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a: \(=2ab\cdot\dfrac{-15}{b^2a}=\dfrac{-30}{b}\)
b: \(=\dfrac{2}{3}\cdot\left(1-a\right)=\dfrac{2}{3}-\dfrac{2}{3}a\)
c: \(=\dfrac{\left|3a-1\right|}{\left|b\right|}=\dfrac{3a-1}{b}\)
d: \(=\left(a-2\right)\cdot\dfrac{a}{-\left(a-2\right)}=-a\)
a)\(\sqrt{\dfrac{9+12a+4a^2}{b^2}}=\sqrt{\dfrac{\left(2a+3\right)^2}{b^2}}=\dfrac{\left|2a+3\right|}{\left|b\right|}=\dfrac{-\left(2a+3\right)}{b}\)
b) \(\left(a-b\right).\sqrt{\dfrac{ab}{\left(a-b\right)^2}}\)
\(\Leftrightarrow\left(a-b\right).\dfrac{\left|ab\right|}{\left|a-b\right|}=-ab\)
a,\(ab^2\sqrt{\dfrac{3}{a^2b^4}}=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}=ab^2.\dfrac{\sqrt{3}}{ab^2}=\sqrt{3}\)
b,\(\sqrt{\dfrac{27\left(a-3\right)^2}{48}}=\dfrac{3\sqrt{3}\left(a-3\right)}{4\sqrt{3}}=\dfrac{3}{4}\left(a-3\right)\)
c,\(\sqrt{\dfrac{9+12a+4a^2}{b^2}}=\dfrac{\sqrt{\left(3+2a\right)^2}}{\sqrt{b^2}}=\dfrac{3+2a}{b}\)
d, \(\left(a-b\right).\sqrt{\dfrac{ab}{\left(a-b\right)^2}}=\left(a-b\right).\dfrac{\sqrt{ab}}{\sqrt{\left(a-b\right)^2}}=\left(a-b\right).\dfrac{\sqrt{ab}}{\left(a-b\right)}=\sqrt{ab}\)
b: \(=\left|b\cdot\left(b-1\right)\right|=b\cdot\left|b-1\right|\)
c: \(=\left|a\right|\cdot\left|a+1\right|=a\left(a+1\right)=a^2+a\)
d: \(=1-2a-4a=-6a+1\)
Lời giải:
a)
\(\sqrt{36(b-2)^2}=\sqrt{6^2(b-2)^2}=6\sqrt{(b-2)^2}=6|b-2|=6(2-b)\) do \(b<2\)
b)
\(\sqrt{b^2(b-1)^2}=\sqrt{b^2}\sqrt{(b-1)^2}=|b||b-1|\)
Do \(b< 0\Rightarrow b,b-1< 0\)
\(\Rightarrow \sqrt{b^2(b-1)^2}=|b||b-1|=-b(1-b)=b(b-1)\)
c) \(\sqrt{a^2(a+1)^2}=\sqrt{a^2}\sqrt{(a+1)^2}=|a||a+1|\)
\(=a(a+1)\) do \(a>0\)
d) \(\sqrt{(2a-1)^2}-4a=|2a-1|-4a\)
Vì \(a< \frac{1}{2}\Rightarrow 2a-1< 0\)
\(\Rightarrow \sqrt{(2a-1)^2}-4a=|2a-1|-4a=(1-2a)-4a=1-6a\)
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: \(=\left(\sqrt{a}+1\right):\dfrac{\left(a-1\right)^2}{\left(1-\sqrt{a}\right)}=\dfrac{-\left(a-1\right)}{\left(a-1\right)^2}=\dfrac{-1}{a-1}\)
a: \(=\sqrt{\left(2-a\right)^2\cdot\dfrac{2a}{a-2}}=\sqrt{2a\left(a-2\right)}\)
b: \(=\sqrt{\left(x-5\right)^2\cdot\dfrac{x}{\left(5-x\right)\left(5+x\right)}}\)
\(=\sqrt{\left(x-5\right)\cdot\dfrac{x}{x+5}}\)
c: \(=\sqrt{\left(a-b\right)^2\cdot\dfrac{3a}{\left(b-a\right)\left(b+a\right)}}=\sqrt{\dfrac{3a\left(b-a\right)}{b+a}}\)
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 :
\(\dfrac{a}{b}+\dfrac{b}{c}\ge2\sqrt{\dfrac{a}{c}}\\ \dfrac{b}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{b}{a}}\\ \dfrac{c}{a}+\dfrac{a}{b}\ge2\sqrt{\dfrac{c}{b}}\\ \Leftrightarrow2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\ge2\left(\sqrt{\dfrac{a}{c}}+\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}\right)\\ \Rightarrowđpcm\)
a) \(ab^2\cdot\sqrt{\dfrac{3}{a^2b^4}}=ab^2\cdot\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}=ab^2\cdot\dfrac{\sqrt{3}}{ab^2}\)
= \(\sqrt{3}\)
b) b. \(\sqrt{\dfrac{27\cdot\left(a-3\right)^2}{48}=}\dfrac{\sqrt{27}\cdot\sqrt{\left(a-3\right)^2}}{\sqrt{48}}\)
= \(\dfrac{3\cdot\sqrt{3}\cdot\left(a-3\right)}{\sqrt{3}\cdot\sqrt{16}}=\dfrac{3\cdot\left(a-3\right)}{4}\)
= 0.75*(a-3)
Lời giải:
Ta có: \(A-B=\frac{a+b}{2}-\sqrt{ab}=\frac{a+b-2\sqrt{ab}}{2}=\frac{(\sqrt{a}-\sqrt{b})^2}{2}\)
Khi đó:
\(\frac{(a-b)^2}{8(A-B)}=\frac{(a-b)^2}{4(\sqrt{a}-\sqrt{b})^2}=\frac{(\sqrt{a}+\sqrt{b})^2}{4}\)
Ta cần cm: \(B< \frac{(\sqrt{a}+\sqrt{b})^2}{4}< A\)
Thật vậy:
\(B-\frac{(\sqrt{a}+\sqrt{b})^2}{4}=\frac{4\sqrt{ab}-(\sqrt{a}+\sqrt{b})^2}{4}=\frac{-(\sqrt{a}-\sqrt{b})^2}{4}< 0, \forall a\neq b\)
\(\Rightarrow B< \frac{(\sqrt{a}+\sqrt{b})^2}{4}\)
\(A-\frac{(\sqrt{a}+\sqrt{b})^2}{4}=\frac{a+b}{2}-\frac{(\sqrt{a}+\sqrt{b})^2}{4}=\frac{a+b-2\sqrt{ab}}{4}=\frac{(\sqrt{a}-\sqrt{b})^2}{4}>0,\forall a\neq b\)
\(\Rightarrow A> \frac{(\sqrt{a}+\sqrt{b})^2}{4}\)
Ta có đpcm.