Ta có: \(6x+5y+18=2xy\)

\(\Leftrightarrow6x+5y-2xy=-18\)

\(\Leftrightarrow2x\left(3-y\right)+5y=-18\)

\(\Leftrightarrow2x\left(3-y\right)+5y-15=-18-15\)

\(\Leftrightarrow2x\left(3-y\right)+5\left(y-3\right)=-33\)

\(\Leftrightarrow2x\left(3-y\right)-5\left(3-y\right)=-33\)

\(\Leftrightarrow\left(3-y\right)\left(2x-5\right)=-33\)

Dễ rồi

a) khai triển được 2sin²+2cos²=2(sin²+cos²=2.1=2

b)cot²-cos².cot²=cot²(1-cos²)=cot².sin²=cos²/sin².sin²=cos²

c)sin.cos(tan+cot)=sin.cos.tan+sin.cos.cot=sin.cos.sin/cos+sin.cos.cos/sin=sin²+cos²=1

d)tan²-tan².sin²=tan²(1-sin²)=tan².cos²=sin²/cos².cos²=sin²

\(\sqrt{x}+\sqrt{y}+\sqrt{xy}+1\)

\(=\sqrt{x}+\sqrt{y}+\sqrt{x}.\sqrt{y}+1\)

\(=\sqrt{x}\left(\sqrt{y}+1\right)+\left(\sqrt{y}+1\right)\)

\(=\left(\sqrt{x}+1\right)\left(\sqrt{y}+1\right)\)

( căn x + 1 ) + ( căn y + căn xy )
( căn x + 1 ) + căn y.( căn x +  1)
( căn x +1 )(căn y + 1 )

\(100^2-99^2+98^2-97^2+...+2^2-1^2\)

\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)

\(=100+99+98+97+...+2+1\)

\(=\frac{100.\left(100+1\right)}{2}\)

\(=5050\)

  \(100^2-99^2+98^2-...+2^2-1^2\)

=\(\left(100-99\right).\left(100+99\right)+\left(98-97\right).\left(98+97\right)+...+\left(2-1\right).\left(2+1\right)\)

\(=100+99+98+98+...+2+1\)

\(=5050\)

giúp mình với ạ

bạn kiểm tra lại thử đề có sai không ạ?

\(A=\left(15.\sqrt{200}-4\sqrt{450}+2\sqrt{20}\right):\sqrt{10}-10\sqrt{20}\)      \(=\left(15.10\sqrt{2}-4.15\sqrt{2}+2.2\sqrt{5}\right):\sqrt{10}-10.2\sqrt{5}\)

   \(=\left(150\sqrt{2}-60\sqrt{2}+4\sqrt{5}\right):\sqrt{10}-20\sqrt{5}\)

   \(=\left(90\sqrt{2}+4\sqrt{5}\right):\sqrt{10}-20\sqrt{5}\)

   \(=\frac{90\sqrt{2}}{\sqrt{10}}+\frac{4\sqrt{5}}{\sqrt{10}}-20\sqrt{5}\)

     \(=18\sqrt{5}+2\sqrt{2}-20\sqrt{5}\)

    \(=2\sqrt{2}-2\sqrt{5}\)