mk chỉ lm đk với đề như này th à

\(\sqrt{28-16\sqrt{3}}-\sqrt{28+16\sqrt{3}}\)

Đặt A = \(\sqrt{28-16\sqrt{3}}-\sqrt{28+16\sqrt{3}}\)

nhận xét : A < 0, bình phương hai vế ta được :

\(A^2=\left(\sqrt{28-16\sqrt{3}}-\sqrt{28+16\sqrt{3}}\right)^2\)

\(\Rightarrow A^2=\left(\sqrt{28-16\sqrt{3}}\right)^2+\left(\sqrt{28+16\sqrt{3}}\right)^2-2\sqrt{\left(28-16\sqrt{3}\right)\left(28+16\sqrt{3}\right)}\)

=> \(A^2=28-16\sqrt{3}+28+16\sqrt{3}-2\sqrt{28^2-\left(16\sqrt{3}\right)^2}\)

=>\(A^2=56-2\sqrt{784-768}\)

=> \(A^2=56-2\sqrt{16}=56-2.4\)

=> \(A^2=48\)

=> \(A=\pm\sqrt{48}\) mà A < 0 nên

\(A=-\sqrt{48}\)

Bài 2:

\(B=\sqrt{28-16\sqrt{3}}+\sqrt{13-4\sqrt{3}}\)

\(=\sqrt{\left(4-2\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}-1\right)^2}\)

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

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

\(=3\)

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

\(=\sqrt{2}.\sqrt{4+\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\)

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

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

\(=5-3=2\)

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

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

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

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

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

1) \(\Leftrightarrow x^2-7x+8+\sqrt{x^2-7x+8}-20=0\)

Đặt \(t=\sqrt{x^2-7x+8}\ge0\)

Phương trình tương đương

\(t^2+t-20=0\)

\(\left[{}\begin{matrix}t=4\left(TM\right)\\t=-5\left(KTM\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2-7x+8}=4\)

Bạn đọc tự giải quyết tiếp bài toán.

2.

A=\(\sqrt{\sqrt{\left(\sqrt{16}-\sqrt{12}\right)^2}}-\sqrt{\sqrt{\left(\sqrt{16}+\sqrt{12}\right)^2}}\)

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

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

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

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

\(=-2\)

B= \(\sqrt{5-2\sqrt{2+\sqrt{\left(\sqrt{8}+\sqrt{1}\right)^2}}}\)

\(=\sqrt{5-2\sqrt{2+\sqrt{8}+1}}\)

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

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

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

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

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

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

a) \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)

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

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

b) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{4}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

= \(\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(\sqrt{2}+1\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}+1\)

a: \(4\sqrt{7}=\sqrt{4^2\cdot7}=\sqrt{112}\)

\(3\sqrt{13}=\sqrt{3^2\cdot13}=\sqrt{117}\)

mà 112<117

nên \(4\sqrt{7}< 3\sqrt{13}\)

b: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)

\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)

mà 108>64

nên \(3\sqrt{12}>2\sqrt{16}\)

c: \(\dfrac{1}{4}\sqrt{84}=\sqrt{\dfrac{1}{16}\cdot84}=\sqrt{\dfrac{21}{4}}\)

\(6\sqrt{\dfrac{1}{7}}=\sqrt{36\cdot\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)

mà \(\dfrac{21}{4}>\dfrac{36}{7}\)

nên \(\dfrac{1}{4}\sqrt{84}>6\sqrt{\dfrac{1}{7}}\)

d: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)

\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)

mà 108>64

nên \(3\sqrt{12}>2\sqrt{16}\)

a) Ta có: \(4\sqrt{28}+3\sqrt{63}-3\sqrt{112}-2\sqrt{175}\)

\(=8\sqrt{7}+9\sqrt{7}-12\sqrt{7}-10\sqrt{7}\)

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

b) Ta có: \(\sqrt{5}\left(\sqrt{5}-3\sqrt{20}+2\sqrt{80}\right)\)

\(=\sqrt{5}\left(\sqrt{5}-6\sqrt{5}+8\sqrt{5}\right)\)

\(=\sqrt{5}\cdot3\sqrt{5}=15\)

c) Ta có: \(\left(\sqrt{\dfrac{16}{3}}-\sqrt{\dfrac{25}{3}}\right)\cdot\sqrt{3}\)

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

=-1

e) Ta có: \(\left(\sqrt{\dfrac{32}{3}}-\sqrt{54}+\sqrt{\dfrac{50}{3}}\right)\cdot\sqrt{6}\)

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

\(=\dfrac{9\sqrt{12}}{\sqrt{3}}-18\)

\(=0\)

f) Ta có:  \(\left(\sqrt{6}-2\right)\left(\sqrt{3}+\sqrt{2}\right)\)

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

\(=\sqrt{2}\)