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23 tháng 7 2023
a: =(x-y)^2+2(x-y)
=(x-y)(x-y+2)
c: =(x-3)(x+3)+(x-3)^2
=(x-3)(x+3+x-3)
=2x(x-3)
d: =(x+3)(x^2-3x+9)-4x(x+3)
=(x+3)(x^2-7x+9)
e: =(x^2-8x+7)(x^2-8x+15)-20
=(x^2-8x)^2+22(x^2-8x)+85
=(x^2-8x+17)(x^2-8x+5)
16 tháng 3 2023
a:=>x^2-1-x=2x-1
=>x^2-x-1=2x-1
=>x^2-3x=0
=>x=0(loại) hoặc x=3(nhận)
b:=>x+2=0 hoặc 5-3x=0
=>x=-2 hoặc x=5/3
c:=>20(1-2x)+6x=9(x-5)-24
=>20-40x+6x=9x-45-24
=>-34x+20=9x-69
=>-43x=-89
=>x=89/43
d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3
=>2x^2+4x-19=-2x+7
=>2x^2+6x-26=0
=>x^2+3x-13=0
=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)
e: =>(2x-3)(2x-3-x-1)=0
=>(2x-3)(x-4)=0
=>x=4 hoặc x=3/2
TA
1
SM
12 tháng 9 2018
1 ) 2x2 - 5x + 4x - 10 = 0
=> 2x2 + 4x - 5x - 10 = 0
=> 2x ( x + 2 ) - 5. ( x + 2 ) = 0
=> ( x + 2 ) . ( 2x - 5 ) = 0
=> \(\orbr{\begin{cases}x+2=0\\2x-5=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-2\\x=\frac{5}{2}\end{cases}}\)
Vậy \(x\in\left\{-2;\frac{5}{2}\right\}\)
2 ) x2 ( 2x - 3 ) + 3 - 2x = 0
=> x2 ( 2x - 3 ) - ( 2x - 3 ) = 0
=> ( 2x - 3 ) . ( x2 - 1 ) = 0
=> \(\orbr{\begin{cases}2x-3=0\\x^2-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x=3\\x^2=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=\pm1\end{cases}}\)
Vậy \(x\in\left\{\frac{3}{2};\pm1\right\}\)
H
16 tháng 3 2023
\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)
\(\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x\left(x+1\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)
Ta có : `(x-1)/x -1/(x+1) =(2x-1)/(x(x+1))`
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}=\dfrac{2x-1}{x\left(x+1\right)}\)
`=> x^2 +x -x-1 -x-2x+1=0`
`<=> x^2 -3x =0`
`<=> x(x-3)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=3\end{matrix}\right.\)
__
`(x+2)(5-3x)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\5-3x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{5}{3}\end{matrix}\right.\)
__
\(\dfrac{5\left(1-2x\right)}{3}+\dfrac{x}{2}=\dfrac{3\left(x-5\right)}{4}-2\)
\(\Leftrightarrow\dfrac{20\left(1-2x\right)}{12}+\dfrac{6x}{12}=\dfrac{9\left(x-5\right)}{12}-\dfrac{24}{12}\)
`<=> 2x- 40x + 6x = 9x - 45 -24`
`<=> 2x- 40x + 6x-9x + 45 +24=0`
`<=>-41x+69=0`
`<=>-41x=-69`
`<=> x=69/41`
CL
1
10 tháng 1 2018
\(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)=18\)
\(\Leftrightarrow\left(4x^2+8x+3\right)\left(x^2+2x+1\right)-18=0\)
\(\Leftrightarrow\left[4\left(x^2+2x\right)+3\right]\left(x^2+2x+1\right)-18=0\)
Đặt \(t=x^2+2x\)ta có
\(\left(4t+3\right)\left(t+1\right)-18=0\)
\(\Leftrightarrow4t^2+7x-15=0\)
\(\Leftrightarrow4t^2+12t-5t-15=0\)
\(\Leftrightarrow4t\left(t+3\right)-5\left(t+3\right)=0\)
\(\Leftrightarrow\left(t+3\right)\left(4t-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t+3=0\\4t-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}t=-3\\t=\frac{5}{4}\end{cases}}}\)
Nếu \(t=-3\Rightarrow x^2+2x=-3\)
\(\Leftrightarrow x^2+2x+3=0\)
\(\Rightarrow\)x vô nghiệm vì \(x^2+2x+3>0\)với mọi x
Nếu \(t=\frac{5}{4}\Rightarrow x^2+2x=\frac{5}{4}\)
\(\Leftrightarrow x^2+2x-\frac{5}{4}=0\)
\(\Leftrightarrow4x^2+8x-5=0\)
\(\Leftrightarrow4x^2-2x+10x-5=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{5}{2}\end{cases}}}\)
Vậy \(S=\left\{-\frac{5}{2};\frac{1}{2}\right\}\)
P/s tham khảo nha
23 tháng 6 2023
a: =>2x^2-2x+2x-2-2x^2-x-4x-2=0
=>-5x-4=0
=>x=-4/5
b: =>6x^2-9x+2x-3-6x^2-12x=16
=>-19x=19
=>x=-1
c: =>48x^2-12x-20x+5+3x-48x^2-7+112x=81
=>83x=83
=>x=1
3 tháng 1 2023
a. 12xy2 - 8x2y = 4xy . (3y - 2x)
b. 3x + 3y - x2 - xy = (3x + 3y) - (x2 + xy) = 3 . (x + y) - x . (x + y) = (x + y)(3 - x)
Ta có: \(\left(x^2+2x+3\right)^2-9\left(x^2+2x+3\right)+18=0\)
\(\Leftrightarrow\left(x^2+2x+3\right)^2-3\left(x^2+2x+3\right)-6\left(x^2+2x+3\right)+18=0\)
\(\Leftrightarrow\left(x^2+2x+3\right)\left(x^2+2x+3-3\right)-6\left(x^2+2x+3-3\right)=0\)
\(\Leftrightarrow\left(x^2+2x+3-6\right)\left(x^2+2x\right)=0\)
\(\Leftrightarrow\left(x^2+2x-3\right)\cdot x\cdot\left(x+2\right)=0\)
\(\Leftrightarrow\left(x^2+2x+1-4\right)\cdot x\cdot\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+1-2\right)\left(x+1+2\right)\cdot x\cdot\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)\cdot x\cdot\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\\x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\\x=0\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{1;-3;0;-2\right\}\)