a: \(=\sqrt{5}-3\sqrt{5}-4\sqrt{3}+15\sqrt{3}=-2\sqrt{5}+11\sqrt{3}\)

b: \(=3\sqrt{10}-\sqrt{5}+6-\sqrt{2}\)

c; \(=15\sqrt{2}-10\sqrt{3}-12\sqrt{2}-\sqrt{3}=-11\sqrt{3}+3\sqrt{2}\)

d: \(=3-\sqrt{3}+\sqrt{3}-1=2\)

f: \(=\sqrt{10}-\sqrt{10}-2-2\sqrt{10}=-2-2\sqrt{10}\)

a: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)

b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)

c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)

a) \(\Leftrightarrow A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)

b) \(\Leftrightarrow B=\sqrt{7-2\sqrt{12}}+\sqrt{12+2\sqrt{27}}=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}=2-\sqrt{3}+3+\sqrt{3}=5\)

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

d) \(\Leftrightarrow D=3-\left(-2\right)-5=0\)

câu g

(câu cuối) đề nhiều trôi hết nhìn thấy mỗi câu (g)

\(G=0,1\sqrt{200}+2\sqrt{0,08}+0,4\sqrt{50}\)

\(G=0,1.10\sqrt{2}+\dfrac{2.2}{10}\sqrt{2}+0,4.5\sqrt{2}\)

\(G=\sqrt{2}\left(1+\dfrac{2}{5}+2\right)=\dfrac{\sqrt{2}\left(5+2+10\right)}{5}=\dfrac{17\sqrt{2}}{5}\)

a) \(2\sqrt{75}-4\sqrt{12}-3\sqrt{50}-\sqrt{72}\)

= \(2.5\sqrt{3}-4.2\sqrt{3}-3.5\sqrt{2}-6\sqrt{2}\)

= \(10\sqrt{3}-8\sqrt{3}-15\sqrt{2}-6\sqrt{2}\)

= \(\sqrt{3}\left(10-8\right)-\sqrt{2}\left(15+6\right)\)

= \(2\sqrt{3}-21\sqrt{2}\)

Bài 1:

a) Ta có: \(\left(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}-\dfrac{5}{4}\sqrt{\dfrac{4}{5}}+\sqrt{5}\right)\)

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

\(=3\sqrt{5}-\dfrac{1}{2}\sqrt{5}\)

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

c) Ta có: \(\dfrac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)

\(=\dfrac{\sqrt{35}\left(\sqrt{5}-\sqrt{7}+2\sqrt{2}\right)}{\sqrt{35}}\)

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

Bài 2:

e) ĐKXĐ: \(\dfrac{4}{3}\le x\le6\)

Ta có: \(\sqrt{6-x}=3x-4\)

\(\Leftrightarrow6-x=\left(3x-4\right)^2\)

\(\Leftrightarrow9x^2-24x+16+6-x=0\)

\(\Leftrightarrow9x^2-25x+22=0\)

\(\Delta=\left(-25\right)^2-4\cdot9\cdot22=625-792< 0\)

Vậy: Phương trình vô nghiệm

Đề không khó, mỗi tội dài

Bài 2:

a: \(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=14\sqrt{2x}\)

b: \(=6-4\sqrt{3}+4\sqrt{3}-8=-2\)

c: \(=\sqrt{2}+1+2-\sqrt{2}=3\)

d: \(=\dfrac{1}{\sqrt{2}}\left(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\right)+\sqrt{6}\)

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

=0

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

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

=5-9

=-4

f: \(=\dfrac{1}{5}\cdot5\sqrt{2}-2\cdot4\sqrt{6}-\sqrt{2}+2\sqrt{6}\)

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

a: ĐKXĐ: x-5>=0

=>x>=5

\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)

=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

=>\(2\sqrt{x-5}=4\)

=>x-5=4

=>x=9(nhận)

b: ĐKXĐ: x-1>=0

=>x>=1

\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}=4\)

=>\(\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=4\)

=>\(-2\sqrt{x-1}=4\)

=>\(\sqrt{x-1}=-2\)(vô lý)

Vậy: Phương trình vô nghiệm

c: ĐKXĐ: x-2>=0

=>x>=2

\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot\sqrt{9x-18}+6\cdot\sqrt{\dfrac{x-2}{81}}=-4\)

=>\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)

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

=>\(-\sqrt{x-2}=-4\)

=>x-2=16

=>x=18(nhận)

d: ĐKXĐ: x+3>=0

=>x>=-3

\(\sqrt{9x+27}+4\sqrt{x+3}-\dfrac{3}{4}\cdot\sqrt{16x+48}=0\)

=>\(3\sqrt{x+3}+4\sqrt{x+3}-\dfrac{3}{4}\cdot4\sqrt{x+3}=0\)

=>\(4\sqrt{x+3}=0\)

=>x+3=0

=>x=-3(nhận)

a) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)

= \(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\)

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

= \(2\sqrt{x-5}=4\)

= \(\sqrt{x-5}=2\)

= \(\left|x-5\right|=4\)

=> \(x-5=\pm4\)

\(x=\pm4+5\)

\(x=9;x=1\)

Vậy x=9; x=1

b) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}=4\)

\(\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=4\)

\(-2\sqrt{x-1}=4\)

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

=>\(\left|x-1\right|=-2\)

\(x-1=\mp2\)

\(x=-3;x=1\)

Vậy x=-3; x=1

c: Ta có: \(C=\left(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}-\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\right):\sqrt{48}\)

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

\(=\dfrac{1}{2}\)