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\(1.\) Gỉa sử : \(\sqrt{25-16}< \sqrt{25}-\sqrt{16}\)
\(\Leftrightarrow3< 1\) ( Vô lý )
\(\Rightarrow\sqrt{25-16}>\sqrt{25}-\sqrt{16}\)
\(2.\sqrt{a}-\sqrt{b}< \sqrt{a-b}\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2< a-b\)
\(\Leftrightarrow a-2\sqrt{ab}+b< a-b\)
\(\Leftrightarrow2b-2\sqrt{ab}< 0\)
\(\Leftrightarrow2\left(b-\sqrt{ab}\right)< 0\)
Ta có :\(a>b\Leftrightarrow ab>b^2\Leftrightarrow\sqrt{ab}>b\)
\(\RightarrowĐpcm.\)
\(2a.\) Áp dụng BĐT Cauchy , ta có :
\(a+b\ge2\sqrt{ab}\left(a;b\ge0\right)\)
\(\Leftrightarrow\dfrac{a+b}{2}\ge\sqrt{ab}\)
\(b.\) Áp dụng BĐT Cauchy cho các số dương , ta có :
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\left(x,y>0\right)\left(1\right)\)
\(\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{yz}}\left(y,z>0\right)\left(2\right)\)
\(\dfrac{1}{x}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{xz}}\left(x,z>0\right)\left(3\right)\)
Cộng từng vế của ( 1 ; 2 ; 3 ) , ta được :
\(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge2\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\)
\(3a.\sqrt{x-4}=a\left(a\in R\right)\left(x\ge4;a\ge0\right)\)
\(\Leftrightarrow x-4=a^2\)
\(\Leftrightarrow x=a^2+4\left(TM\right)\)
\(3b.\sqrt{x+4}=x+2\left(x\ge-2\right)\)
\(\Leftrightarrow x+4=x^2+4x+4\)
\(\Leftrightarrow x^2+3x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-3\left(KTM\right)\end{matrix}\right.\)
KL....
làm rõ \(\sum_{cyc}\frac{a}{a+b}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{a+b}-\frac{1}{2}\right)=\sum_{cyc}\frac{a-b}{2(a+b)}\)
\(=\sum_{cyc}\frac{(a-b)(c^2+ab+ac+bc)}{2\prod\limits_{cyc}(a+b)}=\sum_{cyc}\frac{c^2a-c^2b}{2\prod\limits_{cyc}(a+b)}\)
\(=\sum_{cyc}\frac{a^2b-a^2c}{2\prod\limits_{cyc}(a+b)}=\frac{(a-b)(a-c)(b-c)}{2\prod\limits_{cyc}(a+b)}\geq0\) (đúng)
ok thỏa thuận rồi tui làm nửa sau thui nhé :D
Đặt \(a^2=x;b^2=y;c^2=z\) thì ta có:
\(VT=\sqrt{\dfrac{x}{x+y}}+\sqrt{\dfrac{y}{y+z}}+\sqrt{\dfrac{z}{x+z}}\)
Lại có: \(\sqrt{\dfrac{x}{x+y}}=\sqrt{\dfrac{x}{\left(x+y\right)\left(x+z\right)}\cdot\sqrt{x+z}}\)
Tương tự cộng theo vế rồi áp dụng BĐT C-S ta có:
\(VT^2\le2\left(x+y+z\right)\left[\dfrac{x}{\left(x+y\right)\left(x+z\right)}+\dfrac{y}{\left(y+z\right)\left(y+x\right)}+\dfrac{z}{\left(z+x\right)\left(z+y\right)}\right]\)
\(\Leftrightarrow VT^2\le\dfrac{4\left(x+y+z\right)\left(xy+yz+xz\right)}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)
Vì \(VP^2=\dfrac{9}{2}\) nên cần cm \(VT\le \frac{9}{2}\)
\(\Leftrightarrow9\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge8\left(x+y+z\right)\left(xy+yz+xz\right)\)
Can you continue
b) \(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)
\(=\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-\sqrt{ab}+b-2b}{a-b}\)
\(=\dfrac{a}{a-b}\)
a: \(\dfrac{5}{3\sqrt{8}}=\dfrac{5\sqrt{2}}{3\cdot4}=\dfrac{5\sqrt{2}}{12}\)
\(\dfrac{2}{\sqrt{b}}=\dfrac{2\sqrt{b}}{b}\)
b: \(\dfrac{5}{5-2\sqrt{3}}=\dfrac{25+10\sqrt{3}}{13}\)
\(\dfrac{2a}{1-\sqrt{a}}=\dfrac{2a\left(1+\sqrt{a}\right)}{1-a}\)
c: \(\dfrac{4}{\sqrt{7}+\sqrt{5}}=\dfrac{4\left(\sqrt{7}-\sqrt{5}\right)}{2}=2\sqrt{7}-2\sqrt{5}\)
\(\dfrac{6a}{2\sqrt{a}-\sqrt{b}}=\dfrac{6a\left(2\sqrt{a}+\sqrt{b}\right)}{4a-b}\)
a: \(=\dfrac{\sqrt{a}-1}{\sqrt{a}\left(a-\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{1}\)
\(=a-1\)
b: \(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\dfrac{\sqrt{ab}+b+\sqrt{ab}-b}{\sqrt{a}\left(a-b\right)}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{1}{\sqrt{a}}\)
c: \(=\dfrac{a\sqrt{b}+b}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\cdot\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\sqrt{\dfrac{b\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(a+\sqrt{b}\right)^2}}\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}}=b\)
đặt \(\sqrt{a}=x;\sqrt{b}=y\)
=> bđt : x^3+y^3>=xy(x+y)
ta có: x^3 + y^3 =(x+y) ( x^2-xy+y^2)>=(x+y)(2xy-xy)=xy(x+y)
\(\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}=\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{ab}}=\left(\sqrt{a}+\sqrt{b}\right)\left(\dfrac{a+b-\sqrt{ab}}{\sqrt{ab}}\right)=\left(\sqrt{a}+\sqrt{b}\right)\left(\dfrac{a+b}{\sqrt{ab}}-1\right)\) Áp dụng BĐT Cauchy cho các số dương , ta có :
\(a+b\) ≥ \(2\sqrt{ab}\) ⇔ \(\dfrac{a+b}{\sqrt{ab}}\text{≥}2\) ⇔ \(\dfrac{a+b}{\sqrt{ab}}-1\text{≥}1\)
⇒\(\left(\sqrt{a}+\sqrt{b}\right)\left(\dfrac{a+b}{\sqrt{ab}-1}\right)\text{≥}\sqrt{a}+\sqrt{b}\)
⇔ \(\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}\text{≥}\sqrt{a}+\sqrt{b}\)
P/s : Đề như này mới đúng nhé :3
a: \(=4\left|a-3\right|=4\left(a-3\right)=4a-12\)
b: \(=9\cdot\left|a-9\right|=9\left(9-a\right)=81-9a\)
c: \(a^3b^6\cdot\sqrt{\dfrac{3}{a^6b^4}}=a^3b^6\cdot\dfrac{\sqrt{3}}{-a^3b^2}=-b^4\sqrt{3}\)
d: \(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{a-b}\)
\(=\dfrac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\)
2, a, \(a+\dfrac{1}{a}\ge2\)
\(\Leftrightarrow\dfrac{a^2+1}{a}\ge2\)
\(\Rightarrow a^2-2a+1\ge0\left(a>0\right)\)
\(\Leftrightarrow\left(a-1\right)^2\ge0\)( là đt đúng vs mọi a)
vậy...................
Câu 1:
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}=3\)
\(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)
Ta có VT =\(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)
=\(\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}-\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\) =\(\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
=\(\dfrac{a-b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
=\(\dfrac{a-b}{a-b}=1=VP\)
a) CM:\(\sqrt{\left(n+1\right)^2}+\sqrt{n^2}=\left(n+1\right)^2-n^2\)
\(\Leftrightarrow n+1+n=\left(n+1-n\right)\left(n+1+n\right)\)
\(\Leftrightarrow2n+1=1\left(2n+1\right)\)
\(\Leftrightarrow2n+1=2n+1\)
\(\Rightarrow\sqrt{\left(n+1\right)^2}+\sqrt{n^2}=\left(n+1\right)^2-n^2\)
Câu b) ý 2:
Áp dụng BĐT cô si ta có :
\(\dfrac{a}{b}+\dfrac{b}{c}\ge2\sqrt{\dfrac{a}{c}}\\ \dfrac{b}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{b}{a}}\\ \dfrac{c}{a}+\dfrac{a}{b}\ge2\sqrt{\dfrac{c}{b}}\\ \Leftrightarrow2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\ge2\left(\sqrt{\dfrac{a}{c}}+\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}\right)\\ \Rightarrowđpcm\)