cậu dùng bất đẳng thức bunhia cópki cho e bộ số là (x²,y²,z²)và (1,1,1) là được

CÓ gì ko hiểu hỏi riêng mình nha

ta có a+b+c=0=>(a+b+c)^2=0
=>a^2+b^2+c^2+2ab+2ac+2bc=0
=>1+2(ab+bc+ac)=0(vì a^2+b^2+c^2=1)
=>ab+bc+cd=-1/2
=>(ab+bc+cd)^2=1/4
=>a^2b^2+a^2c^2+b^2c^2+2a^2bc+2ab^2c+2abc^2=1/4
=>a^2b^2+a^2c^2+b^2c^2+2abc(a+b+c)=1/4
=>a^2b^2 +a^2c^2+b^2c^2=1/4(vì a+b+c=0)*
mặt khác a^2+b^2+c^2=1
=>(a^2+b^2+c^2)^2=1
=>a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2=1
=>a^4+b^4+c^4+2(a^2b^2+a^2c^2+b^2c^2)=1
=>a^4+b^4+c^4+2.1/4=1(theo *)
=>a^4+b^4+c^4=1- 1/2=1/2(dpcm)

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

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

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

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ca}{abc}=0\Leftrightarrow ab+bc+ca=0\)

\(\left(a+b+c\right)^2=1\Leftrightarrow a^2+b^2+c^2+2.\left(ab+bc+ca\right)=1\)

\(\Leftrightarrow a^2+b^2+c^2+2.0=1\)

\(\Leftrightarrow a^2+b^2+c^2=1\)

a) \(x^2-5=0\)

\(\Leftrightarrow x^2=5\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=5\\x=-5\end{array}\right.\)

b) \(\left(2x-1\right)^2-\left(3x+1\right)^2=0\)

\(\Leftrightarrow\left(2x-1+3x+1\right)\left(2x-1-3x-1\right)=0\)

\(\Leftrightarrow5x\left(-x-2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\-x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-2\end{array}\right.\)

c) \(\frac{4}{9}\cdot x^2=-4x-9\)

\(\Leftrightarrow\left(\frac{2}{3}x\right)^2+4x+9=0\)

\(\Leftrightarrow\left(\frac{2}{3}x+3\right)^2=0\)

\(\Leftrightarrow\)\(\frac{2}{3}x+3=0\Leftrightarrow x=-\frac{9}{2}\)

 \(a+b=10\) và \(ab=4\)

1. Có: \(A=a^2+b^2=\left(a+b\right)^2-2ab=10^2-2.4=92\)

2. \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=10^3-3.4.10=880\)

3. \(a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2=92^2-2.4^2=8432\)

4. \(a^5+b^5=\left(a^2+b^2\right)\left(a^3+b^3\right)-a^2b^2\left(a+b\right)=92.880-4^2.10=80800\)