a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)

\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)

\(=33\sqrt{3}\cdot\sqrt{3}\)

=99

b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)

\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)

\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)

c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)

\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)

\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)

\(=36-36\sqrt{2}+18\sqrt{3}\)

d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)

\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)

\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)

\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)

a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)

   \(=28.3+9.3-4.3=99\)

b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)

  \(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)

d,Ta có:\(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)

        \(=3\sqrt{75\sqrt{2}}+5\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)

        \(=15\sqrt{3\sqrt{2}}+20\sqrt{3\sqrt{2}}-16\sqrt{3\sqrt{2}}\)

        \(=19\sqrt{3\sqrt{2}}\)

a: \(2\sqrt{8\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\sqrt{12}}\)

\(=2\sqrt{4\cdot2\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\cdot2\sqrt{3}}\)

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

=0

b: \(\sqrt{3}+\sqrt{7-4\sqrt{3}}\)

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

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

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

=2

c: \(\sqrt{\left(\sqrt{7}-4\right)^2}-\sqrt{28}+\sqrt{63}\)

\(=\left|\sqrt{7}-4\right|-2\sqrt{7}+3\sqrt{7}\)

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

=4

d: \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)

\(=\dfrac{\sqrt{10}\left(15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\right)}{\sqrt{10}}\)

\(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)

\(=15\sqrt{5}+5\cdot2\sqrt{5}-3\cdot3\sqrt{5}\)

\(=16\sqrt{5}\)

e: \(\sqrt{3}-2\sqrt{48}+3\sqrt{75}-4\sqrt{108}\)

\(=\sqrt{3}-2\cdot4\sqrt{3}+3\cdot5\sqrt{3}-4\cdot6\sqrt{3}\)

\(=\sqrt{3}-8\sqrt{3}+15\sqrt{3}-24\sqrt{3}\)

\(=-16\sqrt{3}\)

e , \(\sqrt{11^2-\left(6\sqrt{2}\right)^2}\)

g, h. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath

Làm ngắn gọn nha, có gì ko hiểu bạn cmt nhé :3

Và ko viết lại đề nha (lười ~~)

\(1,\left(\sqrt{4\cdot7}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\\ =\left(2\sqrt{7}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{4\cdot2}\\ =\left(3\sqrt{7}-2\sqrt{14}\right)\cdot\sqrt{7}+7\sqrt{8}\\ =3\cdot7-2\cdot7\cdot\sqrt{2}+7\cdot2\sqrt{2} =21\)

\(2,\left(\sqrt{4\cdot2}-3\sqrt{2}+\sqrt{10}\right)\cdot\left(\sqrt{2}-3\sqrt{4\cdot0,1}\right)\\ =\left(2\sqrt{2}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-6\sqrt{0,1}\right)\\ =\left(\sqrt{10}-\sqrt{2}\right)\left(\sqrt{2}-6\sqrt{0,1}\right)\\ =2\sqrt{5}-2-6+6\sqrt{0,2}=2\sqrt{5}-8+6\sqrt{0,2}\)

(khúc này mình chỉ biết rút gọn đến đây thui ._.)

\(3,\left(15\sqrt{25\cdot2}+5\sqrt{100\cdot2}-3\sqrt{225\cdot2}\right):\sqrt{10}\\ =\left(75\sqrt{2}+50\sqrt{2}-45\sqrt{2}\right):\sqrt{10}\\ =80\sqrt{2}:\sqrt{10}\\ =80\sqrt{2}:\left(\sqrt{2}\cdot\sqrt{5}\right)=16\sqrt{5}\)

\(4,\sqrt{5+2\cdot\sqrt{5}+1}+\sqrt{5-2\cdot\sqrt{5}\cdot1+1}\\ =\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\\ =\left|\sqrt{5}+1\right|+\left|\sqrt{5}-1\right|\\ =\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)

\(5,\sqrt{9+2\cdot3\cdot\sqrt{2}+2}-\sqrt{9-2\cdot3\cdot\sqrt{2}+2}\\ =\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\\ =\left|3+\sqrt{2}\right|-\left|3-\sqrt{2}\right|\\ =3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)

b: Ta có: \(\left(\sqrt{7-3\sqrt{5}}\right)\cdot\left(7+3\sqrt{5}\right)\cdot\left(3\sqrt{2}+\sqrt{10}\right)\)

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

\(=4\left(7+3\sqrt{5}\right)\)

\(=28+12\sqrt{5}\)

Lời giải:

a. 

$A=\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}$
$\sqrt{2}A=\sqrt{16+2\sqrt{55}}-\sqrt{16-2\sqrt{55}}-\sqrt{250}$

$=\sqrt{(\sqrt{11}+\sqrt{5})^2}-\sqrt{(\sqrt{11}-\sqrt{5})^2}-5\sqrt{10}$

$=|\sqrt{11}+\sqrt{5}|-|\sqrt{11}-\sqrt{5}|-5\sqrt{10}$

$=2\sqrt{5}-5\sqrt{10}$

$\Rightarrow A=\sqrt{10}-5\sqrt{5}$

b.

$B=\sqrt{7-3\sqrt{5}}.(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$

$B\sqrt{2}=\sqrt{14-6\sqrt{5}}(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$

$=\sqrt{(3-\sqrt{5})^2}(7+3\sqrt{5}).\sqrt{2}(3+\sqrt{5})$

$=(3-\sqrt{5})(7\sqrt{2}+3\sqrt{10})(3+\sqrt{5})$

$=(3^2-5)(7\sqrt{2}+3\sqrt{10})$

$=4(7\sqrt{2}+3\sqrt{10})=28\sqrt{2}+12\sqrt{10}$

$\Rightarrow B=28+12\sqrt{5}$

c.

$C=\sqrt{2}(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{6+\sqrt{35}}$

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{12+2\sqrt{35}}$

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{(\sqrt{7}+\sqrt{5})^2}

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})(\sqrt{7}+\sqrt{5})$

$=(7-5)(6-\sqrt{35})$

$=2(6-\sqrt{35})=12-2\sqrt{35}$

@.@ Trời ơi, nhiều thế ^^

a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)=\left(2\sqrt{2}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-\frac{3\sqrt{2}}{\sqrt{5}}\right)\)

\(=\left(\sqrt{2}.\sqrt{5}-\sqrt{2}\right)\left(\sqrt{2}-\frac{3\sqrt{2}}{\sqrt{5}}\right)=2\sqrt{5}-2-6+\frac{6}{\sqrt{5}}=\frac{16\sqrt{5}}{5}-8\)

b) \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}=\frac{75\sqrt{2}+50\sqrt{2}-45\sqrt{2}}{\sqrt{10}}=\frac{80\sqrt{2}}{\sqrt{10}}=\frac{80}{\sqrt{5}}=16\sqrt{5}\)c) \(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(2-\sqrt{2}\right)^3}\)

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

d) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)}^2\)

\(=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)

e) \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)

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

f)\(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\sqrt[3]{\left(\sqrt{2}-1\right)^3}=1+\sqrt{2}-\sqrt{2}+1=2\)g) \(\sqrt[3]{26+15\sqrt{3}}-\sqrt[3]{26-15\sqrt{3}}=\sqrt[3]{\left(2+\sqrt{3}\right)^3}-\sqrt[3]{\left(2-\sqrt{3}\right)^3}\)

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

Bài 1: Rút gọn

a) Ta có: \(\left(\sqrt{3}-\sqrt{2}+1\right)\cdot\left(\sqrt{3}-1\right)\)

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

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

\(=2-\sqrt{2}-\sqrt{6}\)

b) Ta có: \(0.2\cdot\sqrt{\left(-10\right)^2\cdot3}+2\cdot\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\)

\(=0.2\cdot\sqrt{\left(-10\right)^2}\cdot\sqrt{3}+2\cdot\left(\sqrt{5}-\sqrt{3}\right)\)

\(=0.2\cdot10\cdot\sqrt{3}+2\sqrt{5}-2\sqrt{3}\)

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

\(=2\sqrt{5}\)

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

\(=\sqrt{196}-2\cdot\sqrt{98}+\sqrt{49}+7\sqrt{8}\)

\(=14-\sqrt{392}+7+\sqrt{392}\)

=21

d) Ta có: \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)

\(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)

\(=\sqrt{5}\left(15+5\cdot2-3\cdot3\right)\)

\(=16\sqrt{5}\)

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

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

b, \(=\sqrt{300.0,04}+2\left|\sqrt{3}-\sqrt{5}\right|\)

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

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

c, \(=\sqrt{196}-2\sqrt{98}+\sqrt{49}+7\sqrt{8}\)

\(=14-14\sqrt{2}+7+14\sqrt{2}=21\)

d, \(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)

\(=15\sqrt{5}+10\sqrt{5}-9\sqrt{5}=16\sqrt{5}\)