Ta có :

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

Do \(a+b=a^3+b^3\)

\(\Rightarrow a+b=\left(a+b\right)\left(a^2-ab+b^2\right)\)

\(\Rightarrow a^2-ab+b^2=1\)

Mà \(a^2=b^2=a+b\) ,ta có :

\(a+b-ab=1\)

\(\Rightarrow a+b-ab-1=0\)

\(\Rightarrow\left(a-1\right)-\left(ab-b\right)=0\)

\(\Rightarrow\left(a-1\right)-b\left(a-1\right)=0\)

\(\Rightarrow\left(a-1\right)\left(1-b\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}a-1=0\\1-b=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)

Thay vaò biểu thức ,có :

\(1^{2015}+1^{2015}=1+1=2\)

a+b+c=0 => (a+b+c)^2=0 <=> a^2+b^2+c^2+2(ab+bc+ca)=0

<=> 2+2(ab+bc+ca)=0 => ab+bc+ca=-1

(ab+bc+ca)^2=(ab)^2+(bc)^2+(ca)^2+2ab^2c+2abc^2+2a^2bc=(ab)^2+(bc)^2+(ca)^2+2abc(a+b+c)

=> (ab)^2+(bc)^2+(ca)^2 = (-1)^2 = 1

(a^2+b^2+c^2)^2 = a^4+b^4+c^4+2[(ab)^2+(bc)^2+(ca)^2] = a^4+b^4+c^4 + 2

<=>4=a^4+b^4+c^4+2 => a^4+b^4+c^4 = 2

Bạn kiểm tra lại có sai chỗ nào không nhé

1/2 nhá

\(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2abc\left(a+b+c\right)\)

thay \(\hept{\begin{cases}a+b+c=0\\a^2+b^2+c^2=0\end{cases}}\)

=>\(a^4+b^4+c^4=14^2-2abc.0=196\)

ta có x-y = 7 => (x-y)^2 = 49 <=> x^2 + y^2 - 2xy = 49 <=> x^2+y^2 - 2*60 = 49 <=> x^2+y^2 = 49+ 120 <=> x^2+y^2 = 169 => \(\left(x^2+y^2\right)^2=169^2\)<=> x^4+y^4 + \(2x^2y^2\)= 28561 (1)

từ xy = 60 => x^2 * y^2 = 360 => 2x^2 * y^2 = 720 thay vào 1 tính được A= x^4 + y^4 = 27841 

Chúc bạn học tốt!

to cam on

theo bai ra, ta co:

\(\frac{a}{b}=\frac{132}{143}\Leftrightarrow\frac{a}{132}=\frac{b}{143}=k\)

\(\Rightarrow a=132k;b=143k\)

ta co: BCNN(a,b)=BCNN(132k;143k)=156k

\(\Rightarrow\)156k=1092\(\Leftrightarrow\)k=7

\(\Rightarrow\)a=132.k=924

a = 84

\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=\frac{ayz+bxz+cxy}{xyz}=0\Rightarrow ayz+bxz+cxy=0\)

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

\(=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{cxy+bxz+ayz}{abc}\right)=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\cdot\frac{0}{abc}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2\)mà \(\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1^2=1\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\)