-11)(x+11+11)=0
<=>x(x+22)=0
<=>x=0 hoặc x+22=0
<=>x=0 hoặc x=-22
vậy x=0 hoặc x=-22
c/ x.(x+7)-6x-42=0
<=>x2+7x-6x-42=0
<=>x2+x-42=0
<=>x2-6x+7x-42=0
<=>x(x-6)+7(x-6)=0
<=>(x-6)(x-7)=0
<=>x-6=0 hoặc x-7=0
<=>x=6 hoặc x=7
vậy x=6;7
d/ x^4-2x^3+10x^2-20x=0
<=>x3(x-2)+10x(x-2)=0
<=>(x-2)(x3+10x)=0
<=>(x-2)x(x2+10)=0
<=>x-2=0 hoặc x=0 hoặc x2+10=0
<=>x=2 hoặc x=0 hoặc x2=-10(vô lí)
vậy x=2;0
AM
2 tháng 7 2015
a)(x-5)2-49=0
<=>(x-5-7)(x-5+7)=0
<=>(x-12)(x+2)=0
<=>x-12=0 hoặc x+2=0
<=>x=12 hoặc x=-2
b)(x+11)2=121
<=>(x+11)2-121=0
<=>(x+11-11)(x+11+11)=0
<=>x(x+22)=0
<=>x=0 hoặc x+22=0
<=>x=0 hoặc x=-22
c)x(x+7)-6x-42=0
<=>x(x+7)-(6x+42)=0
<=>x(x+7)-6(x+7)=0
<=>(x+7)(x-6)=0
<=>x+7=0 hoặc x-6=0
<=>x=-7 hoặc x=6
d)x4-2x3+10x2-20x=0
<=>x(x3-2x2+10x-20)=0
<=>x[(x3-2x2)+(10x-20)]=0
<=>x[x2(x-2)+10(x-2)]=0
<=>x(x-2)(x2+10)=0
Do x2>0=>x2+10>0
=>x(x-2)=0
<=>x=0 hoặc x-2=0
<=>x=0 hoặc x=2
23 tháng 9 2021
a: \(x^2+3y^2-4x+6y+7=0\)
\(\Leftrightarrow x^2-4x+4+3y^2+6y+3=0\)
\(\Leftrightarrow\left(x-2\right)^2+3\left(y+1\right)^2=0\)
\(\Leftrightarrow\left(x,y\right)=\left(-2;1\right)\)
CM
11 tháng 9 2019
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
25 tháng 3 2020
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
29 tháng 10 2021
a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
25 tháng 6 2023
a: x^2+10x+100
=x^2+10x+25+75=(x+5)^2+75>0 với mọi x
b: -x^2+4x-100
=-(x^2-4x+100)
=-(x^2-4x+4+96)
=-(x-2)^2-96<0 với mọi x
c: x^2-5x+6
=x^2-5x+25/4-1/4
=(x-5/2)^2-1/4 chưa chắc lớn hơn 0 đâu nha bạn
a) \(x^2+10x=0\)
\(x\left(x+10\right)=0\)
\(\left[{}\begin{matrix}x=0\\x=-10\end{matrix}\right.\)
b) \(\left(x-7\right)^3=x-7\)
\(\left(x-7\right)^3-\left(x-7\right)=0\)
\(\left(x-7\right)\left[\left(x-7\right)^2-1\right]=0\)
\(\left(x-7\right)\left(x-7-1\right)\left(x-7+1\right)=0\)
\(\left(x-7\right)\left(x-8\right)\left(x-6\right)=0\)
\(\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)
c) \(x^2-20x+100=0\)
\(x^2-10x-10x+100=0\)
\(x\left(x-10\right)-10\left(x-10\right)=0\)
\(\left(x-10\right)\left(x-10\right)=0\)
\(\left(x-10\right)^2=0\)
=> x = 10
a) \(x^2+10x=0\)
\(\Leftrightarrow x\left(x+10\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+10=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-10\end{matrix}\right.\)
Vậy..
b) \(\left(x-7\right)^3=\left(x-7\right)\)
\(\Leftrightarrow\left(x-7\right)^3-\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left[\left(x-7\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-8\right)\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-7=0\\x-8=0\\x-6=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)
Vậy..
c) \(x^2-20x+100=0\)
\(\Leftrightarrow\left(x-10\right)^2=0\)
\(\Rightarrow x-10=0\)
\(\Rightarrow x=10\)