a: Ta có: \(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}\)

\(=6\sqrt{7}-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}\)

\(=0\)

b: Ta có: \(\left(\sqrt{125}+\sqrt{245}-\sqrt{5}\right):\sqrt{5}\)

\(=5+7-1\)

=11

\(\left(\sqrt{12}+3\sqrt{15}-4\sqrt{135}\right)\sqrt{3}\)

\(=\left(2\sqrt{3}+3\sqrt{15}-12\sqrt{15}\right)\sqrt{3}\)

\(=\left(2\sqrt{3}-9\sqrt{15}\right)\sqrt{3}\)

\(=6-9\sqrt{45}\)

\(a.\left(\sqrt{12}+3\sqrt{15}-4\sqrt{135}\right)\sqrt{3}=\left(2\sqrt{3}+3\sqrt{15}-12\sqrt{15}\right)\sqrt{3}=2.3-9\sqrt{9.5}=6-27\sqrt{5}\) \(b.\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}=\sqrt{36.7}-\sqrt{100.7}+\sqrt{144.7}-\sqrt{64.7}=6\sqrt{7}-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}=0\)

Bấm máy tính là ra thui mà bn

a/ \(=2\sqrt{3}-3\sqrt{3}+\sqrt{3}=0\)

b/ \(=\left(2\sqrt{3}-10\sqrt{3}\right)\sqrt{3}=-24\)

c/ \(=15-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}=15-6\sqrt{7}\)

d/ \(=\sqrt{3}\left(2\sqrt{3}+3\sqrt{3}-\sqrt{3}\right)=12\)

Bài 1:

a: \(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}\)

\(=6\sqrt{7}-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}\)

\(=8\sqrt{7}\)

Bài 3: 

a: \(\sqrt{27^2-23^2}=10\sqrt{2}\)

b: \(\sqrt{37^2-35^2}=12\)

c: \(\sqrt{65^2-63^2}=16\)

d: \(\sqrt{117^2-108^2}=45\)

c.

=(6-10+12-8)\(\sqrt{7}\)

=0

d.

Tiếp nè bạn

\(\dfrac{\sqrt{405}+3\sqrt{27}}{3\sqrt{3}+3\sqrt{5}}\)

\(=\dfrac{9\sqrt{5}+9\sqrt{3}}{3\sqrt{3}+3\sqrt{5}}\)

\(=\dfrac{9+9}{3+3}\)

\(=\dfrac{18}{6}=3\)

a)

\(\sqrt{12}-\sqrt{27}+\sqrt{3}=\sqrt{4}.\sqrt{3}-\sqrt{9}.\sqrt{3}+\sqrt{3}=2\sqrt{3}-3\sqrt{3}+\sqrt{3}\)

\(=\sqrt{3}(2-3+1)=0\)

b)

\(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}=\sqrt{4}.\sqrt{63}-\sqrt{4}.\sqrt{175}+\sqrt{4}.\sqrt{252}-\sqrt{4}.\sqrt{112}\)

\(=2(\sqrt{63}-\sqrt{175}+\sqrt{252}-\sqrt{112})\)

\(=2(\sqrt{9}.\sqrt{7}-\sqrt{25}.\sqrt{7}+\sqrt{36}.\sqrt{7}-\sqrt{16}.\sqrt{7})\)

\(=2(3\sqrt{7}-5\sqrt{7}+6\sqrt{7}-4\sqrt{7})=2\sqrt{7}(3-5+6-4)=0\)

------------------

\(\sqrt{3}(\sqrt{12}+\sqrt{27}-\sqrt{3})=\sqrt{36}+\sqrt{81}-\sqrt{9}\)

\(=\sqrt{6^2}+\sqrt{9^2}-\sqrt{3^2}=6+9-3=12\)

c)

\(\frac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}=\frac{\sqrt{2}.\sqrt{3}+\sqrt{2}.\sqrt{5}}{\sqrt{7}.\sqrt{3}+\sqrt{7}.\sqrt{5}}=\frac{\sqrt{2}(\sqrt{3}+\sqrt{5})}{\sqrt{7}(\sqrt{3}+\sqrt{5})}=\frac{\sqrt{2}}{\sqrt{7}}\)

\(\frac{\sqrt{405}+3\sqrt{27}}{3\sqrt{3}+\sqrt{45}}=\frac{\sqrt{81}.\sqrt{5}+3\sqrt{9}.\sqrt{3}}{3\sqrt{3}+\sqrt{9}.\sqrt{5}}=\frac{9\sqrt{5}+9\sqrt{3}}{3\sqrt{3}+3\sqrt{5}}\)

\(=\frac{3(3\sqrt{5}+3\sqrt{3})}{3\sqrt{3}+3\sqrt{5}}=3\)

d)

\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-(\sqrt{6}+\sqrt{9}+\sqrt{12})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

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

Bài 1:

a) Sửa đề: \(\left(\sqrt{12}+3\sqrt{5}-4\sqrt{135}\right)\cdot\sqrt{3}\)

Ta có: \(\left(\sqrt{12}+3\sqrt{5}-4\sqrt{135}\right)\cdot\sqrt{3}\)

\(=\sqrt{12}\cdot\sqrt{3}+3\sqrt{5}\cdot\sqrt{3}-4\sqrt{135}\cdot\sqrt{3}\)

\(=6+3\sqrt{15}-36\sqrt{5}\)

b) Ta có: \(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}\)

\(=3\sqrt{28}-5\sqrt{28}+3\sqrt{112}-2\sqrt{112}\)

\(=-2\sqrt{28}+\sqrt{112}=-\sqrt{112}+\sqrt{112}=0\)

c) Ta có: \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)

\(=2\cdot4\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}-2\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}-3\cdot2\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}\)

\(=8\sqrt{5}\cdot\sqrt{\sqrt{3}}-2\sqrt{5}\sqrt{\sqrt{3}}-6\sqrt{5}\sqrt{\sqrt{3}}\)

=0

Bài 2:

a) Ta có: \(A=\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)

\(=\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}\)

\(=\frac{1}{\sqrt{2}}\)

b) Ta có: \(B=\frac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}\)

\(=\frac{\sqrt{405}+\sqrt{243}}{\sqrt{5}+\sqrt{3}}\)

\(=\frac{9\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=9\)

c) Ta có: \(C=\frac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)

\(=\frac{\sqrt{72}-\sqrt{48}+\sqrt{20}}{\sqrt{162}-\sqrt{108}+\sqrt{45}}\)

\(=\frac{2\left(\sqrt{18}-\sqrt{12}+\sqrt{5}\right)}{3\left(\sqrt{18}-\sqrt{12}+\sqrt{5}\right)}=\frac{2}{3}\)

 \(a,\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)

\(=-13\sqrt{3}\)

\(b,2\sqrt{3}.\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)

\(=2\sqrt{3}.6\sqrt{3}=36\)

\(c,\left(2\sqrt{2}-\sqrt{3}\right)^2=8-4\sqrt{6}+3\)

\(=11-4\sqrt{6}\)

\(d,\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)=1+2\sqrt{3}+3-2\)

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