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a: Ta có: \(x\left(2-x\right)+\left(x^2+x\right)=7\)
\(\Leftrightarrow2x-x^2+x^2+x=7\)
\(\Leftrightarrow3x=7\)
hay \(x=\dfrac{7}{3}\)
b: Ta có: \(\left(2x+1\right)^2-x\left(4-5x\right)=17\)
\(\Leftrightarrow4x^2+4x+1-4x+5x^2=17\)
\(\Leftrightarrow9x^2=16\)
\(\Leftrightarrow x^2=\dfrac{16}{9}\)
hay \(x\in\left\{\dfrac{4}{3};-\dfrac{4}{3}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x\right)^2-\left(3x+3\right)^2=0\Leftrightarrow\left(4x-3x-3\right)\left(4x+2x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(6x+3\right)=0\Leftrightarrow x=-\frac{1}{2};x=3\)
b, \(\left(5x-4\right)^2-49x^2=0\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\)
\(\Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\Leftrightarrow x=-2;x=\frac{1}{3}\)
c, \(5x^3-20x=0\Leftrightarrow5x\left(x^2-4\right)=0\)
\(\Leftrightarrow5x\left(x-2\right)\left(x+2\right)=0\Leftrightarrow x=0;x=\pm2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: Ta có: \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{7}{3}\end{matrix}\right.\)
2: Ta có: \(\left(5x-4\right)^2-49x^2=0\)
\(\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\)
\(\Leftrightarrow\left(2x+4\right)\left(12x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3: Ta có: \(5x^3-20x=0\)
\(\Leftrightarrow5x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-5x^2+16x-3=0\)
\(-5x^2+x+15x-3=0\)
\(x\cdot\left(-5x+1\right)-3\cdot\left(-5x-1\right)=0\)
\(\left(-5x-1\right)\cdot\left(x-3\right)=0\)
\(\hept{\begin{cases}-5x-1=0\\x-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{1}{5}\\x=3\end{cases}}}\)
Vậy.......
-5x^2 +15x +x-3=0
5x(-x +3)-(-x+3)=
(5x-1)(-x+3)=0
5x-1 =0 hoặc -x+3=0
x=1/5 hoặc x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
giải
5x-(4-2x+x^2)(x+2)+x(x-1)(x+1)=0
5x-(4x+8-2x^2-4x+x^3+2x^2)+x(x^2-1)=0
5x-4x-8+2x^2+4x-x^3-2x^2+x^3-1x=0
(5x-4x+4x-1x)+(-8)+(2x^2-2x^2)+(-x^3+x^3)=0
4x+(-8)=0
4x=0+8
4x=8
x=8:4
x=2
D)(4x+1)(16x^2-4x+1)-16x(4x^2-5)=17
64x^3-16x^2+4x+16x^2-4x+1-64x^3+80x=17
80x+1=17
80x=17-1
80x=16
x=1/5
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2: Tính giá trị của biểu thức sau:
\(16x^2-y^2=\left(4x+y\right)\left(4x-y\right)\)
Thay \(\hept{\begin{cases}x=87\\y=13\end{cases}}\)
\(\Rightarrow\left(4.87+13\right)\left(4.87-13\right)=361.335=120935\)
Bài 4: Tìm x
a) \(9x^2+x=0\)
\(\Rightarrow x\left(9x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\9x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{9}\end{cases}}\)
b) \(27x^3+x=0\)
\(\Rightarrow x\left(27x^2+1=0\right)\)
\(\Rightarrow\orbr{\begin{cases}x=0\\27x^2+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\27x^2=\left(-1\right)\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=\frac{-1}{27}\end{cases}}\)
Ta có: \(\frac{-1}{27}\) loại vì \(x^2\ge0\forall x\)
Vậy \(x=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(5x-4\right)^2-16x^2=0\)
\(\Leftrightarrow\left(5x-4\right)^2-\left(4x\right)^2=0\)
\(\Leftrightarrow\left(5x-4-4x\right)\left(5x-4+4x\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(9x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{4}{9}\end{matrix}\right.\)
\(\left(5x-4\right)^2-16x^2=0\\ \Leftrightarrow\left(5x-4\right)^2-\left(4x\right)^2=0\\ \Leftrightarrow\left(5x-4-4x\right).\left(5x-4+4x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x-4-4x=0\\5x-4+4x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{4}{9}\end{matrix}\right.\\ \Rightarrow S=\left\{\dfrac{4}{9};4\right\}\)