cứu cứu xong tui tăng sp cho
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![](https://rs.olm.vn/images/avt/0.png?1311)
Em tách nhỏ ra để hỏi, không đăng cả đề như thế này, em nhé!
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Ta có \(\dfrac{a^3}{a^2+b^2}=a-\dfrac{ab^2}{a^2+b^2}\ge a-\dfrac{ab^2}{2ab}=a-\dfrac{b}{2}=\dfrac{2a-b}{2}\)(áp dụng cosi cho \(a^2+b^2\ge2ab\))
\(\dfrac{b^3}{b^2+1}=b-\dfrac{b}{b^2+1}\ge b-\dfrac{b}{2b}=b-\dfrac{1}{2}=\dfrac{2b-1}{2}\)(áp dụng cosi cho\(b^2+1\ge2b\))
\(\dfrac{1}{a^2+1}=1-\dfrac{a^2}{a^2+1}\ge1-\dfrac{a^2}{2a}=1-\dfrac{a}{2}=\dfrac{2-a}{2}\)( áp dụng cosi cho \(a^2+1\ge2a\))
Cộng vế theo vế
\(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+1}+\dfrac{1}{a^2+1}\ge\dfrac{2a-b+2b-1+2-a}{2}\)\(\ge\dfrac{a+b+1}{2}\left(đpcm\right)\)
Dấu "=" xảy ra <=> a=b=1
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Ta có:
\(36=2^2.3^2\)
\(54=2.3^3\)
\(ƯCLN\left(36;54\right)=2.3=6\)
\(BCNN\left(36;54\right)=2^2.3^3=4.27=108\)
![](https://rs.olm.vn/images/avt/0.png?1311)
10: \(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)
\(=\left(\sqrt{3-\sqrt{5}}\right)^2+\left(\sqrt{3+\sqrt{5}}\right)^2+2\cdot\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
\(=3-\sqrt{5}+3+\sqrt{5}+2\cdot\sqrt{9-5}\)
\(=6+2\cdot2=10\)
11: \(\left(\sqrt{\sqrt{7}+\sqrt{3}}+\sqrt{\sqrt{7}-\sqrt{3}}\right)^2\)
\(=\left(\sqrt{\sqrt{7}+\sqrt{3}}\right)^2+\left(\sqrt{\sqrt{7}-\sqrt{3}}\right)^2+2\cdot\sqrt{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}\)
\(=\sqrt{7}+\sqrt{3}+\sqrt{7}-\sqrt{3}+2\cdot\sqrt{7-3}\)
\(=2\sqrt{7}+2\cdot2=2\sqrt{7}+4\)
12: \(\left(\sqrt{\sqrt{11}+\sqrt{7}}-\sqrt{\sqrt{11}-\sqrt{7}}\right)^2\)
\(=\left(\sqrt{\sqrt{11}+\sqrt{7}}\right)^2+\left(\sqrt{\sqrt{11}-\sqrt{7}}\right)^2-2\cdot\sqrt{\left(\sqrt{11}-\sqrt{7}\right)\left(\sqrt{11}+\sqrt{7}\right)}\)
\(=\sqrt{11}+\sqrt{7}+\sqrt{11}-\sqrt{7}-2\cdot\sqrt{11-7}\)
\(=2\sqrt{11}-4\)
13:
\(\sqrt{\sqrt{2}-1}\cdot\sqrt{2-\sqrt{3-\sqrt{2}}}\cdot\sqrt{2+\sqrt{3-\sqrt{2}}}\)
\(=\sqrt{\sqrt{2}-1}\cdot\sqrt{4-\left(3-\sqrt{2}\right)}\)
\(=\sqrt{\sqrt{2}-1}\cdot\sqrt{\sqrt{2}+1}\)
\(=\sqrt{2-1}=1\)
14:
\(\sqrt{4+\sqrt{8}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{\left(2+\sqrt{2+\sqrt{2}}\right)\left(2-\sqrt{2+\sqrt{2}}\right)}\)
\(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{4-2-\sqrt{2}}\)
\(=\sqrt{\left(4+2\sqrt{2}\right)\left(2-\sqrt{2}\right)}\)
\(=\sqrt{8-4\sqrt{2}+4\sqrt{2}-4}=\sqrt{4}=2\)
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\(B=\dfrac{3}{1x2}+\dfrac{3}{2x3}+...+\dfrac{3}{50x51}\)
\(B=3x\left(\dfrac{1}{1x2}+\dfrac{1}{2x3}+...+\dfrac{1}{50x51}\right)\)
\(B=3x\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{50}-\dfrac{1}{51}\right)\)
\(B=3x\left(1-\dfrac{1}{51}\right)=3x\dfrac{50}{51}=\dfrac{150}{51}\)
làm được câu nào thì làm cho tui nhé thanks
Bài 1
a. 9/4
b.-1/5
Bài 2.
a.6
b. 22/15