a)x⁵+y⁵=(x+y)(x⁴y-x³y²+x²y³-xy⁴) 

b)tự làm nhá 

sao bé đi hỏi toán 8 chi zợ

25

cho mìn ****

x³ +27 - (x³ +9x² +3x+27)=0

=>x³ +27 -x³ -9x²-3x-27=0

=>-9x² -3x =0

=> -3x(3x+1)=0

=>x=0 ; x =-1/3

\(\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2.\frac{xy}{ab}+2.\frac{xz}{ac}+2.\frac{yz}{bc}=1\)

Ta có: \(2.\frac{xy}{ab}+2.\frac{xz}{ac}+2.\frac{yz}{bc}=2.\left(\frac{xy}{ab}+\frac{xz}{ac}+\frac{yz}{bc}\right)\)

Mặt khác, \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\) => \(\frac{ayz+bxz+cxy}{xyz}=0\)=> ayz + bxz + cxy = 0 

=> \(\frac{ayz+bxz+cxy}{abc}=0\) => \(\frac{yz}{bc}+\frac{xz}{ac}+\frac{xy}{ab}=0\)

Do đó, \(2.\frac{xy}{ab}+2.\frac{xz}{ac}+2.\frac{yz}{bc}=2.\left(\frac{xy}{ab}+\frac{xz}{ac}+\frac{yz}{bc}\right)=0\)

=> đpcm

x>/5

Hai vế không âm bình phương 2 vế

x+1 = x² -10x +25

x² -11x +24 =0

(x-3)(x-8) =0

=> x = 3 (loại)  hoặc x =8 (TM)

Vậy x =8

sorry, em mới học lớp 6 thui à

x²+y²=5

=>x²+2xy+y²=5+2xy

=>(x+y)²=5+2xy

=>3²=5+2xy

=>4=2xy =>xy=2

Suy ra: x³+y³=(x+y)(x²-xy+y²)=3.(5-2)=9

\(\frac{x^2+x+1}{\left(x+1\right)^2}=\frac{x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}}{\left(x+1\right)^2}=\frac{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}{\left(x+1\right)^2}\)

\(\text{Đặt }y=x+1\Rightarrow y-\frac{1}{2}=x+\frac{1}{2}\text{ ta được:}\)

\(\frac{\left(y-\frac{1}{2}\right)^2+\frac{3}{4}}{y^2}=\frac{y^2-y+\frac{1}{4}+\frac{3}{4}}{y^2}\)

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

\(=\frac{1}{4}-2.\frac{1}{y}.\frac{1}{2}+\frac{1}{y^2}+\frac{3}{4}\)

\(=\left(\frac{1}{y}-\frac{1}{2}\right)^2+\frac{3}{4}\)

Vậy GTNN của \(\frac{x^2+x+1}{\left(x+1\right)^2}\) là 3/4 tại: 

\(\frac{1}{y}=\frac{1}{2}\Rightarrow y=2\Rightarrow x=y-1=2-1=1\)

\(\frac{3}{4}\) tick nha