`@` `\text {Ans}`

`\downarrow`

`a)`

`A + (x^2 - 4xy^2 + 2xz - 3y^2) = 2xz - 5xy^2 - x^2`

`=> A = (2xz - 5xy^2 - x^2) - (x^2 - 4xy^2 + 2xz - 3y^2)`

`= 2xz - 5xy^2 - x^2 - x^2 + 4xy^2 - 2xz + 3y^2`

`= (2xz - 2xz) + (-5xy^2 + 4xy^2) + (-x^2 - x^2) + 3y^2`

`= -xy^2 - 2x^2 + 3y^2`

Vậy, `A= -xy^2 - 2x^2 - 3y^2`

`b)`

`B - (xy+y^2-x^2) = x^2 + y^2`

`=> B = x^2 + y^2 + xy + y^2 - x^2`

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

`= 2y^2 + xy`

Vậy, `B = 2y^2 + xy.`

a: A=2xz-5xy^2-x^2-x^2+4xy^2-2xz+3y^2

=-2x^2-xy^2+3y^2

b: B=x^2+y^2+xy+y^2-x^2

=2y^2+xy

a) \(\dfrac{2x^2-2xy}{x^2+x-xy-y}\) \(\left(x\ne y;x\ne-1\right)\)

\(=\dfrac{2x\left(x-y\right)}{x\left(x+1\right)-y\left(x+1\right)}\)

\(=\dfrac{2x\left(x-y\right)}{\left(x-y\right)\left(x+1\right)}\)

\(=\dfrac{2x}{x+1}\)

b) \(\dfrac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}\)

\(=\dfrac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}\)

\(=\dfrac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)

\(=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{\left(x-y+z\right)\left(x+y+z\right)}\)

\(=\dfrac{x+y-z}{x-y+z}\)

Hướng dẫn :\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\)

Thay vào:\(x^2+2yz=x^2+yz+yz=x^2+yz-xy-zx=x\left(x-y\right)-z\left(x-y\right)=\left(x-y\right)\left(x-z\right)\)

Tương tự thay vào mà quy đồng

tao đẹp trai thì có gì sai

bài này mà là âm nhạc???

Bài 3:

a, (\(x\)+y+z)²

=((\(x\)+y) +z)²

= (\(x\) + y)² + 2(\(x\) + y)z + z²

= \(x^2\) + 2\(xy\) + y² + 2\(xz\) + 2yz + z²

=\(x^2\) + y² + z² + 2\(xy\) + 2\(xz\) + 2yz

b, (\(x-y\))(\(x^2\) + y² + z² - \(xy\) - yz - \(xz\))

= \(x^3\) + \(xy^2\) + \(xz^2\) - \(x^2\)y - \(xyz\) - \(x^2\)z - y³ 

Đến dây ta thấy xuất hiện \(x^3\) - y³ khác với đề bài, em xem lại đề bài nhé

c,

(\(x\) + y + z)³ 

=(\(x\) + y)³ + 3(\(x\) + y)²z + 3(\(x\)+y)z² + z³

= \(x^3\) + 3\(x^2\)y + 3\(xy^{2^{ }}\) + y³ +  3(\(x\)+y)z(\(x\) + y + z) + z³

= \(x^3\) + y³ + z³ + 3\(xy\)(\(x\) + y) + 3(\(x+y\))z(\(x+y+z\))

= \(x^3\) + y³ + z³ + 3(\(x\) + y)( \(xy\) + z\(x\) + yz + z²)

= \(x^3\) + y³ + z³ + 3(\(x\) + y){(\(xy+xz\)) + (yz + z²)}

= \(x^3\) + y³ + z³ + 3(\(x\) + y){ \(x\)( y +z) + z(y+z)}

= \(x^3\) + y³ + z³ + 3(\(x\) + y)(y+z)(\(x+z\)) (đpcm)

g. G(x)=2x²+2y²+z²+2xy-2xz-2yz-2x-4y

           = [x²+2x(y-z)+(y²-2yz+z²)]+(x²-2x+1)+(y²-4y+4)-5

          = (x+y-z)²+(x-1)²+(y-2)²-5

Vì (x+y-z)²≥0∀x,y,z

     (x-1)²≥0∀x

      (y-2)²≥0∀y

⇒ G  = (x+y-z)²+(x-1)²+(y-2)²-5 ≥ -5

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+y-z=0\\x-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z=3\\x=1\\y=2\end{matrix}\right.\)

h,H(x)=x² + y²-xy-x+y+1

⇔ 2H=2x²+2y²-2xy-2x-2y+2

         = (x²-2xy+y²)+(x²-2x+1)+(y²-2y+1)

         = (x-y)²+(x-1)²+(y-1)²

Vì (x-y)²≥0 ∀x,y

    (x-1)²≥0 ∀x

     (y-1)² ≥0 ∀y

⇒ 2H≥0 ⇒ H≥0

Dấu "=" xảy ra ⇔ x=y=1