Ta có : x² + 4x - y² + 4

= ( x² + 4x + 4 ) - y²

= ( x + 2 )² - y²

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

\(x^2+4x-y^2+4\)

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

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

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

Ta có : \(A=x^2+2y^2+2xy-4y-3\)

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

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

Có \(\hept{\begin{cases}\left(x+y\right)^2\ge0\forall x,y\\\left(y-2\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x+y\right)^2+\left(y-2\right)^2-7\ge-7\forall x,y\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+y=0\\y-2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\y=2\end{cases}}\)

=> MinA = -7 <=> x = -2 ; y = 2

? tìm min

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

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

Trong tam giác vuông, đường phân giác ứng với cạnh huyền và bằng nửa cạnh huyền

PT <=> \(x^2+8x+16+8x\left(x^2+8x+16\right)+15x^2\)

\(=x^2+8x+16+8x^3+64x^2+112+15x^2\)

\(=80x^2+8x^3+128=8\left(10x^2+x^3+16\right)\)

Số tự nhiên a chia 3 dư 2 => a có dạng 3k + 2 ( k ∈ N )

a² = ( 3k + 2 )² = 9k² + 12k + 4 = 9k² + 12k + 4

Ta có : 9k² chia hết cho 3 ; 12k chia hết cho 3 ; 4 chia 3 dư 1

=> 9k² + 12k + 4 chia 3 dư 1

hay a² chia 3 dư 1 ( đpcm )

\(a^2+b^2+c^2=ab+bc+ca\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)\)

\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

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

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Vì \(\left(a-b\right)^2\ge0\), \(\left(b-c\right)^2\ge0\), \(\left(c-a\right)^2\ge0\)\(\forall a,b,c\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Leftrightarrow a=b=c\)

Vậy \(a=b=c\)( đpcm )

\(a^2+b^2+c^2=ab+bc+ca\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)\)

\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

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

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Ta có : ( a - b )² ≥ 0 ∀ a, b ; ( b - c )² ≥ 0 ∀ b, c ; ( c - a )² ≥ 0 ∀ c, a

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)

Dấu "=" xảy ra khi và chỉ khi a = b = c ( đpcm )