Ta có: \(\frac{a}{abc+ab+a+1}=\frac{acd}{\left(abc+ab+a+1\right)cd}=\frac{acd}{abc^2d+abcd+acd+cd}\)

\(=\frac{acd}{c+1+acd+cd}\left(abcd=1\right)\)

\(\frac{b}{bcd+bc+b+1}=\frac{b}{bcd+bc+b+abcd}=\frac{1}{acd+cd+c+1}\)

\(\frac{d}{dab+da+d+1}=\frac{dc}{\left(dab+da+d+1\right)c}=\frac{dc}{abcd+acd+cd+c}=\frac{cd}{1+acd+cd+c}\)

=> \(\frac{acd}{acd+cd+c+1}+\frac{1}{acd+cd+c+1}+\frac{c}{acd+cd+c+1}+\frac{cd}{acd+cd+c+1}\)

=> đpcm

cả hình nữa nhé mn

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

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

Đặt \(A=x\left(x-2\right)\left(x+2\right)\left(x+4\right)+16\)

\(\Rightarrow A=x\left(x+2\right)\left(x-2\right)\left(x+4\right)+16\)

\(=\left(x^2+2x\right)\left(x^2+2x-8\right)+16\)

Đặt \(x^2+2x-4=t\)

\(\Rightarrow A=\left(t+4\right)\left(t-4\right)+16=t^2-16+16\)

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

\(\frac{1}{2}\left(x^2+y^2\right)^2-2x^2y^2=\frac{1}{2}.\left[\left(x^2+y^2\right)^2-4x^2y^2\right]\)

\(=\frac{1}{2}.\left[\left(x^2+y^2\right)^2-\left(2xy\right)^2\right]\)

\(=\frac{1}{2}.\left(x^2+y^2-2xy\right)\left(x^2+y^2+2xy\right)\)

\(=\frac{1}{2}\left(x-y\right)^2.\left(x+y\right)^2\)

4x³+1=4x³+1³=(4x+1)(16x-4x+1)

Đề sai rồi thì phải bạn, mik vẽ nó ra thế đấy =.='

hình đây 

Đặt \(A=x\left(x-2\right)\left(x+2\right)\left(x+4\right)+16\)

\(\Rightarrow A=x\left(x+2\right)\left(x-2\right)\left(x+4\right)+16\)

\(=\left(x^2+2x\right)\left(x^2+2x-8\right)+16\)

Đặt \(x^2+2x-4=t\)

\(\Rightarrow A=\left(t+4\right)\left(t-4\right)+16\)

\(=t^2-16+16=t^2=\left(x^2+2x-4\right)^2\)