a)2x²-7x-2x²-2x=7

-9x=7

x=-7/9

b)3x²+24x-x²-2x²-2x=2

22x=2

x=1/11

c: \(3x\left(x-7\right)-2\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)

d: \(7x^2-28=0\)

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

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

`c)-x^2+7x-2=-(x^2-7x)-2`

`=-(x^2-7x+49/4-49/4)-2`

`=-(x-7/2)^2+49/4-2`

`=-(x-7/2)^2+41/4<=41/4`

Dấu "=" xảy ra khi `x=7/2`

`d)-4x^2+8x-9=-(4x^2-8x)-9`

`=-(4x^2-8x+4-4)-9`

`=-(2x-2)^2-5<=-5`

Dấu "=" xảy ra khi `x=1`

`e)-3x^2+5x+10`

`=-3(x^2-5/3x)+10`

`=-3(x^2-5/3x+25/36-25/36)+10`

`=-3(x-5/6)^2+25/12+10`

`=-3(x-5/6)^2+145/12<=145/12`

Dấu "=" xảy ra khi`x=5/6`

b. -x²-2x+15

= -(x-1)²+14

= 14-(x-1)²

Do (x-1)² ≥0∀x nên 14-(x-1)²≤ 14

Dấu bằng xảy ra khi x=1

Vậy max=14 khi x=1

b: Ta có: \(N=a^3+b^3+3ab\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\)

\(=1-3ab+3ab\)

=1

\(=a^3b-a^3c+b^3c-ab^3+c^3a-bc^3\)

\(=\left(a^3b-ab^3\right)+\left(a^3c-ac^3\right)+\left(b^3c-bc^3\right)\)

\(=ab\left(a^2-b^2\right)+ac\left(a^2-c^2\right)+bc\left(b^2-c^2\right)\)

chị làm CTV đi chị .-.

Bài 4: 

1: 

a: \(=35x^3-14x^2+7x+14x^2=35x^3+7x\)

b: \(=\dfrac{6x^3-9x^2-10x^2+15x+8x-12}{2x-3}=3x^2-5x+4\)

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

=>(2x+1)(6x-1)=0

=>x=-1/2 hoặc x=1/6

\(a,A=0,2\left(5x-1\right)-\dfrac{1}{2}\left(\dfrac{2}{3}x+4\right)+\dfrac{2}{3}\left(3-x\right)\)

\(=x-0,2-\dfrac{1}{3}x-2+2-\dfrac{2}{3}x\)

\(=\left(-0,2-2+2\right)+\left(x-\dfrac{1}{3}x-\dfrac{2}{3}x\right)\)

\(=-0,2\)

\(b,B=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x^3-8y^3+10\right)\)

\(=x^3-8y^3-x^3+8y^3-10\)

\(=-10\)

\(c,C=4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)-4x\)

\(=4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-4x\)

\(=4x^2+8x+4+4x^2-4x+1-8x^2+8-4x\)

\(=13\)

a) \(A=0,2\left(5x-1\right)-\dfrac{1}{2}\left(\dfrac{2}{3}x+4\right)+\dfrac{2}{3}\left(3-x\right)\)

\(A=x-\dfrac{1}{5}-\dfrac{1}{3}x-2+2-\dfrac{2}{3}x\)

\(A=\left(x-\dfrac{1}{3}x-\dfrac{2}{3}x\right)-\left(\dfrac{1}{5}+2-2\right)\)

\(A=-\dfrac{1}{5}\)

Vậy: ...

b) \(B=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x^3-8y^3+10\right)\)

\(B=\left[x^3-\left(2y\right)^3\right]-\left[x^3-\left(2y\right)^3\right]-10\)

\(B=-10\)

Vậy: ...

c) \(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x+1\right)\left(x-1\right)-4x\)

\(=4\left(x^2+2x+4\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-4x\)

\(=4x^2+8x+4+4x^2-4x+1-8x^2+8-4x\)

\(=\left(4x^2+4x^2-8x^2\right)+\left(8x-4x-4x\right)+\left(4+1+8\right)\)

\(=13\)

Vậy:...