xác định hàm số 

a, \(y=\sqrt{x^2+x-4}\)

b , \(y=\frac{1}{x^2+1}\)

c, y= l 2x - 3 l 

d , \(y=\frac{1}{x^2-3x}\)

e , \(y=\sqrt{1-x}+\frac{1}{x\sqrt{1}+x}\)

f , \(y=\frac{2x-1}{\sqrt{x\sqrt{\left(x-4\right)}}}\)

g , \(y=\sqrt{3+x}+\frac{1}{x^2-1}\)

h , \(y=\frac{1}{\sqrt{2x^2-4x+4}}\)

i, \(y=\sqrt{6-x}+2x\sqrt{2x+1}\)

j, \(y=\frac{x^2+1}{\sqrt{2-5}}+x\sqrt{1+x}\)

k, \(y=\frac{1}{x^2+3x+3}+\left(x+2\right)\sqrt{x+3}\)

l, \(y=\sqrt[3]{\frac{3x+5}{x^2-1}}\)

Tìm tập xác định của các hàm số sau:

a) y = \(\frac{\sqrt{x+1}}{x^2-x-6}\)

b) y = \(\sqrt{6-3x}\) - \(\sqrt{x-1}\)

c) y = \(\frac{\sqrt{2-x}+\sqrt{x+2}}{x}\)

d) y = \(\frac{\sqrt{3x-2}+6x}{\sqrt{4-3x}}\)

e) y = \(\sqrt{6-x}\) + \(\frac{2x+1}{1+\sqrt{x-1}}\)

f) y = \(\frac{2x+9}{\left(x+4\right)\sqrt{x+3}}\)

g) y = \(\frac{\sqrt{x^2-2x+3}}{x-3\sqrt{x}+2}\)

h) f(x) = \(\frac{1}{\sqrt{1-\sqrt{1+4x}}}\)

i) y = \(\frac{2x^2}{\sqrt{x^2-3x+2}}\)

tập xác định hàm số

a, y=\(\sqrt{x^2+x-4}\)

b , y = \(\frac{1}{x^2+1}\)

c , y=\(\frac{\left|2x-3\right|}{x^2+x+6}\)

d , y =\(\frac{1}{x^2-3x}\)

e , y =\(\sqrt{1-x}\) +\(\frac{1}{x\sqrt{1}+x}\)

f , \(\frac{2x-1}{\sqrt{x\sqrt{\left(x-4\right)}}}\)

g, y = \(\frac{x^2+1}{\sqrt{2-5}}\) + \(\frac{1}{x^2-1}\)

h , y= \(\frac{1}{\sqrt{2x^2-4x+4}}\)

i, \(\sqrt{6-x}\)+2x\(\sqrt{2x+1}\)

j, y = \(\sqrt{3+x}\) +\(\frac{1}{x^2-1}\)

k, y = \(\frac{1}{x^2+3x+3}\)+(x+2)\(\sqrt{x+3}\)

l, y =\(\sqrt[3]{\frac{3x+5}{x^2-1}}\)

Tìm tập xác định của các hàm số sau:

a) y=\(\sqrt{2x-3}\) b) y= \(\sqrt{\left|2x-3\right|}\) c) y= \(\sqrt{4-x}+\sqrt{x+1}\) d) y=\(\sqrt{x-1}+\frac{1}{x-3}\) e) y=\(\frac{1}{\left(x+2\right)\sqrt{x-1}}\)

f) y=\(\sqrt{x+3-2\sqrt{x+2}}\) g) y=\(\frac{\sqrt{5-2x}}{\left(x-2\right)\sqrt{x-1}}\) h) y=\(\sqrt{2x-1}+\sqrt{\frac{1}{3-x}}\) i) y= \(\sqrt{x+3}+\frac{1}{x^2-4}\)

 \(x+y=4\Rightarrow\frac{x+y}{2}=2\Rightarrow\sqrt{\frac{x+y}{2}}=\sqrt{2}\)

\(P.\sqrt{\frac{x+y}{2}}=\sqrt{2}\sqrt{x^2+\frac{1}{x^2}}+\sqrt{2}\sqrt{x^2+\frac{1}{x^2}}\)

\(\Leftrightarrow\sqrt{2}P=\sqrt{1+1}\sqrt{x^2+\frac{1}{x^2}}+\sqrt{1+1}\sqrt{x^2+\frac{1}{x^2}}\)

\(\Leftrightarrow\sqrt{2}P\ge x+\frac{1}{x}+y+\frac{1}{y}\)

\(x+\frac{1}{x}=\left(\frac{1}{x}+4x\right)-3x\ge4-3x\)

\(y+\frac{1}{y}=\left(\frac{1}{y}+4y\right)-3y\ge4-3y\)

\(\Rightarrow\sqrt{2}P\ge8-3\left(x+y\right)=8-3.4=-4\)

đến đay sau răng

Rút gọn
a.\(\left(2\sqrt{x}+\sqrt{2x}\right)\left(\sqrt{x}-\sqrt{2x}\right)\)
b. \(\left(\sqrt{3x}+\sqrt{2x}\right)\left(3\sqrt{x}-\sqrt{6x}\right)\)
c.\(\left(\frac{4}{3}\sqrt{3}+\sqrt{2}\sqrt{3\frac{1}{3}}\right)\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{3}}\right)-2\)
d.\(\left(2\sqrt{x}+\sqrt{y}\right)\left(3\sqrt{x}-2\sqrt{y}\right)\)(x,y lớn hơn hoặc bằng 0)
e.\(\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{x}\sqrt{y}+\sqrt{y}\right)\) (x,y lớn hơn hoặc bằng 0)