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x(x-5).(x+5)-(x+2).(x^2-2x+4)=17
\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3-2x^2+4x+2x^2-4x+8\right)=17\)
\(\Leftrightarrow x^3-25x-x^3+2x^2-4x-2x^2+4x-8=17\)
\(\Leftrightarrow-25x=17+8\)
\(\Leftrightarrow-25x=25\)
\(\Leftrightarrow x=-1\)
#)Giải :
\(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)
\(\Rightarrow x\left(x^2-25\right)-\left(x^3-2x^2+4x+2x^2-4x+8\right)=17\)
\(\Rightarrow x^3-25-\left(x^3+8\right)=17\)
\(\Rightarrow x^3-25x-x^3-8=17\)
\(\Rightarrow-25x=25\Rightarrow x=-1\)
Vậy x = -1
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a) \(\left(x+2\right)^2=4\left(2x-1\right)^2\)
\(\left(x+2\right)^2-4\left(2x-1\right)^2=0\)
\(\left(x+2\right)^2-\left[2\left(2x-1\right)\right]^2=0\)
\(\left(x+2\right)^2-\left(4x-2\right)^2=0\)
\(\left(x+2-4x+2\right)\left(x+2+4x-2\right)=0\)
\(6x\left(-3x+4\right)=0\)
\(\Rightarrow6x=0\) hoặc \(-3x+4=0\)
*) \(6x=0\)
\(x=0\)
*) \(-3x+4=0\)
\(3x=4\)
\(x=\dfrac{4}{3}\)
Vậy \(x=0;x=\dfrac{4}{3}\)
b) \(4x\left(x-2019\right)-x+2019=0\)
\(4x\left(x-2019\right)-\left(x-2019\right)=0\)
\(\left(x-2019\right)\left(4x-1\right)=0\)
\(\Rightarrow x-2019=0\) hoặc \(4x-1=0\)
*) \(x-2019=0\)
\(x=2019\)
*) \(4x-1=0\)
\(4x=1\)
\(x=\dfrac{1}{4}\)
Vậy \(x=\dfrac{1}{4};x=2019\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x(x - 5)(x + 5) - (x + 2)(x2 - 2x + 4) = 17
=> x(x2 - 25) - (x3 + 23) = 17
=> x3 - 25x - x3 - 8 = 17
=> 25x - 8 = 17
=> 25x = 17 + 8
=> 25x = 25
=> x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x^2+1\right)\left(x-2\right)+2x=4\Leftrightarrow x^3-2x^2+x-2+2x-4=0\Leftrightarrow x^3-2x^2+3x-6=0\Leftrightarrow\left(x-2\right)\left(x^2+3\right)=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)(do \(x^2+3\ge3>0\))
![](https://rs.olm.vn/images/avt/0.png?1311)
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a) Ta có: \(8x\left(2x-3\right)-4x\left(4x+3\right)=72\)
\(\Leftrightarrow16x^2-24x-16x^2-12x=72\)
\(\Leftrightarrow-36x=72\)
hay x=-2
b) Ta có: \(\left(x+2\right)\left(x+4\right)-x\left(x+2\right)=104\)
\(\Leftrightarrow x^2+6x+8-x^2-2x=104\)
\(\Leftrightarrow4x=96\)
hay x=24
c) Ta có: \(\left(x-1\right)\left(x+4\right)-x\left(x-1\right)=308\)
\(\Leftrightarrow x^2+3x-4-x^2+x=308\)
\(\Leftrightarrow4x=312\)
hay x=78
d) Ta có: \(15x\left(2x-3\right)-\left(5x+2\right)\left(6x-5\right)=-22\)
\(\Leftrightarrow30x^2-45x-30x^2+25x-12x+10=-22\)
\(\Leftrightarrow-32x=-32\)
hay x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
2:
a: =>x-1=0 hoặc 3x+1=0
=>x=1 hoặc x=-1/3
b: =>x-5=0 hoặc 7-x=0
=>x=5 hoặc x=7
c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)
d: =>x=0 hoặc x^2-1=0
=>\(x\in\left\{0;1;-1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+5x-10=3x^2-12x-5x+20\)
\(\Leftrightarrow-2x-22+17x-20=0\)
\(\Leftrightarrow15x=42\)
hay \(x=\dfrac{14}{5}\)
PT <=> \(\left(2x+5\right)\left(x-4\right)-\left(x-4\right)\left(5-x\right)=0\)
<=> \(\left(x-4\right)\left(2x+5-5+x\right)=0\)
<=> \(\left(x-4\right).3x=0\)
<=> \(\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
KL: \(S=\left\{4;0\right\}\)