Tìm x biết:
4x2-8x+4=2(1-x)(x+1)
4x2-25-(2x-5)(2x+7)=0
8x2+30x+7=0
x2+3x-18=0
K
Khách
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13 tháng 11 2021
\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
C
1
7 tháng 7 2021
\(a,\left(3x-7\right)^2=\left(2-2x\right)^2\)
a,\(=>\left(3x-7\right)^2-\left(2-2x\right)^2=0\)
\(< =>\left(3x-7+2-2x\right)\left(3x-7-2+2x\right)=0\)
\(< =>\left(x-5\right)\left(5x-9\right)=0=>\left[{}\begin{matrix}x=5\\x=1,8\end{matrix}\right.\)
b, \(x^2-8x+6=0< =>x^2-2.4x+16-10=0\)
\(< =>\left(x-4\right)^2-\sqrt{10}^2=0\)
\(=>\left(x-4+\sqrt{10}\right)\left(x-4-\sqrt{10}\right)=0\)
\(=>\left[{}\begin{matrix}x=4-\sqrt{10}\\x=4+\sqrt{10}\end{matrix}\right.\)
c, \(4x^2-2x-1=0\)
\(< =>\left(2x\right)^2-2.2.\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{5}{4}=0\)
\(=>\left(2x-\dfrac{1}{2}\right)^2-\left(\dfrac{\sqrt{5}}{2}\right)^2=0\)
\(=>\left(2x+\dfrac{-1+\sqrt{5}}{2}\right)\left(2x-\dfrac{1+\sqrt{5}}{2}\right)=0\)
\(=>\left[{}\begin{matrix}x=\dfrac{1-\sqrt{5}}{4}\\x=\dfrac{1+\sqrt{5}}{4}\end{matrix}\right.\)
d,\(x^4-4x^2-32=0\)
đặt \(t=x^2\left(t\ge0\right)=>t^2-4t-32=0\)
\(< =>t^2-2.2t+4-6^2=0\)
\(=>\left(t-2\right)^2-6^2=0=>\left(t-8\right)\left(t+4\right)=0\)
\(=>\left[{}\begin{matrix}t=8\left(tm\right)\\t=-4\left(loai\right)\end{matrix}\right.\)\(=>x=\pm\sqrt{8}\)
M
1
30 tháng 10 2021
a) \(\Leftrightarrow4x^2-10x-4x^2-3x=19\\ \Leftrightarrow-13x=19\\ \Leftrightarrow x=-\dfrac{19}{13}\)
b) \(\Leftrightarrow\left(x-7\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{5}{3}\end{matrix}\right.\)
LN
2
7 tháng 8 2021
1, \(\left(x-1\right)\left(x+2\right)-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left[x+2-\left(x-1\right)\right]=0\)
\(\Leftrightarrow3\left(x-1\right)=0\Leftrightarrow x=1\)
2, \(\left(x-2\right)^2-3\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x-2-3\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(-2x-5\right)=0\Leftrightarrow x=-\dfrac{5}{2};x=2\)
3, \(\left(5-2x\right)\left(2x+7\right)=4x^2-25=\left(2x-5\right)\left(2x+5\right)\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(5-2x\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7+2x+5\right)=0\Leftrightarrow\left(4x+12\right)\left(5-2x\right)=0\Leftrightarrow x=-3;x=\dfrac{5}{2}\)
7 tháng 8 2021
1) Ta có: \(\left(x-1\right)\left(x+2\right)-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2-x+1\right)=0\)
\(\Leftrightarrow x-1=0\)
hay x=1
2) Ta có: \(\left(x-2\right)^2-3\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2-3x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-5}{2}\end{matrix}\right.\)
K
1 tháng 12 2019
\(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)
\(\Leftrightarrow-\left(2x-5\right)\left(2x+7\right)=\left(2x-5\right)\left(2x+5\right)\)
\(\Leftrightarrow-2x-7=2x+5\)
\(\Leftrightarrow-4x=12\Leftrightarrow x=-3\)
\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left[2\left(2x+7\right)\right]^2-\left[3\left(x+3\right)\right]^2=0\)
\(\Leftrightarrow\left[2\left(2x+7\right)-3\left(x+3\right)\right].\left[2\left(2x+7\right)+3\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\7x+23=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\frac{23}{7}\end{matrix}\right.\)
18 tháng 8 2021
a) \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\Rightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\Rightarrow\left(2x-3\right)\left(7x-2x+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-3=0\\5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
b) \(\left(2x-7\right).\left(x-2\right)\left(x^2-4\right)=0\Rightarrow\left(2x-7\right)\left(x-2\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}2x-7=0\\\left(x-2\right)^2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)
c)\(\left(9x^2-25\right)-\left(6x-10\right)=0\Rightarrow\left(3x-5\right)\left(3x+5\right)-2\left(3x-5\right)=0\Rightarrow\left(3x-5\right)\left(3x+5-2\right)=0\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=1\end{matrix}\right.\)
18 tháng 8 2021
a: Ta có: \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\)
\(\Leftrightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)
b: Ta có: \(\left(2x-7\right)\left(x-2\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)^2\cdot\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)
c: Ta có: \(\left(9x^2-25\right)-\left(6x-10\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+5-2\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)
13 tháng 12 2021
c: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
CM
0
8x2+30x+7=0
8x2+16x+14x+7=0
8x(x+2) +7(x+2)=0
(8x+7)(x+2)=0
=>\(\orbr{\begin{cases}8x+7=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{7}{8}\\x=-2\end{cases}}}\)
a)
4x2-8x+4=2(1-x)(x+1)
4x2-8x+4-2+2x2=0
6x2-8x+2=0
2(3x2-4x+1)=0
3x2-3x-x+1=0
3x(x-1) -(x-1)=0
(3x-1)(x-1)=0
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}}\)