Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=\left(x^2+y^2-2xy\right)\left(x^2+y^2+2xy\right)\)
\(=\left(x+y\right)^2\cdot\left(x-y\right)^2\)
\(a,VT=\left(a^2-1\right)^2+4a^2\\ =a^4-2a^2+1+4a^2\\ =a^4+2a^2+1\\ =\left(a^2+1\right)^2 =VP\\ b,VT=\left(x-y\right)^2+\left(x+y\right)^2+2\left(x^2-y^2\right)\\ =x^2-2xy+y^2+x^2+y^2+2xy+2x^2-2y^2\\ =4x^2=VP\)
\(\left(x+y\right)^2+\left(x-y\right)^2=2\left(x^2+y^2\right)\)
\(\Leftrightarrow x^2+2xy+y^2+x^2-2xy=2\left(x^2+y^2\right)\)
\(\Leftrightarrow2x^2+2y^2=2\left(x^2+y^2\right)\left(đúng\right)\)
a: Ta có: \(\left(x+y\right)^2\)
\(=x^2+2xy+y^2\)
\(\Leftrightarrow x^2+y^2=\dfrac{\left(x+y\right)^2}{2xy}\ge\dfrac{\left(x+y\right)^2}{2}\forall x,y>0\)
a) Ta có:
\(VT=\left(a-b\right)^2\)
\(=a^2-2ab+b^2\)
\(=a^2+2ab+b^2-4ab\)
\(=\left(a+b\right)^2-4ab=VP\left(dpcm\right)\)
b) Ta có:
\(VT=\left(x+y\right)^2+\left(x-y\right)^2\)
\(=x^2+2xy+y^2+x^2-2xy+y^2\)
\(=\left(x^2+y^2\right)+\left(x^2+y^2\right)\)
\(=2\left(x^2+y^2\right)=VP\left(dpcm\right)\)
\(\dfrac{\left(a+b\right)^2-\left(a-b\right)^2}{4}=\dfrac{a^2+2ab+b^2-a^2+2ab-b^2}{4}=\dfrac{4ab}{4}=ab\left(đpcm\right)\)
\(\left(x+y\right)^2+\left(x-y\right)^2=x^2+2xy+y^2+x^2-2xy+y^2=2x^2+2y^2=2\left(x^2+y^2\right)\left(dpcm\right)\)
a) VT = ( a + b + a − b ) ( a + b − a + b ) 4 = 2 a . 2 b 4 = 4 = VP => đpcm.
b) VP = x 2 + 2 xy + y 2 + x 2 – 2 xy + y 2 = 2 ( x 2 + y 2 ) = VT => đpcm.
CMR : a) Có thể tìm được số có dạng 199119911991...19910...0 chia hết cho 1992
Help
\(\left(x+y\right)\left(x+y+2\right)\)
\(=\left(x+y\right)^2+2\left(x+y\right)\)
\(=x^2+y^2+2xy+2\left(x+y\right)\)
\(=2+2xy+2\left(x+y\right)\)
\(=2\left(xy+x+y+1\right)\)
\(=2\left(x+1\right)\left(y+1\right)\)