1/a/
\(A=\frac{2}{xy}+\frac{3}{x^2+y^2}=\left(\frac{1}{xy}+\frac{1}{xy}+\frac{4}{x^2+y^2}\right)-\frac{1}{x^2+y^2}\)

\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}-\frac{1}{\frac{\left(x+y\right)^2}{2}}=16-2=14\)

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)

b/

\(4B=\frac{4}{x^2+y^2}+\frac{8}{xy}+16xy=\left(\frac{4}{x^2+y^2}+\frac{1}{xy}+\frac{1}{xy}\right)+\left(\frac{1}{xy}+16xy\right)+\frac{5}{xy}\)

\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}+2\sqrt{\frac{1}{xy}.16xy}+\frac{5}{\frac{\left(x+y\right)^2}{4}}\)

\(=16+8+20=44\)

\(\Rightarrow B\ge11\)

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)

2/

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

\(\ge\frac{\left(x+\frac{1}{x}+y+\frac{1}{y}\right)^2}{2}\ge\frac{\left(1+\frac{4}{x+y}\right)^2}{2}=\frac{25}{2}\)

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)

\(\text{Tìm x:}\)

\(a.x\left(x-1\right)-3x+3x=0\)

\(x\left(x-1\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=1\end{cases}}}\)

\(b.3x\left(x-2\right)+10-5x=0\)

\(3x^2-6x+10-5x=0\)

\(3x^2-11x+10=0\)

\(3x^2-11x=-10\)(bn xem lại đề nhé)

\(c.x^3-5x^2+x-5=0\)

\(x^3-5x^2+x=5\)

\(d.x^4-2x^3+10x^2-20x=0\)

bài 1:phân tích thành phân tử

  a> x^2-6x-y^2+9

= (x-3)^2 -y^2

= (x-3 -y) (x-3+y)

b>x^2-xy-8x+8y

= x(x-y) - 8(x-y)

= (x-8) (x-y)

c>25-4x^2-4xy-y^2

= 5^2 - (2x + y)^2 

= (5 - 2x -y) (5 +2x+y) 

d>xy-xz-y+z

= x(y-z) - (y-z)

= (x-1) (y-z)

e>x^2-xz-yz+2xy+y^2

= (x+y)^2 - z(x+y)

= (x+y-z) (x+y)

g>x^2-4xy+4y^2-z^2-4zt-4t^2

= (x-2y)^2 - (z + 2t)^2 

= (x-2y -x-2t) (x-2y + z +2t)

bài 2:tìm X bt 

a>x.(x-1)-3x+3x=0

x (x-1) =0

\(\Rightarrow\hept{\begin{cases}x=0\\x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=1\end{cases}}}\)

Vậy x=0 và x=1

b>3x.(x-2)+10-5x=0

3x(x-2) - 5 (x-2)=0

(3x-5) (x-2) =0

\(\Rightarrow\hept{\begin{cases}3x-5=0\\x-2=0\end{cases}\Rightarrow\hept{\begin{cases}3x=5\\x=2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}}}\)

c>x^3-5x^2+x-5=0

x^2 (x-5) + (x-5) =0

(x^2 +1)(x-5) =0

\(\Rightarrow\hept{\begin{cases}x^2+1=0\\x-5=0\end{cases}\Rightarrow\hept{\begin{cases}x^2=-1\\x=5\end{cases}\Rightarrow}\hept{\begin{cases}x\in\varphi\\x=5\end{cases}}}\)

Vậy x=5

d>x^4-2x^3+10x^2-20x=0

x^3 (x-2) + 10x(x-2) =0 

(x^3 + 10x) (x-2) =0

x(x^2 + 10) (x-2) =0

\(\Rightarrow\hept{\begin{cases}x=0\\x^2+10=0\\x-2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=-10\\x=2\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x\in\varphi\\x=2\end{cases}}}}\)

Vậy x=0 và x=2

Ta có: \(x+y+z=0\)

\(\Rightarrow x+y=-z\)

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

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

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

Chứng minh tương tự ta có:

\(x^2+z^2-y^2=-2xz\)

\(y^2+z^2-x^2=-2yz\)

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

\(=\frac{xy}{-2xy}+\frac{xz}{-2xz}+\frac{yz}{-2yz}\)

\(=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}\)

\(=-\frac{3}{2}\)

Vậy giá trị biểu thức là \(-\frac{3}{2}\)

áp dụng bđt cô si ta có:

\(xy\le\frac{x^2+y^2}{2};yz\le\frac{y^2+z^2}{2};zx\le\frac{z^2+x^2}{2}\)

\(\Rightarrow A\ge\sqrt{\frac{x^2+y^2}{2}}+\sqrt{\frac{y^2+z^2}{2}}+\sqrt{\frac{z^2+x^2}{2}}\)

theo bunhia thì \(2\left(x^2+y^2\right)\ge\left(x+y\right)^2;2\left(y^2+z^2\right)\ge\left(y+z\right)^2;2\left(z^2+x^2\right)\ge\left(z+x\right)^2\)

\(\Rightarrow A\ge\sqrt{\frac{\left(x+y\right)^2}{4}}+\sqrt{\frac{\left(y+z\right)^2}{4}}+\sqrt{\frac{\left(z+x\right)^2}{4}}=\frac{x+y}{2}+\frac{y+z}{2}+\frac{z+x}{2}=x+y+z=1\)

Vậy \(Min_A=1\Leftrightarrow x=y=z=\frac{1}{3}\)

ta có:

\(|x-\frac{1}{2}|=|y+\frac{2}{3}|=|x^2+xz|=0\)

ta tính được x=\(\frac{1}{2}\)

                   y=\(\frac{-2}{3}\)

va \(\frac{1}{2^2}+\frac{1}{2}z=0\)

\(\Rightarrow\)\(z=\frac{-1}{2}\)