a ) \(\sqrt{6+\sqrt{35}}.\sqrt{6-\sqrt{35}}=1\)
\(\Leftrightarrow VT=\sqrt{\left(6+\sqrt{35}\right)\left(6-\sqrt{35}\right)}\)
\(\Leftrightarrow VT=\sqrt{6^2-35}=\sqrt{1}=1=VP\)
b ) \(VT=\left(\sqrt{2}-1\right)^2=2+1-2\sqrt{2}=3-2\sqrt{2}\)
\(VP=\sqrt{9}-\sqrt{8}=3-2\sqrt{2}\)
=> \(VT=VP.\)
Bài 1:
a)\(\sqrt{\left(2\sqrt{6}-4\right)^2}+\sqrt{15-6\sqrt{6}}\)
b) \(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{19+2\sqrt{18}}\)
c) \(\sqrt{9+4\sqrt{5}}-\sqrt{\left(1-\sqrt{5}^2\right)}\)
Bài 2: Biến đổi biểu thức
a) \(\dfrac{1}{\sqrt{7}+3}+\dfrac{1}{\sqrt{7}-3}\)
b) \(\dfrac{3}{\sqrt{2}-1}+\dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{3}+1}\)
c) \(\dfrac{1}{7+4\sqrt{3}}+\dfrac{1}{7-4\sqrt{3}}\)
Tính :
a) \(\dfrac{5+2\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}}-\left(\sqrt{5}+\sqrt{3}\right)\)
b) \(\left(\dfrac{1}{2-\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\right):\dfrac{1}{\sqrt{21+12\sqrt{3}}}\)
c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)
d) \(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}\)
e) \(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
f) \(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{5-2\sqrt{6}}\right)}{9\sqrt{3}-11\sqrt{2}}\)
g) \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)-\dfrac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
a) \(\sqrt{\left(2\sqrt{6}-4\right)^2}+\sqrt{15-6\sqrt{6}}\)
b) \(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{19+2\sqrt{18}}\)
c) \(\sqrt{9+4\sqrt{5}}-\sqrt{\left(1-\sqrt{5}^2\right)}\)
Tính các giá trị của các biểu thức sau:
a. A= \(\dfrac{2}{\sqrt{3}+1}+\dfrac{6}{\sqrt{3}-1}+1\)
b. B= \(\dfrac{\sqrt{\dfrac{7}{2}+\sqrt{6}}.\left(\sqrt{12}-\sqrt{2}\right)}{\sqrt{20}}\)
Rút gọn các biểu thức:
1. A= \(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
2. B= \(\left(2\sqrt{27}-3\sqrt{48}+3\sqrt{75}-\sqrt{192}\right)\)\(\left(1-\sqrt{3}\right)\)
3. C= (\(2\sqrt{7}-2\sqrt{6}\) ) . \(\sqrt{6}\) - \(\sqrt{168}\)
4. D=( \(\sqrt{28}-2\sqrt{8}+\sqrt{7}\) ). \(\sqrt{7}+4\sqrt{14}\)
a chứng minh rằng: \(\dfrac{x+3+2.\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}=\dfrac{\sqrt{x^2-9}}{x-3}\)
b rút gọn biểu thức T = \(\dfrac{x^2+5x+6+x.\sqrt{9-x^2}}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}\)
a.\(\sqrt{17}-6\sqrt{2}+3+\sqrt{2 }\)
b.\(\left(3+\sqrt{ }5\right).\left(\sqrt{ }10.\sqrt{ }2\right).\sqrt{3-\sqrt{ }5}\)
c.\(\left(\sqrt{2}-3\right).\sqrt{11+6\sqrt{2}}\)
d.\(\sqrt{23+8\sqrt{7}}-\sqrt{2}\)
nhanh nha gấp lắm trcs 9h
Tính giá trị biểu thức:
\(a,\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}\)
\(b,\frac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}-\frac{1}{\sqrt{3}+\sqrt{2}+\sqrt{5}}\)
\(c,\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right):\frac{1}{\sqrt{5}-\sqrt{2}}\)
\(d,\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)
Rút gọn các biểu thức:
1. A=\(\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)\)
2. B= \(\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
3. C= \(\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)\)
4. D= \(\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)
5. E= \(\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4\)
6. F= \(\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}\)
