1)

`7x^2 -49x=0`

`<=>x(7x-49)=0`

\(< =>\left[{}\begin{matrix}x=0\\7x-49=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)

2)

`8x^2 -16x=0`

`<=>x(8x-16)=0`

\(< =>\left[{}\begin{matrix}x=0\\8x-16=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

3)

`2x^3 +40x=0`

`<=>x(2x^2 +40)=0`

`<=>x=0` hoặc`2x^2 +40=0`

`<=>x=0` hoặc `2x^2 =-40` (vô lí vì `2x^2` luôn lớn hơn hoặc bằng 0)

`<=>x=0`

4)

`-x^3 +16x=0`

`<=>x^3 -16x=0`

`<=>x(x^2 -16)=0`

\(< =>\left[{}\begin{matrix}x=0\\x^2-16=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x^2=16\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

1.

\(a,\left(-xy\right)\left(-2x^2y+3xy-7x\right)\)

\(=2x^3y^2-3x^2y^2+7x^2y\)

\(b,\left(\dfrac{1}{6}x^2y^2\right)\left(-0,3x^2y-0,4xy+1\right)\)

\(=-\dfrac{1}{20}x^4y^3-\dfrac{1}{15}x^3y^3+\dfrac{1}{6}x^2y^2\)

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

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

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

\(d,\left(x-y\right)\left(x^2-2xy+y^2\right)\)

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

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

2.

\(a,\left(x-y\right)\left(x^2+xy+y^2\right)\)

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

\(b,\left(x+y\right)\left(x^2-xy+y^2\right)\)

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

\(c,\left(4x-1\right)\left(6y+1\right)-3x\left(8y+\dfrac{4}{3}\right)\)

\(=24xy+4x-6y-1-24xy-4x\)

\(=\left(24xy-24xy\right)+\left(4x-4x\right)-6y-1\)

\(=-6y-1\)

#Toru

a: \(=3x^4+3x^2y^2+2x^2y^2+2y^4+y^2\)

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

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

b: \(=7\left(x-y\right)+4a\left(x-y\right)-5=-5\)

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

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

=9-12+1

=-2

\(\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\\ \Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\\ \Leftrightarrow\left(x+2\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)

⇔(5x-4-49x)(5x-4+49x)=0

⇔(-44x-4)(54x-4)=0

⇒-44x-4=0 hoặc 54x-4=0

TH1:-44x-4=0                TH2:54x-4=0

     ⇔x=-1/11                   ⇔x=2/27

Vậy xϵ{-1/11;2/27}

(5x-4-7x)(5x-4+7x)=0

(-2x-4)(12x-4)=0

⇒-2x-4=0 hoặc 12x-4=0

    x=-2                  x=1/3

a) \(3x\left(2x-y\right)+5y\left(y-2x\right)\)

\(=3x\left(2x-y\right)-5y\left(2x-y\right)\)

\(=\left(3x-5y\right)\left(2x-y\right)\)

b) \(\left(x-5\right)^2-9\left(x+y\right)^2\)

\(=\left(x-5\right)^2-3^2\left(x+y\right)^2\)

\(=\left(x-5\right)^2-\left(3x+3y\right)^2\)

\(=\left(x-5+3x+3y\right)\left(x-5-3x-3y\right)\)

\(=\left(4x+3y-5\right)\left(-2x-3y-5\right)\)

a: \(3x\left(2x-y\right)+5y\left(y-2x\right)=\left(2x-y\right)\left(3x-5y\right)\)

e: \(x^2-10x+24=\left(x-4\right)\left(x-6\right)\)

g) \(6x^2+7x-5\)

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

=\(\left(6x^2+10x\right)-\left(3x+5\right)\)

=\(2x\left(3x+5\right)-\left(3x+5\right)\)

=\(\left(2x-1\right)\left(3x+5\right)\)

Bài 2: 

a) Ta có: \(A=\left(7x+5\right)^2+\left(3x-5\right)^2-\left(10-6x\right)\left(5+7x\right)\)

\(=\left(7x+5\right)^2+2\cdot\left(7x+5\right)\cdot\left(3x-5\right)+\left(3x-5\right)^2\)

\(=\left(7x+5+3x-5\right)^2\)

\(=\left(10x\right)^2=100x^2\)

Thay x=-2 vào A, ta được:

\(A=100\cdot\left(-2\right)^2=100\cdot4=400\)

b) Ta có: \(B=\left(2x+y\right)\left(y^2-2xy+4x^2\right)-8x\left(x-1\right)\left(x+1\right)\)

\(=8x^3+y^3-8x\left(x^2-1\right)\)

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

\(=8x+y^3\)

Thay x=-2 và y=3 vào B, ta được:

\(B=-2\cdot8+3^3=-16+27=11\)

Ai help mk vs

Bài 1: 

Ta có: \(A=x\left(4-x\right)\)

\(=4x-x^2\)

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

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

\(=-\left(x-2\right)^2+4\le4\forall x\)

Dấu '=' xảy ra khi x=2

Vậy: \(A_{max}=4\) khi x=2