\(1,=\left|1-\sqrt{2}\right|+\left|\sqrt{2}+3\right|\\ =1-\sqrt{2}+3+\sqrt{2}\\ =4\\ 2,=\left|\sqrt{3}-2\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}-2+\sqrt{3}-1\\ =2\sqrt{3}-3\\ 3,=\left|\sqrt{5}-3\right|+\left|\sqrt{5}-2\right|\\ =\sqrt{5}-3+\sqrt{5}-2\\ =2\sqrt{5}-5\\ 4,=\left|3+\sqrt{2}\right|+\left|3-\sqrt{2}\right|\\ =3+\sqrt{2}+\sqrt{3}-\sqrt{2}\\ =3+\sqrt{3}\\ 5,=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\\ =2-\sqrt{3}-\left(2+\sqrt{3}\right)\\ =2-\sqrt{3}-2-\sqrt{3}\\ =-2\sqrt{3}\)

Bài 1 :

a, ĐKXĐ : \(\dfrac{2x+1}{x^2+1}\ge0\)

Mà \(x^2+1\ge1>0\)

\(\Rightarrow2x+1\ge0\)

\(\Rightarrow x\ge-\dfrac{1}{2}\)

Vậy ...

b, Ta có : \(\sqrt[3]{-27}+\sqrt[3]{64}-\sqrt[3]{-\dfrac{128}{2}}\)

\(=-3+4-\left(-4\right)=-3+4+4=5\)

Bài 2 :

\(a,=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)

\(=\sqrt{5}\left(2+6+5-12\right)=\sqrt{2}\)

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

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

\(c,=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)

\(=\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\)

\(=3\)

\(\sqrt{2-2.\frac{1}{2}\sqrt{2}+\frac{1}{4}}.\sqrt{8-2.2\sqrt{2}.\frac{1}{4}+\frac{1}{16}}=\sqrt{\left(\sqrt{2}-\frac{1}{2}\right)^2}\sqrt{\left(2\sqrt{2}-\frac{1}{4}\right)^2}\)

\(=\left(\sqrt{2}-\frac{1}{2}\right)\left(2\sqrt{2}-\frac{1}{4}\right)=\frac{33-10\sqrt{2}}{8}\)

\(\sqrt{2+2\sqrt{2}+1}.4\sqrt{\frac{288+2\sqrt{288}+1}{16}}=\sqrt{\left(\sqrt{2}+1\right)^2}.4\sqrt{\frac{\left(12\sqrt{2}+1\right)^2}{4^2}}\)

\(=\left(\sqrt{2}+1\right)\left(12\sqrt{2}+1\right)=25+13\sqrt{2}\)

\(\sqrt{28-10\sqrt{3}}=\sqrt{25-2.5\sqrt{3}+3}=\sqrt{\left(5-\sqrt{3}\right)^2}=5-\sqrt{3}\)

a)\(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\sqrt{10\left(4-\sqrt{15}\right)}+\sqrt{6\left(4-\sqrt{15}\right)}\)

\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)

\(=\sqrt{\left(5-\sqrt{15}\right)^2}+\sqrt{\left(3-\sqrt{15}\right)^2}\)

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

\(=2\)

b) \(2\left(\sqrt{10}-\sqrt{2}\right)\left(4+\sqrt{6-2\sqrt{5}}\right)\)

\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(4+\sqrt{\left(1-\sqrt{5}\right)^2}\right)\)

\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(4+\sqrt{5}-1\right)\)

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

\(=6\sqrt{10}+2\sqrt{50}-6\sqrt{2}-2\sqrt{10}\)

\(=6\sqrt{10}+10\sqrt{2}-6\sqrt{2}-2\sqrt{10}\)

\(=4\sqrt{10}+4\sqrt{2}\)

c) \(\left(\sqrt{7}+\sqrt{14}\right)\sqrt{9-2\sqrt{14}}\)

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

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

\(=7\sqrt{7}-7\sqrt{2}+\sqrt{98}-\sqrt{28}\)

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

\(=5\sqrt{7}\)

d) \(\sqrt{\dfrac{289+4\sqrt{72}}{16}}\)

\(=\sqrt{\dfrac{289+42\sqrt{2}}{16}}\)

\(=\dfrac{\sqrt{289+42\sqrt{2}}}{\sqrt{4^2}}\)

\(=\dfrac{\sqrt{\left(1+12\sqrt{2}\right)^2}}{4}\)

\(=\dfrac{1+12\sqrt{2}}{4}\)

e) \(\left(\sqrt{21}+7\right)\sqrt{10-2\sqrt{21}}\)

\(=\left(\sqrt{21}+\sqrt{7}\right)\sqrt{\left(\sqrt{3}-\sqrt{7}\right)^2}\)

\(=\left(\sqrt{21}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{3}\right)\)

\(=\sqrt{147}-\sqrt{63}+7-\sqrt{21}\)

\(=7\sqrt{3}-\sqrt{63}+7-\sqrt{21}\)

f) bạn xem đề lại nhé

Bài 1:

a) Ta có: \(\sqrt{\left(23-15\sqrt{3}\right)^2}\)

\(=\left|23-15\sqrt{3}\right|\)

\(=\left|\sqrt{529}-\sqrt{675}\right|\)

\(=\sqrt{675}-\sqrt{529}\)

\(=15\sqrt{3}-23\)

b) Ta có: \(\sqrt{\left(2-2\sqrt{3}\right)^2}\)

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

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

c) Ta có: \(\sqrt{\left(15-4\sqrt{3}\right)^2}\)

\(=\left|15-4\sqrt{3}\right|\)

\(=15-4\sqrt{3}\)

d) Ta có: \(\sqrt{\left(16-6\sqrt{7}\right)^2}\)

\(=\left|16-6\sqrt{7}\right|\)

\(=\left|\sqrt{256}-\sqrt{252}\right|\)

\(=16-6\sqrt{7}\)

f) Ta có: \(\sqrt{\left(22-8\sqrt{3}\right)^2}\)

\(=\left|22-8\sqrt{3}\right|\)

\(=\left|\sqrt{484}-\sqrt{192}\right|\)

\(=22-8\sqrt{3}\)

g) Ta có: \(\sqrt{\left(9-4\sqrt{2}\right)^2}\)

\(=\left|9-4\sqrt{2}\right|\)

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

h) Ta có: \(\sqrt{\left(13-4\sqrt{3}\right)^2}\)

\(=\left|13-4\sqrt{3}\right|\)

\(=13-4\sqrt{3}\)

i) Ta có: \(\sqrt{\left(7-3\sqrt{3}\right)^2}\)

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

\(=7-3\sqrt{3}\)

Lời giải:

$\sqrt{7+2\sqrt{10}}=\sqrt{2+5+2\sqrt{2.5}}=\sqrt{(\sqrt{2}+\sqrt{5})^2}=\sqrt{2}+\sqrt{5}$

\(\sqrt[3]{3\sqrt[3]{3}-3\sqrt[3]{2}-1}=\sqrt[3]{(1-\sqrt[3]{2})^3}=1-\sqrt[3]{2}\)

Do đó:

\(\text{TS}=\sqrt[3]{2}+\sqrt{2}+\sqrt{5}+1-\sqrt[3]{2}=\sqrt{2}+\sqrt{5}+1=\text{MS}\)

\(A=\frac{\text{TS}}{\text{MS}}=1\)