\(\dfrac{x}{y}=\dfrac{3}{-2}\)

⇒\(\dfrac{x}{3}=\dfrac{y}{-2}\)

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

\(\dfrac{x}{3}=\dfrac{y}{-2}=\dfrac{x-y}{3-\left(-2\right)}=\dfrac{20}{5}=4\)

⇒\(\left\{{}\begin{matrix}x=4.3=12\\y=4.-2=-8\end{matrix}\right.\)

a: \(x^2-5x+10\)

\(=x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{15}{4}\)

\(=\left(x-\dfrac{5}{2}\right)^2+\dfrac{15}{4}>0\forall x\)

b: \(2x^2+8x+15\)

\(=2\left(x^2+4x+\dfrac{15}{2}\right)\)

\(=2\left(x^2+4x+4+\dfrac{7}{2}\right)\)

\(=2\left(x+2\right)^2+7>0\forall x\)

Vì 0xy+yz+xz=0.Nên:X,y,z đều bằng 0 và bằng nhau.

Có : \(2x^2+3x+2\)

\(\Leftrightarrow\) \(\left(x^2+2x+1^2\right)+\left(x^2+x+1^2\right)\)

\(\Leftrightarrow\) \(\left(x^2+2.x.1+1^2\right)\) + \(\left(x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\right)\)

\(\Leftrightarrow\) \(\left(x+1\right)^2+\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(x+1\right)^2\ge0và\left(x+\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow\) \(\left(x+1\right)^2+\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

Vậy \(2x^2+3x+2>0\left(\forall_x\right)\)

a)

\(2x^2+3x+2=\left(x^2+2x+1\right)+\left(x^2+2\cdot\frac{1}{2}x+\frac{1}{4}\right)+\frac{3}{4}\\ =\left(x+1\right)^2+\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

(vì >3/4 nên >0)

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

<=>  \(1-\frac{1}{x^2}-\frac{1}{y^2}+\frac{1}{x^2y^2}=1+\frac{2}{xy}\)

<=>   \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}-\frac{1}{x^2y^2}=0\)

<=>   \(\left(\frac{1}{x}+\frac{1}{y}\right)^2-\frac{1}{x^2y^2}=0\)

<=>  \(\left(\frac{x+y}{xy}\right)^2-\frac{1}{x^2y^2}=0\)

<=>  \(\frac{1}{x^2y^2}-\frac{1}{x^2y^2}=0\)   luôn đúng

=>  đpcm

bạn cảm ơn ai vay có bn ấy có giup bn làm đau

mk chua hok den nen ko co bit lam

cảm ơn b nhé

A = ( x + y )( x + 2y )( x + 3y )( x + 4y ) + y⁴

= [ ( x + y )( x + 4y ) ][ ( x + 2y )( x + 3y ) ] + y⁴

= ( x² + 5xy + 4y² )( x² + 5xy + 6y² ) + y⁴ (1)

Đặt t = x² + 5xy + 5y²

(1) <=> ( t - y² )( t + y² ) + y⁴

       = t² - y⁴ + y⁴

       = t² = ( x² + 5xy + 5y² )²

Vì x, y nguyên => x² nguyên ; 5xy nguyên ; 5y² nguyên

=> x² + 5xy + 5y² nguyên

=> ( x² + 5xy + 5y² )² là một số chính phương

=> đpcm

A = ( x + y )( x + 2y )( x + 3y )( x + 4y ) + y⁴ 

=> A = ( x² + 5xy + 4y² ) ( x² + 5xy + 6y² ) + y⁴

Đặt a = x² + 5xy + 5y² , pt trở thành :

A = ( a - y² ) ( a + y² ) + y⁴

=> A = t² - y⁴ + y⁴ = t² = ( x² + 5xy + 5y² )² là SCP

Vậy A là SCP

Ta có: \(A=\left(x+y\right)\left(x+2y\right)\left(x+3y\right)\left(x+4y\right)+y^4\)

\(A=\left[\left(x+y\right)\left(x+4y\right)\right]\left[\left(x+2y\right)\left(x+3y\right)\right]+y^4\)

\(A=\left(x^2+5xy+4y^2\right)\left(x^2+5xy+6y^2\right)+y^4\)

\(A=\left(x^2+5xy+5y^2-y^2\right)\left(x^2+5xy+5y^2+y^2\right)+y^4\)

\(A=\left(x^2+5xy+5y^2\right)^2-y^4+y^4\)

\(A=\left(x^2+5xy+5y^2\right)^2\) là SCP

=> đpcm

a) VT = x³ + 3x²y + 3xy² + y³ + x³ - 3x²y + 3xy² - y³

          = 2x³ + 6xy²

          = 2x( x² + 3y² ) = VP

=> đpcm

b) VT = x³ + 3x²y + 3xy² + y³ - ( x³ - 3x²y + 3xy² - y³ )

          = x³ + 3x²y + 3xy² + y³ - x³ + 3x²y - 3xy² + y³

          = 3x²y + 2y³

          = 2y( 3x² + y² ) = VP

=> đpcm

a)

 \(VT=\left(x+y+x-y\right)\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)

\(=2x\left[x^2+2xy+y^2-x^2+y^2+x^2-2xy+y^2\right]\)

\(=2x\left(x^2+3y^2\right)=VP\)

b)

\(VT=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)

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

\(=2y\left(3x^2+y^2\right)=VP\)

a) \(\left(x+y\right)^3+\left(x-y\right)^3\)

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

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

\(=2x\left(x^2+3y^2\right)\)

b) \(\left(x+y\right)^3-\left(x-y\right)^3\)

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

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

\(=2y\left(3x^2+y^2\right)\)