pt (1) <=>5x-2x^2-xy+y^2-y-2=0 

giai phuong trinh (1) theo an y ta co: 
y² - (x+1)y - (2x² - 5x+2)=0 
<=>Δ=(x+1)²+4(2x² - 5x+2)=x²+2x+1+8x²-20y+8=9x²-18x+9 
=9(x-1)² 
Δ>=0 => phuong trinh co nghiem 
<=>y=(x+1+3(x-1))/2 hoac y=(x+1-3(x-1))/2 
<=>y=2x-1 hoac y=2-x 
* thay y=2x-1 vao pt 2 ta duoc: 
x²+(2x-1)²+x+(2x-1)=4 
<=>5x²-x-4=0 
giai phuong trinh tren ta tim duoc x=1 va y=1 hoac x=-4/5 va y=-13/5 
*the y=2-x vao pt 2 ta duoc 
x²+(2-x)²+x+(2-x)=4 
<=>2x²-4x+2=0 
<=>x=1 =>y=1 
vay phuong trinh co 2 nghiem (1;1);(-4/5;-13/5)

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

Chia 2 trường hợp vào dùng pp thế, thế xuống pt dưới.

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)

\(\Leftrightarrow\frac{a+b}{ab}=\frac{c-\left(a+b+c\right)}{c\left(a+b+c\right)}\)

\(\Leftrightarrow\frac{a+b}{ab}=\frac{-\left(a+b\right)}{ac+bc+c^2}\)

\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2\right)=-\left(a+b\right)ab\)

\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2\right)+\left(a+b\right)ab=0\)

\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

=> a = - b or b = - c or c = - a

Với a = - b thì :

\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{-b^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{c^3}\)(1)

\(\frac{1}{a^3+b^3+c^3}=\frac{1}{-b^3+b^3+c^3}=\frac{1}{c^3}\)(2)

Từ (1);(2) \(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{a^3+b^3+c^3}\)

Xet 2 TH còn lại (b = - c; c = - a) ta cx có : \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{a^3+b^3+c^3}\)

=> đpcm

:") Làm bừa nhezzz

a) \(Q=\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^2}}\right):\frac{b}{a-\sqrt{a^2}-b^2}\)

\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a+\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}.\frac{a-\sqrt{a^2-b^2}}{b}\)

\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-\left(\sqrt{a^2-b^2}\right)^2}{b.\left(\sqrt{a^2-b^2}\right)}\)

\(=\frac{a}{\sqrt{a^2-b^2}}-\left(\frac{a^2-\left(a^2-b^2\right)}{b.\left(\sqrt{a^2-b^2}\right)}\right)\)

\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{b^2}{b\sqrt{a^2-b^2}}\)

\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{b}{\sqrt{a^2-b^2}}\)

\(=\frac{a-b}{\sqrt{a^2-b^2}}=\frac{a-b}{\sqrt{\left(a-b\right)\left(a+b\right)}}=\frac{\sqrt{a-b}}{\sqrt{a+b}}\)

b) Thay a = 3b vào , ta được :

\(Q=\frac{\sqrt{3b-b}}{\sqrt{3b+b}}=\frac{\sqrt{2b}}{\sqrt{4b}}=\sqrt{\frac{2b}{4b}}=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}\)

khó quá tui ko biết làm..

k cho tui nha

thanks

bạn Nguyễn tũn nếu như bn ko biết làm thì đừng kêu ng khác k cho mik