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\(b,\sqrt{2}.\sqrt{7+3\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{14+6\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\sqrt{5^2}+2.3\sqrt{5}+3^2}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\left(\sqrt{5}+3\right)^2}-\dfrac{4}{\sqrt{5}-1}\\ =\left|\sqrt{5}+3\right|-\dfrac{4}{\sqrt{5}-1}\\ =\dfrac{\left(\sqrt{5}+3\right)\left(\sqrt{5}-1\right)-4}{\sqrt{5}-1}\\ =\dfrac{2+2\sqrt{5}-4}{\sqrt{5}-1}\\ =\dfrac{-2+2\sqrt{5}}{\sqrt{5}-1}\\ =\dfrac{2\left(-1+\sqrt{5}\right)}{\sqrt{5}-1}\\ =2\)
\(c,\sqrt{27}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\\ =3\sqrt{3}-\dfrac{6}{\sqrt{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\)
\(=\dfrac{3\sqrt{3}.\sqrt{3}-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{9-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{\sqrt{3}}{\sqrt{3}}\\ =1\)
\(d,\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}\\ =\dfrac{\left(9-2\sqrt{3}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}\\ =\dfrac{27\sqrt{6}+18\sqrt{2}-18\sqrt{2}-4\sqrt{6}}{\left(3\sqrt{6}\right)^2-\left(2\sqrt{2}\right)^2}\\ =\dfrac{23\sqrt{6}}{54-8}\\ =\dfrac{23\sqrt{6}}{46}\\ =\dfrac{\sqrt{6}}{2}\)
a) \(\Leftrightarrow A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)
b) \(\Leftrightarrow B=\sqrt{7-2\sqrt{12}}+\sqrt{12+2\sqrt{27}}=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}=2-\sqrt{3}+3+\sqrt{3}=5\)
c) \(\Leftrightarrow C=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{6}{4}=\dfrac{3}{2}\)
d) \(\Leftrightarrow D=3-\left(-2\right)-5=0\)
Bài 1:
a) Ta có: \(\left(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}-\dfrac{5}{4}\sqrt{\dfrac{4}{5}}+\sqrt{5}\right)\)
\(=\left(\sqrt{5}+\sqrt{5}-\dfrac{5}{4}\cdot\dfrac{2}{\sqrt{5}}+\sqrt{5}\right)\)
\(=3\sqrt{5}-\dfrac{1}{2}\sqrt{5}\)
\(=\dfrac{5}{2}\sqrt{5}\)
c) Ta có: \(\dfrac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
\(=\dfrac{\sqrt{35}\left(\sqrt{5}-\sqrt{7}+2\sqrt{2}\right)}{\sqrt{35}}\)
\(=2\sqrt{2}+\sqrt{5}-\sqrt{7}\)
Bài 2:
e) ĐKXĐ: \(\dfrac{4}{3}\le x\le6\)
Ta có: \(\sqrt{6-x}=3x-4\)
\(\Leftrightarrow6-x=\left(3x-4\right)^2\)
\(\Leftrightarrow9x^2-24x+16+6-x=0\)
\(\Leftrightarrow9x^2-25x+22=0\)
\(\Delta=\left(-25\right)^2-4\cdot9\cdot22=625-792< 0\)
Vậy: Phương trình vô nghiệm
Bài 1:
a: \(5\sqrt{8}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
\(=5\cdot2\sqrt{2}-4\cdot3\sqrt{3}-2\cdot5\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-16\sqrt{3}\)
b: \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(1-\sqrt{6}\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|1-\sqrt{6}\right|\)
\(=3-\sqrt{6}+\sqrt{6}-1\)
=3-1=2
c: \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\dfrac{1}{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\dfrac{1\left(4-\sqrt{15}\right)}{16-15}\)
\(=\sqrt{15}+4-\sqrt{15}=4\)
d: \(\dfrac{2\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)
\(=\dfrac{\sqrt{3-\sqrt{5}}\cdot\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}+1\right)}{2\left(\sqrt{3}+1\right)}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\dfrac{\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=3+\sqrt{5}-\dfrac{\sqrt{5}}{2}=3+\dfrac{\sqrt{5}}{2}\)
Bài 2:
Vẽ đồ thị:
Phương trình hoành độ giao điểm là:
\(\dfrac{1}{2}x-4=-3x+3\)
=>\(\dfrac{1}{2}x+3x=3+4\)
=>\(\dfrac{7}{2}x=7\)
=>x=2
Thay x=2 vào y=-3x+3, ta được:
\(y=-3\cdot2+3=-3\)
Vậy: (d1) cắt (d2) tại A(2;-3)
a: \(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+\dfrac{9}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)
\(=60-20\sqrt{18}+\dfrac{45}{2}\sqrt{12}\)
\(=60-60\sqrt{2}+45\sqrt{3}\)
b: \(=\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}\)
\(=\dfrac{2\sqrt{5}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}=\dfrac{2\sqrt{5}+3}{9+6\sqrt{2}}\)
a: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)
b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)
c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)
a: \(=\left(-\sqrt{5}-\sqrt{7}\right)\cdot\left(\sqrt{7}-\sqrt{5}\right)\)
\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=-2
b: \(=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}\)
c: \(=\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}-2-\sqrt{10}+3\sqrt{7}+2\)
\(=\sqrt{10}-\sqrt{10}+3\sqrt{7}=3\sqrt{7}\)
\(1.\text{ }\dfrac{1}{\sqrt{k}-\sqrt{k+1}}=\dfrac{\left(\sqrt{k}+\sqrt{k+1}\right)}{\left(\sqrt{k}+\sqrt{k+1}\right)\left(\sqrt{k}-\sqrt{k+1}\right)}\\ =-\left(\sqrt{k}+\sqrt{k+1}\right)\\ \Rightarrow\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-...-\dfrac{1}{\sqrt{8}-\sqrt{9}}\\ =-\left(\sqrt{1}+\sqrt{2}\right)+\left(\sqrt{2}+\sqrt{3}\right)-\left(\sqrt{3}+\sqrt{4}\right)+...+\left(\sqrt{8}+\sqrt{9}\right)\\ =-\sqrt{1}-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+...+\sqrt{8}+\sqrt{9}\\ \\ =\sqrt{9}-\sqrt{1}=2\)
\(2.\text{ }\dfrac{1}{\left(k+1\right)\sqrt{k}+\sqrt{k+1}k}=\dfrac{1}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}+\sqrt{k}\right)}\\ =\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}\\ =\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(k+1-k\right)}=\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}\\ =\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\\ \Rightarrow\text{ }\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{7\sqrt{6}+6\sqrt{7}}\\ =\text{ }\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{6}}-\dfrac{1}{\sqrt{7}}\\ =\text{ }\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{7}}\\ \text{ }1-\dfrac{1}{\sqrt{7}}\)
1.\(\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-\dfrac{1}{\sqrt{4}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{8}}-\dfrac{1}{\sqrt{8}-\sqrt{9}}=\dfrac{1+\sqrt{2}}{1-2}-\dfrac{\sqrt{2}+\sqrt{3}}{2-3}+\dfrac{\sqrt{3}+\sqrt{4}}{3-4}-\dfrac{\sqrt{4}+\sqrt{5}}{4-5}+\dfrac{\sqrt{5}+\sqrt{6}}{5-6}-\dfrac{\sqrt{6}+\sqrt{7}}{6-7}+\dfrac{\sqrt{7}+\sqrt{8}}{7-8}-\dfrac{\sqrt{8}+\sqrt{9}}{8-9}=-1-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+\sqrt{4}+\sqrt{5}-\sqrt{5}-\sqrt{6}+\sqrt{6}+\sqrt{7}-\sqrt{7}-\sqrt{8}+\sqrt{8}+\sqrt{9}=\sqrt{9}-1=3-1=2\)
a: \(=\dfrac{1}{2}\cdot2\sqrt{3}+3\sqrt{3}-5\sqrt{3}=-\sqrt{3}\)
b: \(=2-\sqrt{3}-\sqrt{3}-1=1\)
c: \(=18\sqrt{3}-10\sqrt{3}-\dfrac{1}{2}\cdot10\sqrt{3}=3\sqrt{3}\)
d: \(=\sqrt{10}+\sqrt{3}-\sqrt{5}+\sqrt{2}-2\sqrt{3}=\sqrt{10}+\sqrt{2}-\sqrt{3}-\sqrt{5}\)