\(a,\Leftrightarrow6x^2-6x^2-11x+10=-12\\ \Leftrightarrow-11x=-22\\ \Leftrightarrow x=2\\ b,\Leftrightarrow x^3+27-x^3-2x=12-5x\\ \Leftrightarrow3x=-15\\ \Leftrightarrow x=-5\\ c,\Leftrightarrow x^2-6x-16=0\\ \Leftrightarrow\left(x-8\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

a: ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-12\)

\(\Leftrightarrow6x^2-6x^2+4x-15x+10=-12\)

\(\Leftrightarrow-11x=-22\)

hay x=2

b: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+2\right)=12-5x\)

\(\Leftrightarrow x^3+27-x^3-2x+5x=12\)

\(\Leftrightarrow x=-5\)

\(a.2x\left(x-1\right)-3\left(x^2+4x\right)+x\left(x+2\right)\) 

\(=2x^2-2x-3x^2-12x+x^2+2x\) 

\(=-12x\) 

\(b.\left(2x-3\right)\left(3x+5\right)-\left(x-1\right)\left(6x+2\right)+3-5x\) 

\(=6x+10x-9x^2-15-6x^2-2x-6x-2+3-5x\) 

\(=-15x^2+3x-14\) 

\(c.\left(x-y\right)\left(x^2+xy+y^2\right)-\left(x+y\right)\left(x^2-y^2\right)\) 

\(=x^3-y^3-x^3+y^3+x^2y-y^3\)

\(=y^3+x^2y\) 

`a)(2x-1)^2-0,25=0`

`<=>(2x-1-0,5)(2x-1+0,5)=0`

`<=>(2x-1,5)(2x-0,5)=0`

`<=>[(x=0,75)(x=0,25):}`

`b)x^2+9=6x`

`<=>(x-3)^2=0`

`<=>x-3=0`

`<=>x=3`

`c)(x^2-4)-3x-6=0`

`<=>(x-2)(x+2)-3(x+2)=0`

`<=>(x+2)(x-2-3)=0`

`<=>(x+2)(x-5)=0`

`<=>[(x=-2),(x=5):}`

a: =>(2x-1-0,5)(2x-1+0,5)=0

=>(2x-1,5)(2x-0,5)=0

=>x=0,25 hoặc x=0,75

b: =>x^2-6x+9=0

=>(x-3)^2=0

=>x-3=0

=>x=3

c: =>(x-2)(x+2)-3(x+2)=0

=>(x+2)(x-5)=0

=>x=5 hoặc x=-2

Phân tích đa thức thành nhân tử

1: \(x^2-x-y^2-y\)

\(=\left(x^2-y^2\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-1\right)\)

2: \(x^2-y^2+x-y\)

\(=\left(x^2-y^2\right)+\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y\right)+\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y+1\right)\)

3: \(3x-3y+x^2-y^2\)

\(=\left(3x-3y\right)+\left(x^2-y^2\right)\)

\(=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)

\(=\left(x-y\right)\left(x+y+3\right)\)

4: \(5x-5y+x^2-y^2\)

\(=\left(5x-5y\right)+\left(x^2-y^2\right)\)

\(=5\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)

\(=\left(x-y\right)\left(5+x+y\right)\)

5: \(x^2-5x-y^2-5y\)

\(=\left(x^2-y^2\right)-\left(5x+5y\right)\)

\(=\left(x-y\right)\left(x+y\right)-5\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-5\right)\)

6: \(x^2-y^2+2x-2y\)

\(=\left(x^2-y^2\right)+\left(2x-2y\right)\)

\(=\left(x-y\right)\left(x+y\right)+2\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y+2\right)\)

7: \(x^2-4y^2+x+2y\)

\(=\left(x^2-4y^2\right)+\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x-2y\right)+\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x-2y+1\right)\)

8: \(x^2-y^2-2x-2y\)

\(=\left(x^2-y^2\right)-\left(2x+2y\right)\)

\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-2\right)\)

9: \(x^2-4y^2+2x+4y\)

\(=\left(x^2-4y^2\right)+\left(2x+4y\right)\)

\(=\left(x-2y\right)\left(x+2y\right)+2\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x-2y+2\right)\)

`#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`

a)-(x-y)(x²+xy-1)=-(x³+x²y-x-x²y-xy²+y)

                          =-(x³-xy²-x+y)

                          =-x³+xy²+x-y

b)x²(x-1)-(x³+1)(x-y)=x³-x²-x³+x²y-x+y

                                =-x²+x²y-x+y

c)(3x-2)(2x-1)+(-5x-1)(3x+2)=6x²-3x-4x+2-15x²-10x-3x-2

                                             =-9x²-20x

d) hình như bạn ghi lỗi

Bài 2: C=x(x²-y)-x²(x+y)+y(x²-x)

             =x³-xy-x³-x²y+x²y-xy

             =-2xy

Thay x=1/2,y=-1 vào C, ta có:

        C=-2.1/2.(-1)=1

Vậy C=1 khi x=1/2 và y=-1.

a) Ta có: \(\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)

\(\Leftrightarrow\left(x^2-5x\right)^2+4\left(x^2-5x\right)+6\left(x^2-5x\right)+24=0\)

\(\Leftrightarrow\left(x^2-5x\right)\left(x^2-5x+4\right)+6\left(x^2-5x+4\right)=0\)

\(\Leftrightarrow\left(x^2-5x+6\right)\left(x^2-5x+4\right)=0\)

\(\Leftrightarrow\left(x^2-2x-3x+6\right)\left(x^2-x-4x+4\right)=0\)

\(\Leftrightarrow\left[x\left(x-2\right)-3\left(x-2\right)\right]\left[x\left(x-1\right)-4\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\\x=4\end{matrix}\right.\)

Vậy: S={1;2;3;4}

b) Ta có: \(\left(2x+1\right)^2-2x-1=2\)

\(\Leftrightarrow\left(2x+1\right)^2-\left(2x+1\right)-2=0\)

\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x+1\right)+\left(2x+1\right)-2=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x+1-2\right)+\left(2x+1-2\right)=0\)

\(\Leftrightarrow\left(2x+1+1\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x+2\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{-1;\dfrac{1}{2}\right\}\)

c) Ta có: \(x\left(x-1\right)\left(x^2-x+1\right)-6=0\)

\(\Leftrightarrow x\left(x^3-x^2+x-x^2+x-1\right)-6=0\)

\(\Leftrightarrow x\left(x^3-2x^2+2x-1\right)-6=0\)

\(\Leftrightarrow x^4-2x^3+2x^2-x-6=0\)

\(\Leftrightarrow x^4-2x^3+2x^2-4x+3x-6=0\)

\(\Leftrightarrow x^3\left(x-2\right)+2x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+2x+3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3-x+3x+3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[x\left(x^2-1\right)+3\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left[x\left(x-1\right)\left(x+1\right)+3\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x^2-x+3\right)=0\)

mà \(x^2-x+3>0\forall x\)

nên (x-2)(x+1)=0

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

Vậy: S={2;-1}

d) Ta có: \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)

\(\Leftrightarrow\left(x^2+1\right)^2+2x\left(x^2+1\right)+x\left(x^2+1\right)+2x^2=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(x^2+1+2x\right)+x\left(x^2+1+2x\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x^2+x+1\right)=0\)

mà \(x^2+x+1>0\forall x\)

nên x+1=0

hay x=-1

Vậy: S={-1}

\(\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\)

\(< =>\left(1-x\right)\left(5x+3+3x-7\right)=0\)

\(< =>\left(1-x\right)\left(8x-4\right)=0\)

\(< =>\orbr{\begin{cases}1-x=0\\8x-4=0\end{cases}< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)

\(\left(x-2\right)\left(x+1\right)=x^2-4\)

\(< =>\left(x-2\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)

\(< =>\left(x-2\right)\left(x+1-x-2\right)=0\)

\(< =>-1\left(x-2\right)=0\)

\(< =>2-x=0< =>x=2\)