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a. ĐKXĐ : \(\hept{\begin{cases}x\ge0\\y\ge0\\y-x\ne0\end{cases}}\)<=> \(\hept{\begin{cases}x\ge0\\y\ge0\\x\ne y\end{cases}}\)
b. \(R=\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{x^3}-\sqrt{y^3}}{y-x}\right):\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(\Leftrightarrow R=\left(\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}+\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{y-x}\right):\frac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)
\(\Leftrightarrow R=\left(\sqrt{x}+\sqrt{y}-\frac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right):\frac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)
\(\Leftrightarrow R=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}.\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(\Leftrightarrow R=\frac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{x-\sqrt{xy}+y}\)
\(\Leftrightarrow R=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
c. Với \(\hept{\begin{cases}x\ge0\\y\ge0\\x\ne y\end{cases}}\)thì \(\sqrt{xy}\ge0\) ( 1 )
Ta có : \(x-\sqrt{xy}+y=\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}\)
Mà \(\orbr{\begin{cases}\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\\\left(1\right)\end{cases}}\)=> \(x-\sqrt{xy}+y\ge0\)( 2 )
Từ ( 1 ) và ( 2 ) => \(R\ge0\) ( Đpcm )
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
Bài 1: \(\left(\frac{\sqrt{xy}-\sqrt{y}}{\sqrt{x}-1}+\frac{\sqrt{xy}-\sqrt{x}}{\sqrt{y}-1}\right)\cdot\left(\sqrt{xy}-\sqrt{y}\right)\)
\(=\left(\frac{\sqrt{y}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\frac{\sqrt{x}\left(\sqrt{y}-1\right)}{\sqrt{y}-1}\right)\cdot\left(\sqrt{xy}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{xy}-\sqrt{y}\right)\)
\(\sqrt{9+4\sqrt{2}}-\sqrt{9-4\sqrt{2}}=\sqrt{\left(2\sqrt{2}+1\right)^2}-\sqrt{\left(2\sqrt{2}-1\right)^2}\\ =2\sqrt{2}+1-2\sqrt{2}+1=2\)
Bài 2:
\(Q=\left(\frac{\sqrt{x}}{1-\sqrt{x}}+\frac{\sqrt{x}}{1+\sqrt{x}}\right)+\frac{3-\sqrt{x}}{x-1}\left(ĐK:x\ge0;x\ne1\right)\)
\(=\frac{-\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}\left(\sqrt{x}-1\right)+3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-x-\sqrt{x}+x-\sqrt{x}+3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{-3}{\sqrt{x}+1}\)
Để Q=2
=> \(\frac{-3}{\sqrt{x}+1}=2\)
\(\Leftrightarrow2\left(\sqrt{x}+1\right)=-3\)
\(\Leftrightarrow2\sqrt{x}+2=-3\)
\(\Leftrightarrow2\sqrt{x}=-5\) (vô lí)
Vậy k có giá trị nào của x thỏa mãn Q=2
a) ĐK : tự ghi nha
b)
\(Q=\left(\frac{2\sqrt{xy}}{x-y}-\frac{\sqrt{x}+\sqrt{y}}{2\sqrt{x}-2\sqrt{y}}\right).\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(Q=\left(\frac{4\sqrt{xy}}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}-\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right).\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(Q=\left(\frac{4\sqrt{xy}-\left(\sqrt{x}+\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right).\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(Q=\left(\frac{4\sqrt{xy}-\left(x+y+2\sqrt{xy}\right)}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right).\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(Q=\left(\frac{4\sqrt{xy}-x-y-2\sqrt{xy}}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right).\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(Q=\left(\frac{2\sqrt{xy}-x-y}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right).\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(Q=-\left(\frac{x-2\sqrt{xy}+y}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right).\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(Q=-\left(\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right).\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(Q=-\left(\frac{\sqrt{x}-\sqrt{y}}{2\left(\sqrt{x}+\sqrt{y}\right)}\right).\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(Q=-\frac{1}{2\left(\sqrt{x}+\sqrt{y}\right)}.2\sqrt{x}\)
\(Q=-\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)
P /s : Các bạn tham khảo nhé
Bài 2 :
a) \(ĐKXĐ:\hept{\begin{cases}x;y>0\\x\ne y\end{cases}}\)
b) \(A=\left(\sqrt{x}+\frac{y-\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\right):\frac{x\sqrt{xy}+y\sqrt{xy}}{\sqrt{xy}\left(y-x\right)}\)
\(\Leftrightarrow A=\frac{x-\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}-\sqrt{y}}:\frac{x+y}{y-x}\)
\(\Leftrightarrow A=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}-\sqrt{y}}\cdot\frac{y-x}{x+y}\)
\(\Leftrightarrow A=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(y-x\right)}{x+y}\)
c) Thay \(x=4+2\sqrt{3},y=4-2\sqrt{3}\)vào A, ta được :
\(A=\frac{\left(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\right)\left(4-2\sqrt{3}-4-2\sqrt{3}\right)}{4+2\sqrt{3}+4-2\sqrt{3}}\)
\(\Leftrightarrow A=\frac{\left(\sqrt{\left(1+\sqrt{3}\right)^2}-\sqrt{\left(1-\sqrt{3}\right)^2}\right).\left(-4\sqrt{3}\right)}{8}\)
\(\Leftrightarrow A=\frac{\left(1+\sqrt{3}-\sqrt{3}+1\right).\left(-4\sqrt{3}\right)}{8}=\frac{-8\sqrt{3}}{8}=-\sqrt{3}\)
Vậy ....
Bài 1:
\(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}-\sqrt{2}}=\frac{2\sqrt{2\cdot4}-\sqrt{3\cdot4}}{\sqrt{2\cdot9}-\sqrt{16\cdot3}}-\frac{\sqrt{5}+\sqrt{9\cdot3}}{\sqrt{30}-\sqrt{2}}\)
\(=\frac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\frac{\sqrt{5}+3\sqrt{3}}{\sqrt{30}-\sqrt{2}}=\frac{\left(4\sqrt{2}-2\sqrt{3}\right)\left(\sqrt{30}-\sqrt{2}\right)-\left(\sqrt{5}+3\sqrt{3}\right)\left(3\sqrt{2}-4\sqrt{3}\right)}{\left(3\sqrt{2}-4\sqrt{3}\right)\left(\sqrt{30}-\sqrt{2}\right)}\)
\(=\frac{4\sqrt{60}-8-2\sqrt{90}+2\sqrt{6}-3\sqrt{10}+4\sqrt{15}-9\sqrt{6}+36}{3\sqrt{60}-6-4\sqrt{90}+4\sqrt{6}}\)
\(=\frac{8\sqrt{15}-8-6\sqrt{10}+2\sqrt{6}-3\sqrt{10}+4\sqrt{15}-9\sqrt{6}+36}{6\sqrt{15}-6-12\sqrt{10}+4\sqrt{6}}\)
\(=\frac{12\sqrt{15}-2\sqrt{10}-7\sqrt{6}+28}{6\sqrt{15}-12\sqrt{10}+4\sqrt{6}-6}\)
\(Q=\frac{x-y}{\sqrt{x}-\sqrt{y}}-\frac{\sqrt{x^3}-\sqrt{y^3}}{x-y}\)
\(Q=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-y\right)-x\sqrt{x}+y\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(Q=\frac{x\sqrt{x}-y\sqrt{x}+x\sqrt{y}-y\sqrt{y}-x\sqrt{x}+y\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(Q=\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(Q=\frac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(R=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(R=\left[\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right].\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(R=\left(1+\sqrt{a}+a\right).\frac{\left(1-\sqrt{a}\right)^2}{\left(1-\sqrt{a}\right)^2.\left(1+\sqrt{a}\right)^2}\)
\(=\left(1+\sqrt{a}\right)^2.\frac{1}{\left(1+\sqrt{a}\right)^2}=1\)
\(A=\frac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}+\frac{y}{\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)}-\frac{x+y}{\sqrt{xy}}\)
\(A=\frac{x\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)+y\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)-\left(x+y\right)\left(y-x\right)}{\sqrt{xy}\left(y-x\right)}\)
\(A=\frac{x\sqrt{xy}-x^2+y\sqrt{xy}+y^2-y^2+x^2}{\sqrt{xy}\left(y-x\right)}\)
\(A=\frac{\sqrt{xy}\left(x+y\right)}{\sqrt{xy}\left(y-x\right)}=\frac{y+x}{y-x}\)
KO CÓ GIÁ TRỊ y sao tính đây !!!!!!
CÒN \(x=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\) nhé
\(1A=\frac{x+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\frac{x-y}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}-\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(=1\sqrt{x}-\sqrt{x}-\sqrt{y}=-\sqrt{y}\)