Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(H=\sqrt{a^2+\dfrac{1}{b^2}}+\sqrt{b^2+\dfrac{1}{c^2}}+\sqrt{c^2+\dfrac{1}{a^2}}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\dfrac{81}{\left(a+b+c\right)^2}}\)
\(\ge\sqrt{\left(\dfrac{3}{2}\right)^2+\dfrac{81}{\left(\dfrac{3}{2}\right)^2}}=\dfrac{3\sqrt{17}}{2}\)
1: \(=3\left(x+\dfrac{2}{3}\sqrt{x}+\dfrac{1}{3}\right)\)
\(=3\left(x+2\cdot\sqrt{x}\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{2}{9}\right)\)
\(=3\left(\sqrt{x}+\dfrac{1}{3}\right)^2+\dfrac{2}{3}>=3\cdot\dfrac{1}{9}+\dfrac{2}{3}=1\)
Dấu '=' xảy ra khi x=0
2: \(=x+3\sqrt{x}+\dfrac{9}{4}-\dfrac{21}{4}=\left(\sqrt{x}+\dfrac{3}{2}\right)^2-\dfrac{21}{4}>=-3\)
Dấu '=' xảy ra khi x=0
3: \(A=-2x-3\sqrt{x}+2< =2\)
Dấu '=' xảy ra khi x=0
5: \(=x-2\sqrt{x}+1+1=\left(\sqrt{x}-1\right)^2+1>=1\)
Dấu '=' xảy ra khi x=1
Bài 3:
Áp dụng BĐT Bunhiacopxky ta có:
\((2x+3y)^2\leq (2x^2+3y^2)(2+3)\)
\(\Leftrightarrow A^2\leq 5(2x^2+3y^2)\leq 5.5\)
\(\Leftrightarrow A^2\leq 25\Leftrightarrow A^2-25\leq 0\)
\(\Leftrightarrow (A-5)(A+5)\leq 0\Leftrightarrow -5\leq A\leq 5\)
Vậy \(A_{\min}=-5\Leftrightarrow (x,y)=(-1;-1)\)
\(A_{\max}=5\Leftrightarrow x=y=1\)
Bài 4:
Lời giải:
\(B=\sqrt{x-1}+\sqrt{5-x}\)
\(\Rightarrow B^2=(\sqrt{x-1}+\sqrt{5-x})^2=4+2\sqrt{(x-1)(5-x)}\)
Vì \(\sqrt{(x-1)(5-x)}\geq 0\Rightarrow B^2\geq 4\)
Mặt khác \(B\geq 0\)
Kết hợp cả hai điều trên suy ra \(B\geq 2\)
Vậy \(B_{\min}=2\).
Dấu bằng xảy ra khi \((x-1)(5-x)=0\Leftrightarrow x\in\left\{1;5\right\}\)
---------------------------------------
\(A=\sqrt{x^2+x+1}+\sqrt{x^2-x+1}\)
\(\Rightarrow A^2=2x^2+2+2\sqrt{(x^2+x+1)(x^2-x+1)}\)
\(\Leftrightarrow A^2=2x^2+2+2\sqrt{(x^2+1)^2-x^2}=2x^2+2+2\sqrt{x^4+1+x^2}\)
Vì \(x^2\geq 0\forall x\in\mathbb{R}\)
\(\Rightarrow A^2\geq 2+2\sqrt{1}\Leftrightarrow A^2\geq 4\)
Mà $A$ là một số không âm nên từ \(A^2\geq 4\Rightarrow A\geq 2\)
Vậy \(A_{\min}=2\Leftrightarrow x=0\)
(bài 1) a) \(\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}\) = \(\dfrac{5-2\sqrt{6}-5-2\sqrt{6}}{25-24}\)
= \(\dfrac{-4\sqrt{6}}{1}\) = \(-4\sqrt{6}\)
b) \(\sqrt{6+2\sqrt{5}}-\dfrac{\sqrt{15}-\sqrt{3}}{\sqrt{3}}\) = \(\sqrt{\left(\sqrt{5}+1\right)^2}-\dfrac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{3}}\)
= \(\left(\sqrt{5}+1\right)-\left(\sqrt{5}-1\right)\) = \(\sqrt{5}+1-\sqrt{5}+1\) = \(2\)
c) \(\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{16}}\) = \(\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{16}}\)
= \(\sqrt{6}.\sqrt{16}\) = \(4\sqrt{6}\)
d) \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
= \(\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
= \(\sqrt{3}+2+\sqrt{2}-\dfrac{1}{2-\sqrt{3}}\) = \(\dfrac{\left(\sqrt{3}+2+\sqrt{2}\right)\left(2-\sqrt{3}\right)-1}{2-\sqrt{3}}\)
= \(\dfrac{2\sqrt{3}-3+4-2\sqrt{3}+2\sqrt{2}-\sqrt{6}-1}{2-\sqrt{3}}\)
= \(\dfrac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{3}}\) = \(\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{2}}\) = \(\sqrt{2}\)
e) \(\dfrac{4}{1+\sqrt{3}}-\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}\) = \(\dfrac{4}{1+\sqrt{3}}-\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{1+\sqrt{5}}\)
= \(\dfrac{4}{1+\sqrt{3}}-\sqrt{3}\) = \(\dfrac{4-\sqrt{3}-3}{1+\sqrt{3}}\) = \(\dfrac{1-\sqrt{3}}{1+\sqrt{3}}\)
= \(\dfrac{\left(1-\sqrt{3}\right)\left(1-\sqrt{3}\right)}{1-3}\) = \(\dfrac{1-2\sqrt{3}+3}{-2}\) = \(\dfrac{4-2\sqrt{3}}{-2}\)
= \(\dfrac{-2\left(-2+\sqrt{3}\right)}{-2}\) = \(\sqrt{3}-2\)
bài 2)
a)\(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\dfrac{1}{\sqrt{a}+\sqrt{b}}=\dfrac{\left(a+b-2\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\)
= \(\dfrac{a\sqrt{a}+a\sqrt{b}+b\sqrt{a}+b\sqrt{b}-2a\sqrt{b}-2b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\)
= \(\dfrac{a\sqrt{a}+-a\sqrt{b}+b\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\) = \(\dfrac{a\left(\sqrt{a}-\sqrt{b}\right)-b\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\)
= \(\dfrac{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\) = \(a-b\)
b) \(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{\sqrt{a}\left(a-2\sqrt{a}+1\right)-\sqrt{a}\left(a+2\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{a\sqrt{a}-2a+\sqrt{a}-a\sqrt{a}-2a-\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
= \(\dfrac{2\left(a-1\right)}{4\sqrt{a}}.\dfrac{-4a}{a-1}\) = \(-2\)
Bài 1:
a: \(=\dfrac{1}{mn^2}\cdot\dfrac{n^2\cdot\left(-m\right)}{\sqrt{5}}=\dfrac{-\sqrt{5}}{5}\)
b: \(=\dfrac{m^2}{\left|2m-3\right|}=\dfrac{m^2}{3-2m}\)
c: \(=\left(\sqrt{a}+1\right):\dfrac{\left(a-1\right)^2}{\left(1-\sqrt{a}\right)}=\dfrac{-\left(a-1\right)}{\left(a-1\right)^2}=\dfrac{-1}{a-1}\)
b) \(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)
\(=\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-\sqrt{ab}+b-2b}{a-b}\)
\(=\dfrac{a}{a-b}\)
Bài 1:
a: ĐKXĐ: 2x+3>=0 và x-3>0
=>x>3
b: ĐKXĐ:(2x+3)/(x-3)>=0
=>x>3 hoặc x<-3/2
c: ĐKXĐ: x+2<0
hay x<-2
d: ĐKXĐ: -x>=0 và x+3<>0
=>x<=0 và x<>-3
Bài 3:
a: \(=\dfrac{3+2\sqrt{2}}{1}-\dfrac{\sqrt{2}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}\)
\(=3+2\sqrt{2}-\sqrt{2}=3+\sqrt{2}\)
b: \(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\left(\sqrt{ab}-b\right)}{\left(a+\sqrt{b}\right)^2}\)
\(=\dfrac{\sqrt{b}}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}}=\dfrac{b}{a+\sqrt{b}}\)
c: \(=x+\sqrt{x}-2\sqrt{x}-1+1=x-\sqrt{x}\)
1/ Rút gọn: \(a)3\sqrt{2a}-\sqrt{18a^3}+4\sqrt{\dfrac{a}{2}}-\dfrac{1}{4}\sqrt{128a}\left(a\ge0\right)=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-2\sqrt{2a}=3\sqrt{2a}\left(1-a\right)\)b)\(\dfrac{\sqrt{2}-1}{\sqrt{2}+2}-\dfrac{2}{2+\sqrt{2}}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-1-2}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3+2+1+2\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3}{1+\sqrt{2}}\)c)\(\dfrac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{\left(\sqrt{2}+\sqrt{3+\sqrt{5}}\right)\sqrt{2}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{6+2\sqrt{5}}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{6-2\sqrt{5}}}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{\left(\sqrt{5}-1\right)^2}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{2+\sqrt{5}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}+1}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{3+\sqrt{5}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{3-\sqrt{5}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\sqrt{2}\left(2-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{\sqrt{2}\left(6-2\sqrt{5}+3\sqrt{5}-5+6+2\sqrt{5}-3\sqrt{5}-5\right)}{9-5}=\dfrac{2\sqrt{2}}{4}=\dfrac{1}{\sqrt{2}}\)
Làm nốt nè :3
\(2.a.P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x}=\dfrac{x-1}{x}\left(x>0;x\ne1\right)\)\(b.P>\dfrac{1}{2}\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{x-2}{2x}>0\)
\(\Leftrightarrow x-2>0\left(do:x>0\right)\)
\(\Leftrightarrow x>2\)
\(3.a.A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a-1}=\dfrac{\sqrt{a}-1}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\sqrt{a}-1\left(a>0;a\ne1\right)\)
\(b.Để:A< 0\Leftrightarrow\sqrt{a}-1< 0\Leftrightarrow a< 1\)
Kết hợp với DKXĐ : \(0< a< 1\)
Ta có \(a\sqrt{2-b^2}+b\sqrt{2-a^2}\le\dfrac{a^2+2-b^2}{2}+\dfrac{b^2-2-a^2}{2}=2\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}a=\sqrt{2-b^2}\\b=\sqrt{2-a^2}\end{matrix}\right.\Leftrightarrow a^2+b^2=2\)
Ta có \(P=\dfrac{1}{a}+\dfrac{1}{b}-a-b\ge\dfrac{4}{a+b}-\left(a+b\right)\) (BĐT Schwarz)
= \(\dfrac{4}{a+b}+\left(a+b\right)-2\left(a+b\right)\ge2\sqrt{\dfrac{4}{a+b}.\left(a+b\right)}-2\left(a+b\right)\)
= 4 - 2a - 2b
Lại có 2a \(\le a^2+1\)
<=> -2a \(\ge-a^2-1\)
Tương tự : -2b \(\ge-b^2-1\)
Khi đó P \(\ge4-2a-2b\ge4-a^2-1-b^2-1=2-\left(a^2+b^2\right)=0\)
Dấu "=" xảy ra <=> a = b = 1
Bổ sung : a,b dương