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b: Ta có: \(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)
\(=2-\sqrt{3}+\dfrac{1}{3}\sqrt{3}-1+\dfrac{1}{3}\sqrt{3}\)
\(=\dfrac{3-\sqrt{3}}{3}\)
a: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}\)
\(=\dfrac{1}{\sqrt{2}+1}=\sqrt{2}-1\)
\(a.x=3-2\sqrt{2}\\ \Rightarrow\sqrt{x}=\sqrt{3-2\sqrt{2}}\\ =\sqrt{2-2\sqrt{2}+1}\\ =\sqrt{\left(\sqrt{2}-1\right)^2}\\ =\left|\sqrt{2}-1\right|\\ =\sqrt{2}-1\left(vì\sqrt{2}>1\right)\)
Thay \(\sqrt{x}=\sqrt{2}-1\) vào A ta được
\(A=\dfrac{\sqrt{2}-1}{1+\sqrt{2}-1}=\dfrac{\sqrt{2}-1}{\sqrt{2}}=\dfrac{\sqrt{2}-2}{2}\)
\(b.B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}-\dfrac{10-5\sqrt{x}}{x-5\sqrt{x}+6}\\ B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{10-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{10-5\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ B=\dfrac{x-3\sqrt{x}-\sqrt{x}+3-x+4-10+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{1}{\sqrt{x}-2}\)
\(c,P=A:B\\ P=\dfrac{\sqrt{x}}{1+\sqrt{x}}:\dfrac{1}{\sqrt{x}-2}\\ P=\dfrac{x-2\sqrt{x}}{1+\sqrt{x}}\)
\(P=\dfrac{-\sqrt{x}\left(-\sqrt{x}+2\right)}{\sqrt{x}+1}\)
Có: \(\sqrt{x}\ge0\)
\(\Rightarrow\sqrt{x}+1\ge1\left(I\right)\)
Lại có: \(\sqrt{x}\ge0\)
\(\Rightarrow-\sqrt{x}\le0\\ \Rightarrow-\sqrt{x}+2\le2\)
mà \(-\sqrt{x}\le0\)
\(\Rightarrow-\sqrt{x}\left(-\sqrt{x}+2\right)\ge2\)
Kết hợp với \(\left(I\right)\) \(\Rightarrow\) \(P=\dfrac{-\sqrt{x}\left(-\sqrt{x}+2\right)}{\sqrt{x}+1}\ge2\)
Vậy gtnn của P = \(2\) khi \(x=10+4\sqrt{6}\)
a: Khi \(x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\) thì
\(A=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}}{1+\sqrt{\left(\sqrt{2}-1\right)^2}}=\dfrac{\sqrt{2}-1}{1+\sqrt{2}-1}=\dfrac{\sqrt{2}-1}{\sqrt{2}}=\dfrac{2-\sqrt{2}}{2}\)
b: \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}-\dfrac{10-5\sqrt{x}}{x-5\sqrt{x}+6}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-4\sqrt{x}+3-x+4+5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}-2}\)
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
a) \(E=2\sqrt{40\sqrt{12}}+3\sqrt{5\sqrt{48}}-2\sqrt{\sqrt{75}}-4\sqrt{15\sqrt{27}}.\)
\(=8\sqrt{5\sqrt{3}}+6\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}-12\sqrt{5\sqrt{3}}}\)
\(=0\)
b) \(F=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}.\)
Vì \(=\frac{5}{12}-\frac{1}{\sqrt{6}}=\frac{5-2\sqrt{6}}{12}=\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}\)
\(\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}=\frac{2\sqrt{3}+\sqrt{2}}{6}\)
Nên \(F=\frac{2\sqrt{3}+\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}\)
\(\dfrac{2ab}{\sqrt{a}-\sqrt{b}}=\dfrac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\)
\(\dfrac{1}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}+\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\)
\(\dfrac{3}{\sqrt{10}+\sqrt{7}}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}+\sqrt{7}\right)\left(\sqrt{10}-\sqrt{7}\right)}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{3}=\sqrt{10}-\sqrt{7}\)
\(\dfrac{2}{\sqrt{6}-\sqrt{5}}=\dfrac{2\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}=\dfrac{2\left(\sqrt{6}+\sqrt{5}\right)}{6-5}=2\left(\sqrt{6}+\sqrt{5}\right)\)
a) Đặt \(P=\dfrac{3+4\sqrt{3}}{\sqrt{6}+\sqrt{2}-\sqrt{5}}\)
\(P=\dfrac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}\)
Mà \(H=\left(\sqrt{6}+\sqrt{2}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)\)
\(H=\left(\sqrt{6}+\sqrt{2}\right)^2-\left(\sqrt{5}\right)^2\)
\(H=8+4\sqrt{3}-5\)
\(H=3+4\sqrt{3}\)
Do đó \(P=\dfrac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}-\sqrt{5}\right)}{3+4\sqrt{3}}\)
\(P=\sqrt{6}+\sqrt{2}+\sqrt{5}\)
b) Đặt \(U=\dfrac{1}{2+\sqrt{5}+2\sqrt{2}+\sqrt{10}}\)
Ta có \(O=2+\sqrt{5}+2\sqrt{2}+10\)
\(O=\left(2+\sqrt{5}\right)+\sqrt{2}\left(2+\sqrt{5}\right)\)
\(O=\left(2+\sqrt{5}\right)\left(1+\sqrt{2}\right)\)
Vậy \(U=\dfrac{1}{\left(2+\sqrt{5}\right)\left(1+\sqrt{2}\right)}\)
\(U=\dfrac{\left(\sqrt{2}-1\right)\left(\sqrt{5}-2\right)}{\left[\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)\right]\left[\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)\right]}\)
\(U=\dfrac{\sqrt{10}-2\sqrt{2}-\sqrt{5}+2}{N}\)
Với \(N=\left[\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)\right]\left[\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)\right]\)
\(N=\left[\left(\sqrt{5}\right)^2-2^2\right]\left[\left(\sqrt{2}\right)^2-1\right]\)
\(N=\left(5-4\right)\left(2-1\right)\)
\(N=1\)
Do đó \(U=\sqrt{10}-2\sqrt{2}-\sqrt{5}+2\)