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

a) \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right):\sqrt{3}=7\sqrt{16}+3\sqrt{9}-2\sqrt{4}\)

\(=7.4+3.3-2.2=28+9-4=33\)

b) \(\left(\sqrt{125}+\sqrt{245}-\sqrt{5}\right):\sqrt{5}=\sqrt{25}+\sqrt{49}-1\)

\(=5+7-1=11\)

c) \(\left(\sqrt{\dfrac{1}{7}}-\sqrt{\dfrac{16}{7}}+\sqrt{7}\right):\sqrt{7}=\left(\dfrac{\sqrt{1}}{\sqrt{7}}-\dfrac{\sqrt{16}}{\sqrt{7}}+\sqrt{7}\right):\sqrt{7}\)

\(=\left(\dfrac{1}{\sqrt{7}}-\dfrac{4}{\sqrt{7}}+\sqrt{7}\right):\sqrt{7}=\dfrac{1}{\sqrt{7}.\sqrt{7}}-\dfrac{4}{\sqrt{7}.\sqrt{7}}+1\)

\(=\dfrac{1}{7}-\dfrac{4}{7}+1=\dfrac{1}{7}-\dfrac{4}{7}+\dfrac{7}{7}\Leftrightarrow\dfrac{1-4+7}{7}=\dfrac{4}{7}\)

Bài 2:

a) \(2\sqrt{125}+\dfrac{3}{2}\sqrt{80}-\sqrt{180}-\dfrac{2}{7}\sqrt{245}\)

\(=2\sqrt{5^2\cdot5}+\dfrac{3}{2}\sqrt{4^2\cdot5}-\sqrt{6^2\cdot5}-\dfrac{2}{7}\sqrt{7^2\cdot5}\)

\(=10\sqrt{5}+\dfrac{3\cdot4}{2}\sqrt{5}-6\sqrt{5}-\dfrac{2\cdot7}{7}\sqrt{5}\)

\(=10\sqrt{5}+6\sqrt{6}-6\sqrt{5}-2\sqrt{5}\)

\(=8\sqrt{5}\)

b) \(\sqrt{11-4\sqrt{7}}-\sqrt{16+6\sqrt{7}}\)

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

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

\(=\sqrt{7}-2-\sqrt{7}-3\)

\(=-5\)

\(2a,\\ 2\sqrt{125}+\dfrac{3}{2}.\sqrt{80}-\sqrt{180}-\dfrac{2}{7}\sqrt{245}\\ =2\sqrt{5^2.5}+\dfrac{3}{2}.\sqrt{4^2.5}-\sqrt{6^2.5}-\dfrac{2}{7}.\sqrt{7^2.5}\\ =2.5.\sqrt{5}+\dfrac{3}{2}.4.\sqrt{5}-6\sqrt{5}-\dfrac{2}{7}.7\sqrt{5}\\ =10\sqrt{5}+6\sqrt{5}-6\sqrt{5}-2\sqrt{5}=8\sqrt{5}\)

3: 

a: =>|x-1|=4

=>x-1=4 hoặc x-1=-4

=>x=-3 hoặc x=5

b: =>|6x-5|=4

=>6x-5=4 hoặc 6x-5=-4

=>6x=1 hoặc 6x=9

=>x=1/6 hoặc x=3/2

1,

\(2\sqrt{5}-\sqrt{125}-\sqrt{80}\\ =2\sqrt{5}-\sqrt{25\cdot5}-\sqrt{16\cdot5}\\ =2\sqrt{5}-5\sqrt{5}-4\sqrt{5}\\ =-7\sqrt{5}\)

2,

\(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}\\ =3\sqrt{2}-\sqrt{4\cdot2}+\sqrt{25\cdot2}-4\sqrt{16\cdot2}\\ =3\sqrt{2}-2\sqrt{2}+5\sqrt{2}-16\sqrt{2}\\=-10\sqrt{2}\)

3,

\(\sqrt{18}-3\sqrt{80}-2\sqrt{50}+2\sqrt{45}\\ =\sqrt{9\cdot2}-3\sqrt{16\cdot5}-2\sqrt{25\cdot2}+2\sqrt{9\cdot5}\\ =3\sqrt{2}-12\sqrt{5}-10\sqrt{2}+6\sqrt{5}\\ =-7\sqrt{2}-6\sqrt{5}\)

4,

\(\sqrt{27}-2\sqrt{3}+2\sqrt{48}-3\sqrt{75}\\ =\sqrt{9\cdot3}-2\sqrt{3}+2\sqrt{16\cdot3}-3\sqrt{25\cdot2}\\ =3\sqrt{3}-2\sqrt{3}+8\sqrt{3}-15\sqrt{3}\\ =-6\sqrt{3}\)

5,

\(3\sqrt{2}-4\sqrt{18}+\sqrt{32}-\sqrt{50}\\ =3\sqrt{2}-4\sqrt{9\cdot2}+\sqrt{16\cdot2}-\sqrt{25\cdot2}\\ =3\sqrt{2}-12\sqrt{2}+4\sqrt{2}-5\sqrt{2}\\ =-10\sqrt{2}\)

6,

\(2\sqrt{3}-\sqrt{75}+2\sqrt{12}-\sqrt{147}\\ =2\sqrt{3}-\sqrt{25\cdot3}+2\sqrt{4\cdot3}-\sqrt{49\cdot3}\\ =2\sqrt{3}-5\sqrt{3}+4\sqrt{3}-7\sqrt{3}\\ =-6\sqrt{3}\)

7,

\(\sqrt{20}-2\sqrt{45}-3\sqrt{80}+\sqrt{125}\\ =\sqrt{4\cdot5}-2\sqrt{9\cdot5}-3\sqrt{16\cdot5}+\sqrt{25\cdot5}\\ =2\sqrt{5}-6\sqrt{5}-12\sqrt{5}+5\sqrt{5}\\ =-11\sqrt{5}\)

8,

\(6\sqrt{12}-\sqrt{20}-2\sqrt{27}+\sqrt{125}\\ =6\sqrt{4\cdot3}-\sqrt{4\cdot5}-2\sqrt{9\cdot3}+\sqrt{25\cdot5}\\ =12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}\\ =6\sqrt{3}+3\sqrt{5}\\ =3\left(2\sqrt{3}+\sqrt{5}\right)\)

9,

\(4\sqrt{24}-2\sqrt{54}+3\sqrt{6}-\sqrt{150}\\ =4\sqrt{4\cdot6}-2\sqrt{9\cdot6}+3\sqrt{6}-\sqrt{25\cdot6}\\ =8\sqrt{6}-6\sqrt{6}+3\sqrt{6}-5\sqrt{6}=0\)

10,

\(2\sqrt{18}-3\sqrt{80}-5\sqrt{147}+5\sqrt{245}-3\sqrt{98}\\ =2\sqrt{9\cdot2}-3\sqrt{16\cdot5}-5\sqrt{49\cdot3}+5\sqrt{49\cdot5}-3\sqrt{49\cdot2}\\ =6\sqrt{2}-12\sqrt{5}-35\sqrt{3}+35\sqrt{5}-21\sqrt{2}\\ =-15\sqrt{2}-35\sqrt{3}+23\sqrt{5}\)

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

a: \(=\left(2\sqrt{3}-12\sqrt{3}+15\sqrt{3}\right)\cdot\sqrt{3}=5\sqrt{3}\cdot\sqrt{3}=15\)

b: \(=\left(6\sqrt{2}-16\sqrt{2}+15\sqrt{2}\right):5=\sqrt{2}\)

c: \(=\dfrac{\left(2\sqrt{5}-6\sqrt{5}+15\sqrt{5}\right)}{\sqrt{5}}=17-6=11\)

Ối giời! Bấm máy tính đi bn! Người ta sinh ra cái máy tính là để làm mấy việc này mà. :D

\(1.\sqrt{12}-\sqrt{27}+\sqrt{3}\\ =2\sqrt{3}-3\sqrt{3}+\sqrt{3}\\ =0\)

\(2.\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}\\ =6\sqrt{7}-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}\\ =0\)

\(3\sqrt{112}-7\sqrt{216}+4\sqrt{54}-2\sqrt{252}-3\sqrt{96}\\ =12\sqrt{7}-42\sqrt{6}+12\sqrt{6}-12\sqrt{7}-12\sqrt{6}\\ =-42\sqrt{6}\)

\(4.\sqrt{18}-3\sqrt{80}-2\sqrt{50}+2\sqrt{45}\\ =3\sqrt{2}-12\sqrt{5}-10\sqrt{2}+6\sqrt{5}\\ =-7\sqrt{2}-6\sqrt{5}\)

\(5.6\sqrt{12}-\sqrt{20}-2\sqrt{27}+\sqrt{125}\\ =12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}\\ =6\sqrt{3}+3\sqrt{5}\)

\(6.2\sqrt{18}-3\sqrt{80}-5\sqrt{147}+5\sqrt{245}-3\sqrt{98}\\ =6\sqrt{2}-12\sqrt{5}-35\sqrt{3}+35\sqrt{5}-21\sqrt{2}\\ =-15\sqrt{2}-35\sqrt{3}+23\sqrt{5}\)

a: Ta có: \(A=\sqrt{8}-2\sqrt{18}+3\sqrt{50}\)

\(=2\sqrt{2}-6\sqrt{2}+15\sqrt{2}\)

\(=11\sqrt{2}\)

b: Ta có: \(B=\sqrt{125}-10\sqrt{\dfrac{1}{20}}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)

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

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

\(2\sqrt{5}-\sqrt{125}+\sqrt{80}+\sqrt{605}\)

\(=2\sqrt{5}-5\sqrt{5}+4\sqrt{5}+11\sqrt{5}\)

\(=\left(2-5+4+11\right)\sqrt{5}\)

\(=12\sqrt{5}\)