Giúp mình nha!!!!!

Như nài cx được nè :3 

\(A=\left(x^2-3x+1\right)\left(x^2+2x+1\right)-6x^2\)

\(=\left(x^4+2x^3+x^2-3x^3-6x^2-3x+x^2+2x+1\right)-6x^2\)

\(=\left(x^4-x^3-4x^2-x+1\right)-\left(\sqrt{6}x\right)^2\)

\(=\left(x^4-x^3-4x^2-x+1-\sqrt{6}x\right)\left(x^4-x^3-4x^2-x+1+\sqrt{6}x\right)\)

Ta có : 2a² + 2b² = 5ab

=> 2a² + 2b² - 4ab = 5ab - 4ab

=> 2(a² + b² - 2ab) = ab

=> (a - b)² = ab/2 

Lại có 2a² + 2b² = 5ab

=> 2a² + 2b² + 4ab = 5ab + 4ab

=> 2(a + b)² = 9ab

=> (a + b)² = 9ab/2

Ta có P² = \(\left(\frac{a+b}{a-b}\right)^2=\frac{\left(a+b\right)^2}{\left(a-b\right)^2}=\frac{\frac{9ab}{2}}{\frac{ab}{2}}=9\)

=> P = \(\pm\)3

Vậy P = \(\pm\)3

+)\(x^2-2y^2=xy\)

\(2y^2=x^2-xy\)

\(2y^2=x.\left(x-y\right)\)

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

+)\(x^2-2y^2=xy\)

\(x^2=xy+2y^2\)

\(x^2=xy+2y^2-y^2+y^2\)

\(x^2=xy+y^2+y^2\)

\(x^2=\left(x+y\right).y+y^2\)

\(\Rightarrow x^2-y^2=\left(x+y\right).y\)

\(\Rightarrow x+y=\frac{x^2-y^2}{y}\left(2\right)\)

+)Từ (1) và (2)

\(\Rightarrow A=\frac{x-y}{x+y}=\frac{\frac{2y^2}{x}}{\frac{x^2-y^2}{y}}\)

\(\Rightarrow A=\frac{2y^2}{x}:\frac{x^2-y^2}{y}\)

\(\Rightarrow A=\frac{2y^3}{x^3-x.y^2}\)

Chúc bạn học tốt

\(x-y=5\Leftrightarrow x=5+y\).

\(\frac{x-3y}{5-2y}=\frac{5+y-3y}{5-2y}=\frac{5-2y}{5-2y}=1\)

mình nhầm

đề kiểu gì vậy ?

\(A=x^2-2x.\frac{2^2}{x}-5x+2\)

\(A=x^2-2.2^2-5x+2\)

\(A=x^2-8-5x+2\)

\(A=x^2-5x+2-8\)

\(A=x^2-5x-6\)

\(B=x^2-64.\frac{x}{2}-2x-80\)

\(B=x^2-32x-2x-80\)

\(B=\left(x-32-2\right)x-80\)

\(B=\left(x-30\right)x-80\)

\(B=x^2-30x-80\)

Chúc bạn học tốt

\(\frac{x^2-x-xy+y}{xy-x-y^2+y}\)

ĐKXĐ : \(\hept{\begin{cases}x\ne y\\y\ne1\end{cases}}\)

\(=\frac{\left(x^2-x\right)-\left(xy-y\right)}{\left(xy-x\right)-\left(y^2-y\right)}\)

\(=\frac{x\left(x-1\right)-y\left(x-1\right)}{x\left(y-1\right)-y\left(y-1\right)}\)

\(=\frac{\left(x-1\right)\left(x-y\right)}{\left(y-1\right)\left(x-y\right)}\)

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

Bài làm 

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

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

\(A=x^2-2x+y^2-4y+7=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+2\)

\(=\left(x-1\right)^2+\left(y-2\right)^2+2\)

Vì : \(\left(x-1\right)^2\ge0\forall x;\left(y-2\right)^2\ge0\forall y\)

=)) \(\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\)

Vậy GTNN A = 2 <=> x = 1 ; y = 2

\(A=x^2-2x+y^2-4y+7\)

\(=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+2\)

\(=\left(x-1\right)^2+\left(y-2\right)^2+2\)

Vì \(\left(x-1\right)^2\ge0\forall x\); \(\left(y-2\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\forall x,y\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-1=0\\y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\)

Vậy \(minA=2\)\(\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\)