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a) \(\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\(=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{ab}-1}\)
b) \(\frac{x+\sqrt{xy}}{y+\sqrt{xy}}=\frac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}=\sqrt{\frac{x}{y}}\)
c) \(\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\sqrt{\frac{3}{7}}\)
d) \(\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}=\frac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}\left(\sqrt{1}-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)}\)
\(=\frac{\left(2\sqrt{5}-\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{5}-\sqrt{3}\right)\left(1-\sqrt{2}\right)}=\frac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)
e) \(\frac{-3\sqrt{3}+3}{2\sqrt{3}-2}=\frac{-3\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}-1\right)}=-\frac{3}{2}\)
a/ \(A=\frac{30\left(\sqrt{6}-1\right)}{5}+\frac{2\left(\sqrt{6}+2\right)}{2}-\frac{6\left(3+\sqrt{6}\right)}{3}=6\sqrt{6}-6+\sqrt{6}+2-6-2\sqrt{6}\)
\(A=5\sqrt{6}-10\)
\(B=\sqrt{17-6\sqrt{2}+\sqrt{8+4\sqrt{2}+1}}\)
\(B=\sqrt{17-6\sqrt{2}+\sqrt{\left(2\sqrt{2}+1\right)^2}}=\sqrt{18-4\sqrt{2}}\)
Đến đây ko rút gọn được nữa, nhưng nếu đề là:
\(B=\sqrt{17+6\sqrt{2}+\sqrt{8+4\sqrt{2}+1}}=\sqrt{18+8\sqrt{2}}=4+\sqrt{2}\)
c/
\(C=\sqrt{8-2\sqrt{7}}+\sqrt{8+2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(C=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\)
\(D=\sqrt{a-2\sqrt{a}+1}-\sqrt{a-8\sqrt{a}+16}\)
\(D=\sqrt{\left(\sqrt{a}-1\right)^2}-\sqrt{\left(4-\sqrt{a}\right)^2}=\sqrt{a}-1-\left(4-\sqrt{a}\right)=2\sqrt{a}-5\)
\(E=\sqrt{a-2+2\sqrt{a-2}+1}+\sqrt{a-2-2\sqrt{a-2}+1}\) (\(a\ge2\))
\(E=\sqrt{\left(\sqrt{a-2}+1\right)^2}+\sqrt{\left(\sqrt{a-2}-1\right)^2}\)
\(E=\sqrt{a-2}+1+\left|\sqrt{a-2}-1\right|\)
\(\Rightarrow\left[{}\begin{matrix}E=2\sqrt{a-2}\left(a\ge3\right)\\E=2\left(2\le a\le3\right)\end{matrix}\right.\)
\(F=\sqrt[3]{10+6\sqrt{3}}-\sqrt{3}=\sqrt[3]{1+3.1.\sqrt{3}+3.1.\sqrt{3}^2+\sqrt{3}^3}-\sqrt{3}\)
\(F=\sqrt[3]{\left(1+\sqrt{3}\right)^3}-\sqrt{3}=1+\sqrt{3}-\sqrt{3}=1\)
\(G=\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\Rightarrow G^3=\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)^3\)
\(\Rightarrow G^3=14+3\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)\left(\sqrt[3]{49-50}\right)\)
\(\Rightarrow G^3=14-3G\Rightarrow G^3+3G-14=0\)
\(\Rightarrow G=2\)
a/ \(\frac{\sqrt{a}-\left(\sqrt{a}\right)^2}{\sqrt{a}-1}\)
=\(\frac{\sqrt{a}\left(1-\sqrt{a}\right)}{\sqrt{a}-1}\)
=\(\frac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\)
=\(-\sqrt{a}\)
a: \(=-6\sqrt{b}-\dfrac{1}{3}\cdot3\sqrt{3b}+\dfrac{1}{5}\cdot5\sqrt{6b}\)
\(=-6\sqrt{b}-\sqrt{3}\cdot\sqrt{b}+\sqrt{6}\cdot\sqrt{b}\)
\(=\sqrt{b}\left(-6-\sqrt{3}+\sqrt{6}\right)\)
c: \(=\sqrt{\left(5+2\sqrt{6}\right)^2}+\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=5+2\sqrt{6}+5-2\sqrt{6}=10\)
d: \(A=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
e: \(B=\sqrt{6+2\sqrt{5-2\sqrt{3}-1}}\)
\(=\sqrt{6+2\cdot\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
Ok, ko chép đề nha =))
\(A=\sqrt[3]{\frac{135}{5}}-\sqrt[3]{54\cdot4}\\ =\sqrt[3]{27}-\sqrt[3]{216}=3-6=-3\)
\(B=\frac{1}{2}\cdot\sqrt[3]{2\cdot4}-\frac{1}{4}\cdot\sqrt[3]{16\cdot4}\\ =\frac{1}{2}\cdot\sqrt[3]{8}-\frac{1}{4}\cdot\sqrt[3]{64}\\ =\frac{1}{2}\cdot2-\frac{1}{4}\cdot4=1-1=0\)
\(C=\sqrt[3]{\left(\sqrt{2}+1\right)\left(3+2\sqrt{2}\right)}\\ =\sqrt[3]{\left(\sqrt{2}+1\right)\left(2+2\cdot\sqrt{2}\cdot1+1\right)}\\ =\sqrt[3]{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)^2}\\ =\sqrt[3]{\left(\sqrt{2}+1\right)^3}=\sqrt{2}+1\)
\(D=\sqrt[3]{\frac{3\left(\sqrt[3]{2}-1\right)\left(1+\sqrt[3]{2}+\sqrt[3]{4}\right)}{\sqrt[3]{2}-1}}\\ =\sqrt[3]{\frac{3\left(2-1\right)}{\sqrt[3]{2}-1}}\\ =\sqrt[3]{\frac{3}{\sqrt[3]{2}-1}}\) (chịu, ko bít rút thêm :V)
\(E=\) chịu nốt =))
Chúc bạn học tốt nha
.
\(\frac{\sqrt{6}+\sqrt{3}+\sqrt{2}}{\left(\sqrt{6}+\sqrt{2}-\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{3}\right)}=\frac{\sqrt{6}+\sqrt{3}+\sqrt{2}}{\left(\sqrt{6}+\sqrt{2}\right)^2-3}=\frac{\sqrt{6}+\sqrt{3}+\sqrt{2}}{5+4\sqrt{3}}\)
\(=\frac{\left(4\sqrt{3}-5\right)\left(\sqrt{6}+\sqrt{3}+\sqrt{2}\right)}{\left(4\sqrt{3}+5\right)\left(4\sqrt{3}-5\right)}=\frac{12\sqrt{2}+12+4\sqrt{6}-5\sqrt{6}-5\sqrt{3}-5\sqrt{2}}{23}\)
\(=\frac{12+7\sqrt{2}-\sqrt{6}-5\sqrt{3}}{23}\)