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\(x=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(\dfrac{x}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
\(\dfrac{x}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{2-\sqrt{3}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(\dfrac{x}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{2+\sqrt{3}+1}+\dfrac{2-\sqrt{3}}{2-\sqrt{3}+1}\)
\(\dfrac{x}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(\dfrac{x}{\sqrt{2}}=\dfrac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(3+\sqrt{3}\right)\left(2-\sqrt{3}\right)}{9-3}\)
\(\dfrac{x}{\sqrt{2}}=\dfrac{3+\sqrt{3}+3-\sqrt{3}}{6}=\dfrac{6}{6}=1\)
\(x=\sqrt{2}\)
\(y=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(y\sqrt{2}=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\)
\(y\sqrt{2}=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(y\sqrt{2}=\sqrt{7}+1-\sqrt{7}+1\)
\(y\sqrt{2}=2\)
\(y=\dfrac{2}{\sqrt{2}}\)
Thay \(x=\sqrt{2},y=\dfrac{2}{\sqrt{2}}\) vào A ta có:
\(A=\dfrac{\sqrt{2}.\dfrac{2}{\sqrt{2}}-1}{\sqrt{2}+\dfrac{2}{\sqrt{2}}}-\dfrac{1-\sqrt{2}.\dfrac{2}{\sqrt{2}}}{2\sqrt{2}-\dfrac{2}{\sqrt{2}}}\)
\(=\dfrac{2-1}{2\sqrt{2}}-\dfrac{1-2}{\sqrt{2}}\)
\(=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{\sqrt{2}}\)
\(=\dfrac{3\sqrt{2}}{4}\)
Tự kết luận nha
1.
\(\sqrt{\dfrac{x-1+\sqrt{2x-3}}{x+2-\sqrt{2x+3}}}\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\sqrt{\dfrac{\left(\sqrt{2x-3}+1\right)^2}{\left(\sqrt{2x+3}-1\right)^2}}\end{matrix}\right.\)\(\Leftrightarrow\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{2x-3}+1}{\sqrt{2x+3}-1}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\left(\sqrt{2x-3}+1\right)\left(\sqrt{2x+3}+1\right)}{2\left(x+1\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{4x^2-9}+\sqrt{2x-3}+\sqrt{2x+3}+1}{2\left(x+1\right)}\end{matrix}\right.\)
hết tối giải rồi
\(\dfrac{\left(\sqrt{X}+\sqrt{Y}\right)\left(1+\sqrt{XY}\right)+\left(\sqrt{X}-\sqrt{Y}\right)\left(1-\sqrt{XY}\right)}{1-XY}\cdot\dfrac{1-XY}{1-XY+\sqrt{X}+\sqrt{Y}+2\sqrt{XY}}=\dfrac{\sqrt{X}+X\sqrt{Y}+\sqrt{Y}+Y\sqrt{X}+\sqrt{X}-X\sqrt{Y}-\sqrt{Y}+Y\sqrt{X}}{1-XY}\cdot\dfrac{1-XY}{XY+X+Y+1}=\dfrac{2\sqrt{X}\left(1+Y\right)}{\left(1+Y\right)\left(X+1\right)}=\dfrac{2\sqrt{X}}{X+1}\)
b: Thay \(x=\dfrac{2}{2+\sqrt{3}}=2\left(2-\sqrt{3}\right)=4-2\sqrt{3}\) vào P, ta được:
\(P=\dfrac{2\left(\sqrt{3}-1\right)}{4-2\sqrt{3}+1}=\dfrac{2\sqrt{3}-2}{5-2\sqrt{3}}=\dfrac{6\sqrt{3}+2}{13}\)