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a) Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)\cdot\sqrt{9+2\sqrt{14}}\)

\(=\left(\sqrt{7}-\sqrt{2}\right)\cdot\left(\sqrt{7}+\sqrt{2}\right)\)

=7-2

=5

d) Ta có: \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)

\(=2\sqrt{2}-\sqrt{7}+5\sqrt{7}-\dfrac{2\sqrt{2}\left(3-\sqrt{2}\right)}{3-\sqrt{2}}\)

\(=2\sqrt{2}+4\sqrt{7}-2\sqrt{2}\)

\(=4\sqrt{7}\)

a) Sửa đề: \(A=\sqrt{8+2\sqrt{7}}-\sqrt{7}\)

Ta có: \(A=\sqrt{8+2\sqrt{7}}-\sqrt{7}\)

\(=\sqrt{7+2\cdot\sqrt{7}\cdot1+1}-\sqrt{7}\)

\(=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{7}\)

\(=\left|\sqrt{7}+1\right|-\sqrt{7}\)

\(=\sqrt{7}+1-\sqrt{7}\)

=1

b) Ta có: \(B=\sqrt{7+4\sqrt{3}}-2\sqrt{3}\)

\(=\sqrt{4+2\cdot2\cdot\sqrt{3}+3}-2\sqrt{3}\)

\(=\sqrt{\left(2+\sqrt{3}\right)^2}-2\sqrt{3}\)

\(=\left|2+\sqrt{3}\right|-2\sqrt{3}\)

\(=2+\sqrt{3}-2\sqrt{3}\)

\(=2-\sqrt{3}\)

c) Ta có: \(C=\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}\)

\(=\sqrt{13-2\cdot\sqrt{13}\cdot1+1}+\sqrt{13+2\cdot\sqrt{13}\cdot1+1}\)

\(=\sqrt{\left(\sqrt{13}-1\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}\)

\(=\left|\sqrt{13}-1\right|+\left|\sqrt{13}+1\right|\)

\(=\sqrt{13}-1+\sqrt{13}+1\)

\(=2\sqrt{13}\)

d) Ta có: \(D=\sqrt{22-2\sqrt{21}}-\sqrt{22+2\sqrt{21}}\)

\(=\sqrt{21-2\cdot\sqrt{21}\cdot1+1}-\sqrt{21+2\cdot\sqrt{21}\cdot1+1}\)

\(=\sqrt{\left(\sqrt{21}-1\right)^2}-\sqrt{\left(\sqrt{21}+1\right)^2}\)

\(=\left|\sqrt{21}-1\right|-\left|\sqrt{21}+1\right|\)

\(=\sqrt{21}-1-\left(\sqrt{21}+1\right)\)

\(=\sqrt{21}-1-\sqrt{21}-1\)

=-2

a) Ta có: \(2\sqrt{80}+3\sqrt{45}-\sqrt{245}\)

\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}\)

\(=10\sqrt{5}\)

b) Ta có: \(\dfrac{3}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)

\(=3\left(2-\sqrt{3}\right)+4+\sqrt{3}+2\sqrt{3}\)

\(=6-2\sqrt{3}+4+3\sqrt{3}\)

\(=10+\sqrt{3}\)

c) Ta có: \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

=7-5=2

d) Ta có: \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{28-10\sqrt{3}}\)

\(=2+\sqrt{3}-5+\sqrt{3}\)

\(=-3+2\sqrt{3}\)

6 tháng 7 2021

a. \(2\sqrt{80}+3\sqrt{45}-\sqrt{245}\)

\(=2.4\sqrt{5}+3.3\sqrt{5}-7\sqrt{5}\)

\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}\)

\(=10\sqrt{5}\)

b. \(\dfrac{3}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)

\(=\dfrac{3\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\dfrac{13\left(4+\sqrt{3}\right)}{\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)}+\dfrac{6\sqrt{3}}{\sqrt{3}.\sqrt{3}}\)

\(=\dfrac{3\left(2-\sqrt{3}\right)}{4-3}+\dfrac{13\left(4+\sqrt{3}\right)}{16-3}+\dfrac{6\sqrt{3}}{3}\)

\(=3\left(2-\sqrt{3}\right)+\dfrac{13\left(4+\sqrt{3}\right)}{13}+2\sqrt{3}\)

\(=6-3\sqrt{3}+4+\sqrt{3}+2\sqrt{3}\)

\(=10\)

c. \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

\(=\left(\sqrt{7}+\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

\(=7-5=2\)

d. \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{28-10\sqrt{3}}\)

\(=\left|2+\sqrt{3}\right|-\sqrt{5^2-2.5.\sqrt{3}+\left(\sqrt{3}\right)^2}\)

\(=\left|2+\sqrt{3}\right|-\left(5-\sqrt{3}\right)^2\)

\(=\left|2+\sqrt{3}\right|-\left|5-\sqrt{3}\right|\)

\(=2+\sqrt{3}-\left(5-\sqrt{3}\right)\) (vì \(\left|2+\sqrt{3}\right|\ge0,\left|5-\sqrt{3}\right|\ge0\))

\(=2+\sqrt{3}-5+\sqrt{3}\)

\(=2\sqrt{3}-3\)

AH
Akai Haruma
Giáo viên
17 tháng 9 2021

Lời giải:
a.

\(=2\sqrt{4^2.5}+3\sqrt{3^2.5}-\sqrt{7^2.5}=2.4\sqrt{5}+3.3\sqrt{5}-7\sqrt{5}\)

\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}=10\sqrt{5}\)

b.

\(=\frac{3(2-\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}+\frac{13(4+\sqrt{3})}{(4-\sqrt{3})(4+\sqrt{3})}+\frac{6\sqrt{3}}{3}\)

\(=\frac{6-3\sqrt{3}}{1}+\frac{13(4+\sqrt{3})}{13}+2\sqrt{3}=6-3\sqrt{3}+4+\sqrt{3}+2\sqrt{3}\)

\(=10\)

AH
Akai Haruma
Giáo viên
17 tháng 9 2021

c.

\(=\left[\frac{\sqrt{7}(\sqrt{2}-1)}{\sqrt{2}-1}+\frac{\sqrt{5}(\sqrt{3}-1)}{\sqrt{3}-1}\right].(\sqrt{7}-\sqrt{5})\)

\(=(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})=7-5=2\)

d.

\(=|2+\sqrt{3}|-\sqrt{5^2-2.5\sqrt{3}+3}=|2+\sqrt{3}|-\sqrt{(5-\sqrt{3})^2}\)

\(=|2+\sqrt{3}|-|5-\sqrt{3}|=2+\sqrt{3}-(5-\sqrt{3})=-3+2\sqrt{3}\)

 

a: \(=\sqrt{5}-1\)

b: \(=\sqrt{2}-1\)

c: \(=\sqrt{3}+1\)

d: \(=\sqrt{13}+1\)

8 tháng 1 2023

\(=\left(3\sqrt{2}-2\sqrt{2}+\sqrt{14}\right).\sqrt{2}-\sqrt{7}\\ =\left(\sqrt{2}+\sqrt{14}\right).\sqrt{2}-\sqrt{7}\\ =2+2\sqrt{7}-\sqrt{7}\\ =2+\sqrt{7}\)

21 tháng 10 2023

1:

a: \(\sqrt{25}+\sqrt{49}=5+7=12\)

b: \(\sqrt{121}-\sqrt{81}=11-9=2\)

2: x>-2

=>2x>-4

=>2x+1>-3

=>Với x>-2 thì \(\sqrt{2x+1}\) chưa chắc có nghĩa

3:

a: \(\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)

\(=\left|\sqrt{3}-1\right|-\sqrt{3}\)

\(=\sqrt{3}-1-\sqrt{3}=-1\)

b: \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)

\(=\left(3\sqrt{7}-2\sqrt{14}\right)\cdot\sqrt{7}+14\sqrt{2}\)

\(=21-14\sqrt{2}+14\sqrt{2}=21\)

c:

\(\dfrac{\sqrt{27}-\sqrt{108}+\sqrt{12}}{\sqrt{3}}\)

\(=\dfrac{3\sqrt{3}-6\sqrt{3}+2\sqrt{3}}{\sqrt{3}}=3+2-6=-1\)