\(A=\frac{1}{ab}\cdot\frac{\sqrt{a^2b^4}}{4}\)(a<0;b khác 0)
\(B=\frac{\left(a-3\right)^2}{2}\cdot\sqrt{\frac{25}{a^2-6a+9}}\)(a<3)
K
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Những câu hỏi liên quan
13 tháng 7 2016
a) \(\left(\sqrt{ab}+2\sqrt{\frac{b}{a}}-\sqrt{\frac{a}{b}}+\frac{1}{\sqrt{ab}}\right).\sqrt{ab}\) (ĐK : \(\hept{\begin{cases}a>0\\b>0\end{cases}}\)hoặc \(\hept{\begin{cases}a< 0\\b< 0\end{cases}}\))
\(=ab+2b-a+1\)
b) \(\left(-\frac{am}{b}\sqrt{\frac{n}{m}}-\frac{ab}{n}.\sqrt{mn}+\frac{a^2}{b^2}.\sqrt{\frac{m}{n}}\right)\left(a^2b^2.\sqrt{\frac{n}{m}}\right)\) (ĐK bạn tự xét nhé ^^)
\(=\left(-\frac{a\sqrt{mn}}{b}-\frac{ab\sqrt{m}}{\sqrt{n}}+\frac{a^2}{b^2}.\sqrt{\frac{m}{n}}\right)\left(a^2b^2.\sqrt{\frac{n}{m}}\right)\)
\(=a^2b^2\left(\frac{-an}{b}-ab+\frac{a^2}{b^2}\right)=-a^3bn-a^3b^3+a^4=a^3\left(a-bn-b^3\right)\)
T
31 tháng 5 2020
Bài 1. Ta có: \(a\left(a+2\right)\left(a-1\right)^2\ge0\therefore\frac{1}{4a^2-2a+1}\ge\frac{1}{a^4+a^2+1}\)
Thiết lập tương tự 2 BĐT còn lại và cộng theo vế rồi dùng Vasc (https://olm.vn/hoi-dap/detail/255345443802.html)
Bài 5: Bất đẳng thức này đúng với mọi a, b, c là các số thực. Chứng minh:
Quy đồng và chú ý các mẫu thức đều không âm, ta cần chứng minh:
\(\frac{1}{2}\left(a^2+b^2+c^2-ab-bc-ca\right)\Sigma\left[\left(a^2+b^2\right)+2c^2\right]\left(a-b\right)^2\ge0\)
Đây là điều hiển nhiên.
TH
1
BA
10 tháng 9 2020
\(\frac{a-b}{4b^2}\cdot\sqrt{\frac{4a^2b^4}{a^2-2ab+b^2}}\)
\(=\frac{a-b}{4b^2}\cdot\sqrt{\frac{\left(2ab^2\right)^2}{\left(a-b\right)^2}}\)
\(=\frac{a-b}{4b^2}\cdot\frac{2ab}{a-b}\)
\(=\frac{a}{2b}\)
13 tháng 7 2016
a) \(\left(3+1\sqrt{6}-\sqrt{33}\right)\left(\sqrt{22}+\sqrt{6}+4\right)\)
\(=\sqrt{3}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right).\sqrt{2}\left(\sqrt{11}+\sqrt{3}+2\sqrt{2}\right)\)
\(=\sqrt{6}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right)\left(\sqrt{3}+2\sqrt{2}+\sqrt{11}\right)\)
\(=\sqrt{6}\left[\left(\sqrt{3}+2\sqrt{2}\right)^2-11\right]=\sqrt{6}\left(11+4\sqrt{6}-11\right)=\sqrt{6}.4\sqrt{6}=6.4=24\)
b) \(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)=\left(\frac{5+2\sqrt{6}+10-4\sqrt{6}}{5^2-\left(2\sqrt{6}\right)^2}\right)\left(15+2\sqrt{6}\right)\)
\(=\left(15-2\sqrt{6}\right)\left(15+2\sqrt{6}\right)=15^2-24=201\)
C) \(\left(\frac{4}{3}.\sqrt{3}+\sqrt{2}+\sqrt{3\frac{1}{3}}\right)\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{5}}\right)\)
\(=\left(\frac{4}{\sqrt{3}}+\frac{\sqrt{6}}{\sqrt{3}}+\frac{\sqrt{10}}{\sqrt{3}}\right)\left(\frac{\sqrt{6}}{\sqrt{5}}+\frac{\sqrt{10}}{\sqrt{5}}-\frac{4}{\sqrt{5}}\right)\)
\(=\frac{1}{\sqrt{15}}\left(\sqrt{6}+\sqrt{10}+4\right)\left(\sqrt{6}+\sqrt{10}-4\right)=\frac{1}{\sqrt{15}}\left[\left(\sqrt{6}+\sqrt{10}\right)^2-16\right]\)
\(=\frac{1}{\sqrt{15}}\left(16+4\sqrt{15}-16\right)=\frac{4\sqrt{15}}{\sqrt{15}}=4\)
d) \(\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1990+2\sqrt{1989}}=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1989+2\sqrt{1989}+1}\)
\(=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{\left(\sqrt{1989}+1\right)^2}=\left(\sqrt{1989}-1\right)\left(\sqrt{1989}+1\right)=1989-1=1988\)
e) \(\frac{a-\sqrt{ab}+b}{a\sqrt{a}+b\sqrt{b}}-\frac{1}{a-b}=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}-1}{a-b}\)
A
19 tháng 6 2019
1/ \(\sqrt{\frac{m}{1-2x+x^2}}\cdot\sqrt{\frac{4m-8mx+4mx^2}{81}}\)
\(=\sqrt{\frac{m}{\left(1-x\right)^2}}\cdot\sqrt{\frac{4m\left(1-2x+x^2\right)}{81}}\)
\(=\sqrt{\frac{m}{\left(1-x\right)^2}}\cdot\sqrt{\frac{4m\left(1-x\right)^2}{81}}\)
\(=\sqrt{\frac{m}{\left(1-x\right)^2}\cdot\frac{4m\left(1-x\right)^2}{81}}\)
\(=\sqrt{\frac{4m^2}{81}}=\sqrt{\frac{\left(2m\right)^2}{9^2}}=\frac{2\left|m\right|}{9}\)
3/\(\frac{a+b}{b^2}\sqrt{\frac{a^2b^4}{a^2+2ab+b^2}}\)
\(=\frac{a+b}{b^2}\sqrt{\frac{\left(ab^2\right)^2}{\left(a+b\right)^2}}\)
\(=\frac{a+b}{b^2}\cdot\frac{\left|a\right|b^2}{\left|a+b\right|}\)
TH1: \(\Rightarrow\frac{a+b}{b^2}\cdot\frac{\left|a\right|b^2}{-\left(a+b\right)}=-\left|a\right|\)
TH2: \(\Rightarrow\frac{a+b}{b^2}\cdot\frac{\left|a\right|b^2}{a+b}=\left|a\right|\)
A
19 tháng 6 2019
2/\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\frac{\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\right)\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\frac{\sqrt{a}-a}{1-\sqrt{a}}\right)\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\frac{1-a\sqrt{a}+\sqrt{a}-a}{1-\sqrt{a}}\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\frac{1-a\sqrt{a}+\sqrt{a}-a}{1}\cdot\frac{1-\sqrt{a}}{\left(1-a\right)^2}\)
\(=\frac{\left(1-a\sqrt{a}+\sqrt{a}-a\right)\cdot\left(1-\sqrt{a}\right)}{\left(1-a\right)^2}\)
\(=\frac{1-a\sqrt{a}+\sqrt{a}-a-\sqrt{a}+a^2-a+a\sqrt{a}}{\left(1-a\right)^2}\)
\(=\frac{a^2-2a+1}{\left(1-a\right)^2}\)
\(=\frac{\left(a-1\right)^2}{\left(1-a\right)^2}=\frac{-\left(1-a\right)^2}{\left(1-a\right)^2}=-1\)
A= \(\frac{1}{ab}\). \(\frac{|ab^2|}{4}\)
Vì a<0 nên \(\left|ab^2\right|\)= -a\(b^2\) => A= -b/4
\(A=\frac{1}{ab}.\frac{\left|a\right|b^2}{4}=\frac{1}{a}.\frac{-ab}{4}=\frac{-b}{4}\)
\(B=\frac{\left(a-3\right)^2}{2}.\frac{5}{\left|3-a\right|}=\frac{\left(a-3\right)^2}{2}.\frac{-5}{a-3}=\frac{-5\left(a-3\right)}{2}\)