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![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)
hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)
2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)
hay \(x\in\left\{1;5\right\}\)
3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)
\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)
\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)
hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)
1.
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)
\(\Leftrightarrow x+3=5x-2\)
\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)
2.
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)
\(\Leftrightarrow x^2+x+1=x^2-2x+16\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
3.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`#3107`
`a)`
`(6x - 2)^2 + 4(3x - 1)(2 + y) + (y + 2)^2 - (6x + y)^2`
`= [(6x - 2)^2 - (6x + y)^2] + 4(3x - 1)(2 + y) + (2 + y)^2`
`= (6x - 2 - 6x - y)(6x -2 + 6x + y) + (2 + y)*[ 4(3x - 1) + 2 + y]`
`= (2 - y)(12x + y - 2) + (2 + y)*(12x - 4 + 2 + y)`
`= (2 - y)(12x + y - 2) + (2 + y)*(12x + y - 2)`
`= (12x + y - 2)(2 - y + 2 + y)`
`= (12x + y - 2)*4`
`= 48x + 4y - 8`
`b)`
\(5(2x-1)^2+2(x-1)(x+3)-2(5-2x)^2-2x(7x+12)\)
`= 5(4x^2 - 4x + 1) + 2(x^2 + 2x - 3) - 2(25 - 20x + 4x^2) - 14x^2 - 24x`
`= 20x^2 - 20x + 5 + 2x^2 + 4x - 6 - 50 + 40x - 8x^2 - 14x^2 - 24x`
`= - 51`
`c)`
\(2(5x-1)(x^2-5x+1)+(x^2-5x+1)^2+(5x-1)^2-(x^2-1)(x^2+1)\)
`= [ 2(5x - 1) + x^2 - 5x + 1] * (x^2 - 5x + 1) + (5x - 1)^2 - [ (x^2)^2 - 1]`
`= (10x - 2 + x^2 - 5x + 1) * (x^2 - 5x + 1) + (5x - 1)^2 - x^4 + 1`
`= (x^2 + 5x - 1)(x^2 - 5x + 1) + (5x - 1)^2 - x^4 + 1`
`= x^4 - (5x - 1)^2 + (5x - 1)^2 - x^4 + 1`
`= 1`
`d)`
\((x^2+4)^2-(x^2+4)(x^2-4)(x^2+16)-8(x-4)(x+4)\)
`= (x^2 + 4)*[x^2 + 4 - (x^2 - 4)(x^2 + 16)] - 8(x^2 - 16)`
`= (x^2 + 4)(x^4 + 12x^2 - 64) - 8x^2 + 128`
`= x^6 + 16x^4 - 16x^2 - 256 - 8x^2 + 128`
`= x^6 + 16x^4 - 24x^2 - 128`
![](https://rs.olm.vn/images/avt/0.png?1311)
`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`
`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`
![](https://rs.olm.vn/images/avt/0.png?1311)
1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)
\(=x^3+27-x^3-54\)
=-27
2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=x^2-4-x^2+2x+3=2x-1\\ b,=x^3+3x^2-5x-15+x^2-x^3+4x-4x^2=-x-15\\ c,=2x^2+3x-10x-15-2x^2+6x+x+7=-8\\ d,=\left(2x+1+3x-1\right)^2=25x^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn chú ý đăng lẻ câu hỏi! 1/
a/ \(=x^3-2x^5\)
b/\(=5x^2+5-x^3-x\)
c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)
d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)
e/ \(=x^4-x^2+2x^3-2x\)
f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)
a) \(\left(x-1\right)^3-\left(x^2-3x+9\right)\left(x+3\right)+3\left(x^2-4\right)=2\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3+27+3x^2-12=2\)
\(\Leftrightarrow3x=-12\)
\(\Rightarrow x=-4\)
b) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)
\(\Leftrightarrow-13x=26\)
\(\Rightarrow x=-2\)
c) \(\left(2x-1\right)^2+\left(2x+5\right)\left(2x-5\right)=18\)
\(\Leftrightarrow4x^2-4x+1+4x^2-25=18\)
\(\Leftrightarrow-4x=42\)
\(\Rightarrow x=-\frac{21}{2}\)
d) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow x^2+8x+16-x^2+1=16\)
\(\Leftrightarrow8x=-1\)
\(\Rightarrow x=-\frac{1}{8}\)