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BẠN LÀM CKO CÁI MẪU TRONG DẤU NGOẶC THỨ NHẤT THÀNH HẰNG ĐẲNG THỨC SỐ 3 RỒI LÀM ,..
Ta có: \(\left(\sqrt{x}+\frac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right):\left(\frac{x}{\sqrt{xy}+y}+\frac{y}{\sqrt{xy}-x}-\frac{x+y}{\sqrt{xy}}\right)\)
\(=\frac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\left(\frac{x\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)}{\sqrt{xy}\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{y}-\sqrt{x}\right)}+\frac{y\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{y}-\sqrt{x}\right)}-\frac{\left(x+y\right)\left(y-x\right)}{\sqrt{xy}\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{y}-\sqrt{x}\right)}\right)\)
\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}:\left(\frac{x\sqrt{xy}-x^2+y\sqrt{xy}+y^2-\left(y^2-x^2\right)}{\sqrt{xy}\left(y-x\right)}\right)\)
\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}:\left(\frac{x\sqrt{xy}+y\sqrt{xy}}{\sqrt{xy}\left(y-x\right)}\right)\)
\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}:\frac{\sqrt{xy}\left(x+y\right)}{\sqrt{xy}\left(y-x\right)}\)
\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}:\frac{x+y}{\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\frac{x+y}{\sqrt{y}+\sqrt{x}}\cdot\frac{\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{y}-\sqrt{x}\right)}{x+y}\)
\(=\sqrt{y}-\sqrt{x}\)
Bài 1
a, \(\left(\frac{\sqrt{y}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\frac{\sqrt{x}\left(\sqrt{y}-1\right)}{\sqrt{y}-1}\right).\sqrt{y}\left(\sqrt{x}-1\right)\)
=\(\left(\sqrt{y}+\sqrt{x}\right).\sqrt{y}\left(\sqrt{x}-1\right)\)
b,\(\sqrt{8+2.2\sqrt{2}+1}-\sqrt{8-2.2\sqrt{2}+1}\)
=\(\sqrt{\left(\sqrt{8}+1\right)^2}-\sqrt{\left(\sqrt{8}-1\right)^2}\)
=\(\sqrt{8}+1-\left(\sqrt{8}-1\right)\)
=2
Bài 2
a, ĐKXĐ : x\(\ge\)0, x\(\pm\)1
b, Q=\(\left(\frac{\sqrt{x}\left(1+\sqrt{x}\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}+\frac{\sqrt{x}\left(1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}\right)+\frac{3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\left(\frac{\sqrt{x}\left(1+\sqrt{x}\right)+\sqrt{x}\left(1-\sqrt{x}\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right)+\frac{3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\left(\frac{\sqrt{x}+x+\sqrt{x}-x}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right)+\frac{3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}-\frac{3-\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
=\(\frac{2\sqrt{x}-3+\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
=\(\frac{3\sqrt{x}-3}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
=\(\frac{-3}{1+\sqrt{x}}\)
c, de Q = 2 => \(\frac{-3}{1+\sqrt{x}}\)=2 =>1+\(\sqrt{x}\)=-6 =>\(\sqrt{x}\)=-7 =>x vô nghiệm
Ta có
\(BT=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)=x-y\)