(x-1)x-2=(x-1)x+6
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KB
3
DT
Dang Tung
CTVHS
3 tháng 12 2023
\(\left(x+1\right)^2-3\left(x+1\right)=\left(x+1\right)\left(x+1-3\right)=\left(x+1\right)\left(x-2\right)\)
\(2x\left(x-2\right)-\left(x-2\right)^2=\left(x-2\right)\left[2x-\left(x-2\right)\right]=\left(x-2\right)\left(2x-x+2\right)=\left(x-2\right)\left(x+2\right)\)
\(4x^2-20xy+25y^2=\left(2x\right)^2-2.2x.5y+\left(5y\right)^2=\left(2x-5y\right)^2\)
\(x^2+3x-x-3=x\left(x+3\right)-\left(x+3\right)=\left(x-1\right)\left(x+3\right)\)
\(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)
\(2y\left(x+2\right)-3x-6=2y\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(2y-3\right)\)
QN
0
23 tháng 1 2022
5: \(\Leftrightarrow9\left(x^2-5x-4\right)=36\left(x+1\right)+8\left(x^2-10x\right)\)
\(\Leftrightarrow9x^2-45x-36-36x-36-8x^2+80x=0\)
\(\Leftrightarrow x^2-x-72=0\)
=>(x-9)(x+8)=0
=>x=9 hoặc x=-8
6: \(\Leftrightarrow x^2-9=9x-x^2-9+x\)
\(\Leftrightarrow2x^2-10x=0\)
=>2x(x-5)=0
=>x=0 hoặc x=5
23 tháng 1 2022
5, <=> 9x^2 - 45x - 36 = 36x + 36 + 8x^2 - 80x
<=> x^2 - x - 72 = 0 <=> x = 9 ; x = -8
6, <=> x^2 - 9 = 9x - x^2 - 9 + x = 10x - x^2 - 9
<=> 2x^2 - 10x = 0 <=> x = 0 ; x = 5
7, <=> (x-1)^2 = (3x+3)^2
<=> (x-1-3x-3)(x-1+3x+3) = 0
<=> (-2x-4)(4x+2) = 0 <=> x = -2;x=-1/2
8, = (x^2-10x-15)(x^2-10x+25)
11 tháng 8 2018
theo đề ra ta có \(2x=\frac{1}{7}-\frac{1}{4x}=\frac{1}{6}\)
\(\Rightarrow2x=\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{6}:2=\frac{1}{12}\)
TH
2
Y
9 tháng 2 2023
\(\dfrac{\left(x^2+x+1\right)\left(3x+1\right)}{x+2}=\dfrac{x\left(x^2+x+1\right)}{2\left(x+2\right)}\) \(\left(dkxd:x\ne-2\right)\)
\(\Leftrightarrow\dfrac{\left(x^2+x+1\right)\left(3x+1\right)}{x+2}-\dfrac{x\left(x^2+x+1\right)}{2\left(x+2\right)}=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left[2\left(3x+1\right)-x\right]=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(6x+2-x\right)=0\)
Bỏ vế đằng trước \(x^2+x+1=0\) do vô nghiệm
\(\Leftrightarrow6x+2-x=0\)
\(\Leftrightarrow5x=-2\)
\(\Leftrightarrow x=-\dfrac{2}{5}\left(tmdk\right)\)
Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)
25 tháng 7 2020
\(\frac{5}{3}-\frac{1}{3}:\left(1-x\cdot\frac{1}{3}\right)=\frac{7}{6}\)
=> \(\frac{1}{3}:\left(1-x\cdot\frac{1}{3}\right)=\frac{5}{3}-\frac{7}{6}\)
=> \(\frac{1}{3}:\left(1-x\cdot\frac{1}{3}\right)=\frac{1}{2}\)
=> \(\left(1-x\cdot\frac{1}{3}\right)=\frac{1}{3}:\frac{1}{2}=\frac{1}{3}\cdot2=\frac{2}{3}\)
=> \(1-\frac{x}{3}=\frac{2}{3}\)
=> \(\frac{x}{3}=1-\frac{2}{3}=\frac{1}{3}\)
=> x = 1
\(3-\left(x:\frac{1}{2}+\frac{3}{2}\right)-\frac{1}{4}=\frac{1}{2}\)
=> \(3-\left(x:\frac{1}{2}+\frac{3}{2}\right)=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)
=> \(x:\frac{1}{2}+\frac{3}{2}=3-\frac{3}{4}=\frac{9}{4}\)
=> \(x:\frac{1}{2}=\frac{9}{4}-\frac{3}{2}\)
=> \(x:\frac{1}{2}=\frac{3}{4}\)
=> \(x=\frac{3}{4}\cdot\frac{1}{2}=\frac{3}{8}\)
25 tháng 7 2020
\(\frac{5}{3}-\frac{1}{3}:\left(1-x\times\frac{1}{3}\right)=\frac{7}{6}\)
\(\frac{1}{3}:\left(1-x\times\frac{1}{3}\right)=\frac{5}{3}-\frac{7}{6}\)
\(\frac{1}{3}:\left(1-x\times\frac{1}{3}\right)=\frac{1}{2}\)
\(1-x\times\frac{1}{3}=\frac{1}{3}:\frac{1}{2}\)
\(1-x\times\frac{1}{3}=\frac{2}{3}\)
\(x\times\frac{1}{3}=1-\frac{2}{3}\)
\(x\times\frac{1}{3}=\frac{1}{3}\)
\(x=\frac{1}{3}:\frac{1}{3}\)
\(x=1\)
\(3-\left(x:\frac{1}{2}+\frac{3}{2}\right)-\frac{1}{4}=\frac{1}{2}\)
\(3-\left(x:\frac{1}{2}+\frac{3}{2}\right)=\frac{1}{2}+\frac{1}{4}\)
\(3-\left(x:\frac{1}{2}+\frac{3}{2}\right)=\frac{3}{4}\)
\(x:\frac{1}{2}+\frac{3}{2}=3-\frac{3}{4}\)
\(x:\frac{1}{2}+\frac{3}{2}=\frac{9}{4}\)
\(x:\frac{1}{2}=\frac{9}{4}-\frac{3}{2}\)
\(x:\frac{1}{2}=\frac{3}{4}\)
\(x=\frac{3}{4}\times\frac{1}{2}\)
\(x=\frac{3}{8}\)
VD
11 tháng 12 2016
\(3\frac{3}{11}x27\frac{27}{46}x1\frac{6}{17}x2\frac{4}{9}=\frac{34x1269x23x22}{11x46x17x9}=\frac{2x17x141x9x23x2x11}{11x23x17x9}=282\)
Ta có :
\(\left(x-1\right)^{x-2}=\left(x-1\right)^{x+6}\) \(\Rightarrow\left(x-1\right)^{x+6}-\left(x-1\right)^{x-2}=0\) \(\Leftrightarrow\left(x-1\right)^{x-2}\left[\left(x-1\right)^8-1\right]=0\) \(\Leftrightarrow\)\(\left[{}\begin{matrix}\left(x-1\right)^{x-2}=0\\\left(x-1\right)^8-1=0\end{matrix}\right.\Leftrightarrow\) \(\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^8=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)