bạn ra đề khó quá

4x(x+y)(x+y+z)(x+z)+y²z²=4(x²+xy+xz)(x²+xy+xz+yz)+y²z²=4(x²+xy+xz)²+4yz(x²+xy+xz)+y²z²=(2(x²+xy+xz)+yz)²=(2x²+2xy+2xz+yz)

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

\(=x^3+x^2+x-x^2-x-1=x^3-1\)   đpcm

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

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

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

với x=y=1 thì M =1-32= -31 lấy đâu ra lớn hơn 1 mà CM

a) \(\frac{x+5}{x^2+5x-6}\)

ĐKXĐ: \(x^2+5x-6=x^2+6x-x-6\)

                                       \(x\left(x+6\right)-\left(x+6\right)=\left(x-1\right)\left(x+6\right)\)

            \(\Rightarrow x\ne1\) và \(x\ne-6\)                                                            

b) \(\frac{1-x}{3x^2-7x-6}\)

ĐKXĐ: \(3x^2-7x-6=3x^2-9x+2x-6=\left(x-3\right)\left(3x+2\right)\)

\(\Rightarrow x\ne3\) và \(x\ne\frac{-2}{3}\)

a)  \(x^2-6x+9=\left(x-3\right)^2\)

b)  \(x^2+8x+16=\left(x+4\right)^2\)

c)   \(\left(x-3\right)^2-16=\left(x-3-4\right)\left(x-3+4\right)=\left(x-7\right)\left(x+1\right)\)

d)  \(64+16x+x^2=\left(x+8\right)^2\)

e) \(x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\)

f) mk chỉnh đề

 \(8-36x+54x^2-27x^3=\left(2-3x\right)^3\)

g)  \(8x^3+12x^2y+6xy^2+y^3=\left(2x+y\right)^3\)

\(a^3+b^3+c^3-3abc\)

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

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)

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