Chứng minh các đẳng thức sau

a) \(\left(\frac{2\sqrt{6}-\sqrt{3}}{2\sqrt{2}-1}+\frac{5+2\sqrt{5}}{2+\sqrt{5}}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

b) \(\frac{a-b}{b^2}\sqrt{\frac{a^2b^4}{a^2-2ab+b^2}}=-a\)(Với b<a<0

c)\(\left(\sqrt{a}+\frac{1-a\sqrt{a}}{1-\sqrt{a}}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2=1\)với a\(\ge0\),a khác 1

d) \(\left(\frac{3\sqrt{5}-\sqrt{15}}{\sqrt{27}-3}+\frac{2\sqrt{5}}{\sqrt{3}}\right)40\sqrt{15}=600\)

e) \(\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)=1-x\)với x\(\ge0;x\ne1\)

Chứng minh đẳng thức:

a) \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\sqrt{xy}\left(x\ge0,y\ge0,x^2+y^2\ne0\right)\)

b) \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\left(a\ge0,a\ne1\right)\)

c) \(\sqrt{x+2\sqrt{x-2}-1}\left(\sqrt{x-2}-1\right):\left(\sqrt{x}-\sqrt{3}\right)=\sqrt{x}+\sqrt{3}\left(x\ge2,x\ne3\right)\)

a)\(3\sqrt{40\sqrt{12}}+4\sqrt{\sqrt{75}}-5\)\(\sqrt{5\sqrt{48}}\)

b)\(\sqrt{8\sqrt{3}}+3\sqrt{20\sqrt{3}}-2\sqrt{45\sqrt{3}}\)

c)\(\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)\left(x\ge0;y\ge0\right)\)

d)\(\left(\sqrt{x}+1\right)\left(x+1-\sqrt{x}\right)\left(x\ge0;y\ge0\right)\)

e)\(\left(\sqrt{x}+y\right)\left(x+y^2-y\sqrt{2}\right)\left(x\ge0;y\ge0\right)\)

Bài 2: Xét sự tương đương của các cặp BPT sau

a, \(4x-6+\frac{1}{x-2}\ge2+\frac{1}{x-2}\) và \(4x-8\ge0\)

b, \(3x-2+\frac{1}{x-3}\ge1+\frac{1}{x-3}\) và \(3x-3\ge0\)

c, \(x+4\ge0\) và \(\left(x-1\right)^2\left(x+4\right)>0\)

d,\(\left(x^2-4x+5\right)\left(x-5\right)>0\) và \(x-5>0\)

e, \(x-12\ge0\) và \(\left(x-2\right)^2\ge0\)

f, \(\sqrt{\left(x-1\right)\left(x-2\right)}\ge x\) và \(\sqrt{x-1}.\sqrt{x-2}\ge x\)

Bài 3. Giải bất phương trình

a, \(|5x – 3| &lt; 2\)

b, \(\left|3x-2\right|\ge6\)

c, \(\left|2x-1\right|\le x+2\)

d, \(\left|3x+7\right|>2x+3\)

e, \(\sqrt{x-3}\ge\sqrt{3-x}\)

f, \(\sqrt{x-1}< 3+\sqrt{x-1}\)

g, \(\frac{x-2}{\sqrt{x-4}}\ge\frac{4}{\sqrt{x-4}}\)

h, \(\left(x+5\right)\sqrt{\left(x-3\right)\left(x^2-10x+25\right)}>0\)

Rút gọn

a)\(3\sqrt{40\sqrt{12}}+4\sqrt{\sqrt{75}}-5\)\(\sqrt{5\sqrt{48}}\)

b)\(\sqrt{8\sqrt{3}}+3\sqrt{20\sqrt{3}}-2\sqrt{45\sqrt{3}}\)

c)\(\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)\left(x\ge0;y\ge0\right)\)

d)\(\left(\sqrt{x}+1\right)\left(x+1-\sqrt{x}\right)\left(x\ge0;y\ge0\right)\)

e)\(\left(\sqrt{x}+y\right).\left(x+y^2-y\sqrt{2}\right)\left(x\ge0;y\ge0\right)\)

Rút gọn biểu thức:

a) \(A=\left(\frac{3x-3\sqrt{x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+2}\right):\frac{1}{\sqrt{x}+2}\left(x\ge0,x\ne1\right)\)

b) \(B=\frac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\frac{2\left(\sqrt{x-3}\right)}{\sqrt{x}+1}+\frac{\sqrt{x}+3}{3-\sqrt{x}}\left(x>0,x\ne9\right)\)

c) \(C=\frac{2\sqrt{x}-9}{x-5+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\left(x\ge0,x\ne4,x\ne9\right)\)