nhìn zậy thoy chứ dễ lắm mik làm vd 2 bài còn lại bn làm có gì bí thì hỏi mik

a) biến đổi vế trái ta có : \(\left(x+y\right)^2-y^2=\left(x+y-y\right)\left(x+y+y\right)=x\left(x+2y\right)\)( = vế phải )

b) BĐVT ta có : \(\left(x^2+y^2\right)^2-\left(2xy\right)^2=\left(x^2+y^2-2xy\right)\left(x^2+y^2+2xy\right)=\left(x-y\right)^2\left(x+y\right)^2\)= VP

\(4^{n+1}-3.4^n=4^n.4-3.4^n=4^n.\left(4-3\right)=4^n\)

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

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

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

b) xy+1+x+y = x(y+1)+1+y = (x+1).(y+1)

A = (100+1)² = 100² + 200 + 1 = 10201

=101.101=100.101+101=10100+101=10201

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}\)

\(=\frac{10}{20}-\frac{1}{20}\)

\(=\frac{9}{20}\)

PT <=> \(\frac{4}{5}x^2\left(\frac{x}{3}-\frac{1}{2}\right)-\frac{4}{3}x^2\left(\frac{1}{5}x-\frac{2}{3}\right)-\frac{22}{45}x^2-\left(\frac{1}{5}x-\frac{2}{3}\right)=0\)

<=> \(x^2\left(\frac{4x}{15}-\frac{2}{5}-\frac{4x}{15}+\frac{8}{9}-\frac{22}{45}\right)-\left(\frac{1}{5}x-\frac{2}{3}\right)=0\)

<=> \(x^2.0-\frac{1}{5}x+\frac{2}{3}=0\)

<=> \(\frac{1}{5}x=\frac{2}{3}\Rightarrow x=\frac{2}{3}:\frac{1}{5}=\frac{10}{3}\)

Vậy....

=(x+y)(x-y)+3(x+y)=(x+y)(x-y+3)

=(x+y)(x-y)+3(x+y)=(x+y)(x-y+3)