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![](https://rs.olm.vn/images/avt/0.png?1311)
1) Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
2) Ta có: \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)+\left(x-2\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3) Ta có: \(\left(2x-1\right)^2-\left(2x+5\right)^2=11\)
\(\Leftrightarrow4x^2-4x-1-4x^2-20x-25=11\)
\(\Leftrightarrow-24x=11+1+25=37\)
hay \(x=-\dfrac{37}{24}\)
5) Ta có: \(3x^2-5x-8=0\)
\(\Leftrightarrow3x^2+3x-8x-8=0\)
\(\Leftrightarrow3x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{8}{3}\end{matrix}\right.\)
8) Ta có: \(\left|x-5\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
10) Ta có: \(\left|2x+1\right|=\left|x-1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x-1\\2x+1=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-x=-1-1\\2x+x=1-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
5: =>4x^2-1/9=0
=>(2x-1/3)(2x+1/3)=0
=>x=1/6 hoặc x=-1/6
6: =>x-1=2
=>x=3
7:=>(2x-1)^3=-27
=>2x-1=-3
=>2x=-2
=>x=-1
8: =>1/8(x-1)^3=-125
=>(x-1)^3=-1000
=>x-1=-10
=>x=-9
3: =>(5x-5)^2-4=0
=>(5x-7)(5x-3)=0
=>x=3/5 hoặc x=7/5
4: =>(5x-1)^2=0
=>5x-1=0
=>x=1/5
1: =>(3x-1)(2x-1)=0
=>x=1/3 hoặc x=1/2
2: =>x^2(2x-3)-4(2x-3)=0
=>(2x-3)(x^2-4)=0
=>(2x-3)(x-2)(x+2)=0
=>x=3/2;x=2;x=-2
`@` `\text {Answer}`
`\downarrow`
`1,`
\(2x\left(3x-1\right)+1-3x=0\)
`<=> 2x(3x - 1) - 3x + 1 = 0`
`<=> 2x(3x - 1) - (3x - 1) = 0`
`<=> (2x - 1)(3x-1) = 0`
`<=>`\(\left[{}\begin{matrix}2x-1=0\\3x-1=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}2x=1\\3x=1\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy, `S = {1/2; 1/3}`
`2,`
\(x^2\left(2x-3\right)+12-8x=0\)
`<=> x^2(2x - 3) - 8x + 12 =0`
`<=> x^2(2x - 3) - (8x - 12) = 0`
`<=> x^2(2x - 3) - 4(2x - 3) = 0`
`<=> (x^2 - 4)(2x - 3) = 0`
`<=>`\(\left[{}\begin{matrix}x^2-4=0\\2x-3=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x^2=4\\2x=3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x^2=\left(\pm2\right)^2\\x=\dfrac{3}{2}\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\pm2\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy, `S = {+-2; 3/2}`
`3,`
\(25\left(x-1\right)^2-4=0\)
`<=> 25(x-1)(x-1) - 4 = 0`
`<=> 25(x^2 - 2x + 1) - 4 = 0`
`<=> 25x^2 - 50x + 25 - 4 = 0`
`<=> 25x^2 - 15x - 35x + 21 = 0`
`<=> (25x^2 - 15x) - (35x - 21) = 0`
`<=> 5x(5x - 3) - 7(5x - 3) = 0`
`<=> (5x - 7)(5x - 3) = 0`
`<=>`\(\left[{}\begin{matrix}5x-7=0\\5x-3=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}5x=7\\5x=3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy, `S = {7/5; 3/5}`
`4,`
\(25x^2-10x+1=0\)
`<=> 25x^2 - 5x - 5x + 1 = 0`
`<=> (25x^2 - 5x) - (5x - 1) = 0`
`<=> 5x(5x - 1) - (5x - 1) = 0`
`<=> (5x - 1)(5x-1)=0`
`<=> (5x-1)^2 = 0`
`<=> 5x - 1 = 0`
`<=> 5x = 1`
`<=> x = 1/5`
Vậy,` S = {1/5}.`
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
\(\left(x-5\right)^2+3\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-5+3\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)
2.
\(\left(x^2-9\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
3.
\(\left(2x+1\right)^2+\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1+x-1\right)=0\)
\(\Leftrightarrow\left(2x+1\right).3x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\2x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
4.
\(\left(x-1\right)\left(x+3\right)+\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1+x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)
\(=x^3+3x^2+3x+1+x^3-3x^2+3x-1\)
\(=2x^3+6x\)
b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)
\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)
\(=27x-55\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c: Ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)
\(\Leftrightarrow3x^2+26x=0\)
\(\Leftrightarrow x\left(3x+26\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)
\(a,\Leftrightarrow x^2+8x+16-x^3-12x^2=16\\ \Leftrightarrow x^3+11x^2-8x=0\\ \Leftrightarrow x\left(x^2+11x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+11x-8=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=121+32=153\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11-3\sqrt{17}}{2}\\x=\dfrac{-11+3\sqrt{17}}{2}\end{matrix}\right.\\ S=\left\{0;\dfrac{-11-3\sqrt{17}}{2};\dfrac{-11+3\sqrt{17}}{2}\right\}\)
\(c,\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\\ \Leftrightarrow3x^2+26x=0\\ \Leftrightarrow x\left(3x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\\ d,\Leftrightarrow x^3-6x^2+12x-8-x^3-125-6x^2=11\\ \Leftrightarrow-12x^2+12x-144=0\\ \Leftrightarrow x^2-x+12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: =>x^2+4x-21=0
=>(x+7)(x-3)=0
=>x=3 hoặc x=-7
2: =>(2x-5-4)(2x-5+4)=0
=>(2x-9)(2x-1)=0
=>x=9/2 hoặc x=1/2
3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15
=>-9x^2+27x+9x^2+18x+9=15
=>18x=15-9-27=-21
=>x=-7/6
6: =>4x^2+4x+1-4x^2-16x-16=9
=>-12x-15=9
=>-12x=24
=>x=-2
7: =>x^2+6x+9-x^2-4x+32=1
=>2x+41=1
=>2x=-40
=>x=-20
![](https://rs.olm.vn/images/avt/0.png?1311)
1: Ta có: \(2x\left(x+3\right)-6\left(x-3\right)=0\)
\(\Leftrightarrow2x^2+6x-6x+18=0\)
\(\Leftrightarrow2x^2+18=0\left(loại\right)\)
2: Ta có: \(2x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
hay \(x=-\dfrac{3}{2}\)
3: Ta có: \(\left(x-2\right)\left(x+1\right)-4x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(1-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
4: Ta có: \(2x\left(x-5\right)-3x+15=0\)
\(\Leftrightarrow\left(x-5\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
5: Ta có: \(3x\left(x+4\right)-2x-8=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
6: Ta có: \(x^2\left(2x-6\right)+2x-6=0\)
\(\Leftrightarrow2x-6=0\)
hay x=3
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, \(\left(5-2x\right)\left(2x+7\right)=4x^2-25=\left(2x-5\right)\left(2x+5\right)\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(5-2x\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7+2x+5\right)=0\Leftrightarrow\left(4x+12\right)\left(5-2x\right)=0\Leftrightarrow x=-3;x=\dfrac{5}{2}\)
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: \(\left(x-2\right)^2-3\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2-3x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-5}{2}\end{matrix}\right.\)