VP=x² +xa+xb+ab=x\((x+a)+b(x+a)\)=\((x+a)(x+b)\)=VT                                                                                                                 \(\RightarrowĐPCM\)

k cho mik

kb nữa nha

thanks

hok tốt

=6 nha

/ là gì????

dấu  / ....  /  là dấu giá trị tuyệt đối nha

\(A=3x^3y^4+4xy^3-8y^3+2021-3y^4x^3\)

\(\Rightarrow A=\left(3x^3y^4-3y^4x^3\right)+4xy^3-8y^3+2021\)

\(\Rightarrow A=4xy^3-8y^3+2021\)

Thay x = 2; y = -3 ta có:

\(A=4\cdot2\cdot\left(-3\right)^3-8\cdot\left(-3\right)^3+2021\)

\(\Rightarrow A=-216-\left(-216\right)+2021\)

\(\Rightarrow A=2021\)

~~ Chúc bạn học tốt ~~

d) \(\left|x-1\right|+\left|x-5\right|+\left|2x+5\right|\)

\(=\left|1-x\right|+\left|5-x\right|+\left|2x+5\right|\)

\(\ge\left|1-x+5-x\right|+\left|2x+5\right|\)

\(\ge\left|6-2x+2x+5\right|=11\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(1-x\right)\left(5-x\right)\ge0\\\left(6-2x\right)\left(2x+5\right)\ge0\end{cases}}\Leftrightarrow-\frac{5}{2}\le x\le1\).

e) \(\left|x+2\right|+\left|x-1\right|+\left|x-4\right|+\left|x+5\right|=12\)

\(\Leftrightarrow\left|x+2\right|+\left|1-x\right|+\left|4-x\right|+\left|x+5\right|=12\)

Có \(\left|x+2\right|+\left|1-x\right|+\left|4-x\right|+\left|x+5\right|\ge\left|x+2+1-x\right|+\left|4-x+x+5\right|=3+9=12\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(x+2\right)\left(1-x\right)\ge0\\\left(4-x\right)\left(x+5\right)\ge0\end{cases}}\Leftrightarrow-2\le x\le1\).

f) \(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+\left|3x-10\right|\)

\(\ge\left|x-1+x-2\right|+\left|3-x+3x-10\right|\)

\(=\left|2x-3\right|+\left|2x-7\right|\)

\(\ge\left|2x-3+7-2x\right|=4\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(x-1\right)\left(x-2\right)\ge0\\\left(3-x\right)\left(3x-10\right)\ge0\\\left(2x-3\right)\left(7-2x\right)\ge0\end{cases}}\Leftrightarrow3\le x\le\frac{10}{3}\).

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{x}{2}=\frac{y}{4}=\frac{x^4+y^4}{16+256}=\frac{16}{272}=\frac{1}{17}\)

\(\Rightarrow x=\frac{2}{17};y=\frac{4}{17}\)

Ta có \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

=> b + c = 2a ; c + a = 2b ; a  + b = 2c

Khi đó P = \(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=\frac{2c}{c}+\frac{2a}{a}+\frac{2b}{b}=2+2+2=6\)