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

b)\(x^2+11x+24=x^2+3x+8x+24=x\left(x+3\right)+8\left(x+3\right)=\left(x+3\right)\left(x+8\right)\)

c)\(xy\left(x-y\right)+y^2\left(y-z\right)+zx\left(z-x\right)=x\left[y\left(x-y\right)+z\left(z-x\right)\right]+y^2\left(y-z\right)\)

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

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

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

\(x^4-y^4\)

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

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

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

a,Từ giả thiết ta có

(x²+y²+z²)(x+y+z)²+(xy+yz+zx)²

=(x²+y²+z²)(x²+y²+z²+2xy+2yz+2zx)+(xy+yz+zx)²

Đặt x²+y²+z²=a

xy+yz+zx=b

=>(x²+y²+z²)(x²+y²+z²+2xy+2yz+2zx)+(xy+yz+zx)²

=a(a+2b)+b²

=a²+2ab+b²

=(a+b)²

=(x²+y²+z²+xy+yz+zx)²

câu b hơi dài mình gửi sau nhé

Ta có: 2(x^4+y^4+z^4)-(x^2+y^2+z^2)^2-2(x^2+y^2+z^2)(x+y+z)^2+(x+y+z)^4

Gọi x^4+y^4+z^4=a

x^2+y^2+z^2=b

x+y+z=c

=>2(x^4+y^4+z^4)-(x^2+y^2+z^2)^2-2(x^2+y^2+z^2)(x+y+z)^2+(x+y+z)^4=2a-b^2-2bc^2+c^4

=2a-2b^2+b^2-2bc^2+c^4

=2(a-b^2)+(b+c^2)^2

Ta có

2(a-b²)=2[x^4+y^4+z^4-(x^2+y^2+z^2)²]

=2[x^4+y^4+z^4-x^4-y^4-z^4-2x²y²-2y²z²-2z²x²]

=2.(-2)(x²y²+y²z²+z²x²)

=-4(x²y²+y²z²+z²x²)

Lại có

(b+c^2)^2

=[(x^2+y^2+z^2)+(x+y+z)²]²

=[(x^2+y^2+z^2)-(x^2+y^2+z^2)-2(xy+yz+zx)]²

=4(xy+yz+zx)²

=>2(a-b^2)+(b+c^2)^2

=-4(x²y²+y²z²+z²x²)+4(xy+yz+zx)²

=8xyz(x+y+z)

cauu a cua bn Đen đủi .....lm sai r

Hai câu đầu tham khảo

Câu hỏi của Bangtan Sonyeondan - Toán lớp 8 - Học toán với OnlineMath

c) \(E=\left(x+a\right)\left(x+2a\right)\left(a+3a\right)\left(x+4a\right)+a^4\)

\(=\left(x+a\right)\left(x+4a\right)\left(x+2a\right)\left(a+3a\right)+a^4\)

\(=\left(x^2+5ax+4a^2\right)\left(a^2+5ax+6a^2\right)+a^4\)(1)

Đặt \(x^2+5ax+4a^2=t\)

\(\Rightarrow\left(1\right)=t\left(t+2a^2\right)+a^4\)

\(=t^2+2a^2t+a^4=\left(t+a^2\right)^2\)(2)

Mà \(x^2+5ax+4a^2=t\)

Nên \(\left(2\right)=\left(x^2+5ax+5a^2\right)^2\)

\(a,=\left(4x^2\right)^2\left(x-y\right)-\left(x-y\right)\)

\(=\left[\left(4x^2\right)^2-1^2\right]\left(x-y\right)\)

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

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

b: \(=\dfrac{12\left(y-z\right)^4+3\left(y-z\right)^5}{6\left(y-z\right)^2}=2\left(y-z\right)^2+\dfrac{1}{2}\left(y-z\right)^3\)

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