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25 tháng 9 2021
a) \(=\left(x+y\right)+\left(x+y\right)\left(x-y\right)=\left(x+y\right)\left(1+x-y\right)\)
b) \(=x^2\left(y+1\right)-2y\left(y+1\right)=\left(y+1\right)\left(x^2-2y\right)\)
NT
1
14 tháng 3 2021
Ta có (\(^{x^{2^{ }}^{ }+3x}\)) (\(^{x^{2^{ }}+3x+4}\))
Đặt \(x^{2^{ }^{ }}+3x\) là a ta có
a.(a+4)=-4
4a+\(a^2\) -4=0
\(^{ }\left(a-2\right)^2\)=0
Suy ra a=2
hay \(x^{2^{ }^{ }^{ }}+3x=2\)
\(x^2+3x-2=0\)
𝑥=−3±17√/2
QN
6
25 tháng 11 2021
\(=\left[2\left(x-1\right)-3x\right]\left[2\left(x-1\right)+3x\right]\\ =\left(2x-2-3x\right)\left(2x-2+3x\right)\\ =\left(-2-x\right)\left(5x-2\right)\\ =\left(x+2\right)\left(2-5x\right)\)
19 tháng 2 2022
b: \(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)
=>-6x+16=0
=>-6x=-16
hay x=8/3(nhận)
c: \(\Leftrightarrow\dfrac{x+1+x-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{2}{x+2}\)
\(\Leftrightarrow2x\left(x+2\right)=2\left(x^2-1\right)\)
\(\Leftrightarrow2x^2+4x-2x^2+2=0\)
=>4x+2=0
hay x=-1/2(nhận)
TL
1
19 tháng 2 2022
\(a,\left(3x+1\right)^2-\left(2x-5\right)^2=0\\ \Leftrightarrow\left(3x+1+2x-5\right)\left(3x+1-2x+5\right)=0\\ \Leftrightarrow\left(5x-4\right)\left(x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-6\end{matrix}\right.\\ b,\left(x+3\right)\left(4-3x\right)=x^2+6x+9\\ \Leftrightarrow\left(x+3\right)\left(4-3x\right)-\left(x+3\right)^2=0\\ \Leftrightarrow\left(x+3\right)\left(4-3x-x-3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(1-4x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{4}\end{matrix}\right.\)
HP
11 tháng 1 2022
\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1\right)=\left(3x-2\right)\left(3x+2\right)\left(x+1\right)\)
\(\Leftrightarrow x-1=3x-2\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
11 tháng 1 2022
c: =>x-3=0
hay x=3
d: \(\Leftrightarrow\left(3x-1\right)\cdot\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)
11 tháng 1 2022
\(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right).\)
\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0.\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0.\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(-2x+1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0.\\x+1=0.\\-2x+1=0.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}.\\x=-1.\\x=\dfrac{1}{2}.\end{matrix}\right.\)
11 tháng 1 2022
c: =>(x-3)(x2+3x+5)=0
=>x-3=0
hay x=3
d: =>(3x-1)(x2+2-7x+10)=0
=>(3x-1)(x-3)(x-4)=0
hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)
FS
3
1 tháng 3 2020
\(\Leftrightarrow\left|\left(x-y+1\right)^2+x-2\right|=2x-\left|\left(x-1\right)\left(x-2\right)\right|\)
Có \(\left|\left(x-2\right)\left(x-1\right)\right|\ge0\Rightarrow\left[{}\begin{matrix}x\le1\left(1\right)\\x\ge2\left(2\right)\end{matrix}\right.\)-Trường hợp (1) có PT:
\(x-2\ge0\Rightarrow\left(x-y+1\right)^2+x-2>0\)..PT trở thành
\(\left(x-y+1\right)^2+x-2+4=2x-\left(x^2-3x+2\right)\)
\(\Leftrightarrow2x^2-2xy+y^2-4x-2y+5=0\)
Giải nữa thì nhờ mk nha
T
1
30 tháng 12 2023
a)
\(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
\(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x+1\right)\left(x-1\right)\)
\(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[9x^2-4-\left[\left(3x+2\right)\left(x-1\right)\right]\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left[9x^2-4-\left(3x^2-3x+2x-2\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+3x-2x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(6x^2+x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\6x^2+x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(2x-1\right)\left(3x+2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{1;\dfrac{-2}{3};\dfrac{1}{2}\right\}\)
b)
\(\left(x-1\right)^2-1+x^2=\left(1-x\right)\left(x+3\right)\)
\(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(\Leftrightarrow2x^2-2x=x^2-2x+3\)
\(\Leftrightarrow3x^2=3\)
\(\Leftrightarrow x^2=1\)
\(\Leftrightarrow x=\left(\pm1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy \(x\in\left\{1;-1\right\}\)
`Answer:`
`\frac{2x^2-3x-2}{x^2-4}=2(ĐK:x\ne+-2)`
`<=>\frac{2x^2-3x-2}{(x+2)(x-2)}=\frac{2(x+2)(x-2)}{(x-2)(x+2)}`
`<=>2x^2-3x-2=2(x+2)(x-2)`
`<=>2x^2-3x-2-2(x+2)(x-2)=0`
`<=>2x^2-3x-2-2x^2+8=0`
`<=>-3x+6=0`
`<=>x=2` (Không thoả mãn điều kiện)
Vậy phương trình vô nghiệm.