Bài 1. Tìm điều kiện các BPT sau

a, \(\sqrt{20-x}>\sqrt{3x-6}+1\)

b, \(\frac{\sqrt{9-x^2}}{x-1}>\frac{1}{\sqrt{x}}+1\)

c, \(x+\frac{x+1}{\sqrt{x-4}}>2-\frac{2}{x^2-25}\)

d, \(\sqrt{x}>\sqrt{-x}\)

e, \(3x+\frac{4}{\sqrt{x-5}}\le9+\frac{x}{x-6}\)

f, \(\frac{x+2}{10+3x^2}\ge7+\frac{4}{\left(3x+9\right)^2}\)

g, \(\frac{\sqrt{x+2}}{\sqrt{x-2}}+\frac{1}{\left(x-4\right)\left(x+6\right)}\le\frac{3}{\sqrt{8-x}}\)

h, \(\frac{\sqrt{x+6}}{\left|x\right|-\sqrt{x+6}}\ge\sqrt{16-2x}\)

đặt \(P=\frac{1}{\sqrt{x^5-x^2+3xy+6}}+\frac{1}{\sqrt{y^5-y^2+3yz+6}}+\frac{1}{\sqrt{z^5-z^2+3zx+6}}\)

ta có:\(\left(x^3+2x^2+3x+3\right)\left(x-1\right)^2\ge0\)

\(\Leftrightarrow x^5-x^2\ge3x-3\)

cmtt=>\(y^5-y^2\ge3y-3;z^5-z^2\ge3z-3\)

\(\Rightarrow P\le\frac{1}{\sqrt{3x-3+3xy+6}}+\frac{1}{\sqrt{3y-3+3yz+6}}+\frac{1}{\sqrt{3z-3+3zx+6}}\)

\(=\frac{1}{\sqrt{3\left(x+xy+1\right)}}+\frac{1}{\sqrt{3\left(y+yz+1\right)}}+\frac{1}{\sqrt{3\left(z+zx+1\right)}}\)

áp dụng bunhia ta có:

\(3\left(x+xy+1\right)\ge\left(\sqrt{x}+\sqrt{xy}+1\right)^2\)

cmtt\(\Rightarrow P\le\frac{1}{\sqrt{x}+\sqrt{xy}+1}+\frac{1}{\sqrt{y}+\sqrt{yz}+1}+\frac{1}{\sqrt{z}+\sqrt{zx}+1}\)

đặt \(\sqrt{x}=a;\sqrt{y}=b;\sqrt{z}=c\)

\(\Rightarrow\frac{1}{\sqrt{x}+\sqrt{xy}+1}+\frac{1}{\sqrt{y}+\sqrt{yz}+1}+\frac{1}{\sqrt{z}+\sqrt{zx}+1}=\frac{1}{a+ab+1}+\frac{1}{b+bc+1}+\frac{1}{c+ca+1}\)

\(=\frac{abc}{a+ab+abc}+\frac{1}{b+bc+1}+\frac{b}{bc+abc+b}=\frac{bc}{bc+b+1}+\frac{b}{bc+b+1}+\frac{1}{bc+b+1}=1\)

\(\Rightarrow P\le1\)

tìm GTLN

A=\(3x^2\left(8-x^2\right)\) với \(-2\sqrt{2}\le x\le2\sqrt{2}\)

B=4x(8-5x) với \(0\le x\le\frac{8}{5}\)

C=4(x-1)(8-5x) với \(1\le x\le\frac{8}{5}\)

D=x\(\left(3-\sqrt{3}\right)\) với \(0\le x\le\sqrt{3}\)

Tìm GTNN

A=\(\frac{3x}{2}+\frac{2}{x-1}\) với x>1

B=x+\(\frac{2}{3x-1}\) với x>1/3

1. Chứng minh rằng 

\(S=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}>4\)

2. Chứng minh rằng

\(\frac{\sqrt{1}}{1}+\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{3}+...+\frac{\sqrt{200}}{200}>10+5\sqrt{2}\)

3. Cho a >= 1, b >= 1, chứng minh rằng

\(a\sqrt{b-1}+b\sqrt{a-1}\le ab\)

4. Giải phương trình

\(\sqrt{\left(x^2-2x+5\right)\left(x^2-4x\right)+7}+x^2-3x+6\)

LÀM PHIỀN M.N GIÚP MK. XIN CẢM ƠN !!!

TÌM GIÁ TRỊ LỚN NHẤT (có thể dùng BĐT côsi)

\(y=\left|x\right|\sqrt{25-x^2}Với-5\le x\le5\)

\(f\left(x\right)=\frac{x}{2}+\sqrt{1-x-2x^2}\)

\(E=\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{z-3}}{z}\)

TÍNH

\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)

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

\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+\sqrt{1+\frac{1}{4^2}+\frac{1}{5^2}}+...+\sqrt{1+\frac{1}{2012^2}+\frac{1}{2013^2}}\)

GIÚP EM ĐI Ạ, MAI EM PHẢI KIỂM TRA RỒI

Bài 1: Tìm điều kiện để các phân thức sau có nghĩa

a)\(\frac{x-1}{x+1}b)\frac{2x+1}{-3x+5}c)\frac{3x-1}{x^2-4}d)\frac{x-1}{x^2+4}e)\frac{x-1}{\left(x-2\right)\left(x+3\right)}g)\frac{x-1}{x+2}:\frac{x}{x+1}\)

Bài 2 :Tìm điều kiện để các căn thức sau có nghĩa:\(1)\sqrt{3x}|2)\sqrt{-x}|3)\sqrt{3x+2}|4)\sqrt{5-2x}|5)\sqrt{x^2}|6)\sqrt{-4x^2}|7)\sqrt{x-3}+\sqrt{2x+2}|8)\sqrt{\frac{-3}{x+2}}|9)\frac{3}{2x-4}\)

Mình rút gọn như thế này đúng không nhỉ?

\(P=\left(2-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)

\(P=\left[\frac{2\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right]:\left[\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right]\)

\(P=\left(\frac{4\sqrt{x}-6}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)

\(P=\left(\frac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)

\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}:\frac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\)

\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}.\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}{2x+3\sqrt{x}+1}\)

\(P=\left(3\sqrt{x}-5\right).\frac{\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)

\(P=\frac{3x+3\sqrt{x}-5\sqrt{x}-5}{2x+3\sqrt{x}+1}\)

\(P=\frac{3x-5\sqrt{x}-5}{2x+1}\)

TÌM GIÁ TRỊ LỚN NHẤT ( có thể dùng HĐT côsi)

\(y=\left|x\right|\sqrt{25-x^2}với-5\le x\le5\)

\(f\left(x\right)=\frac{x}{2}+\sqrt{1-x-2x^2}\)

\(E=\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{z-3}}{z}\)

TÍNH

\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+....+\sqrt{1+\frac{1}{2012^2}+\frac{1}{2013^2}}\)