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câu a đáp án bằng 1
câu b đáp án bằng 11
mik hc lớp 9 kb có đk nha bn
a) \(\left(\sqrt[3]{2}-1\right)\left(\sqrt[3]{4}+\sqrt[3]{2}+1\right).\)
\(=\left(\sqrt[3]{2}\right)^3-1^3\)
\(=2-1\)
\(=1\)
b) \(\left(\sqrt[3]{3}+2\right)\left(\sqrt[3]{9}-2\sqrt[3]{3}+4\right)\)
\(=\left(\sqrt[3]{3}\right)^3+2^3\)
\(=3+8\)
\(=11\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)
\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(=4-3\cdot A\)
\(\Leftrightarrow A^3+3A-4=0\)
\(\Leftrightarrow A^3-A+4A-4=0\)
\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)
\(\Leftrightarrow A=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x=\dfrac{3\sqrt[3]{8-3\sqrt{5}}}{\sqrt[3]{57}}.\sqrt[3]{8+3\sqrt{5}}=\dfrac{3\sqrt[3]{\left(8-3\sqrt{5}\right)\left(8+3\sqrt[]{5}\right)}}{\sqrt[3]{57}}=\sqrt[3]{\dfrac{19}{57}}=\dfrac{1}{\sqrt[3]{3}}\)
\(y=\dfrac{\left(\sqrt[3]{3}+\sqrt[4]{2}\right)\left(\sqrt[3]{3}-\sqrt[4]{2}\right)}{\sqrt[3]{3}+\sqrt[4]{2}}+\dfrac{\left(\sqrt[4]{2}-\sqrt[3]{81}\right)\left(\sqrt[4]{2}+\sqrt[3]{81}\right)}{\sqrt[4]{2}-\sqrt[3]{81}}\)
\(=\sqrt[3]{3}-\sqrt[4]{2}+\sqrt[4]{2}+\sqrt[3]{81}=\sqrt[3]{3}+3\sqrt[3]{3}=4\sqrt[3]{3}\)
\(T=xy=\dfrac{4\sqrt[3]{3}}{\sqrt[3]{3}}=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left|a-3\right|-3a=3-a-3a=3-4a\)
\(B=4a+3-\left|2a-1\right|=4a+3-2a+1=2a+4\)
\(C=\dfrac{4}{a^2-4}\left|a-2\right|=\dfrac{-4\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}=\dfrac{-4}{a+2}\)
\(D=\dfrac{a^2-9}{12}:\sqrt{\dfrac{\left(a+3\right)^2}{16}}=\dfrac{a^2-9}{12}:\dfrac{\left|a+3\right|}{4}=\dfrac{\left(a-3\right)\left(a+3\right).4}{-12\left(a+3\right)}=\dfrac{3-a}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=2\sqrt{2}+1-3=2\sqrt{2}-2\)
b: \(=\sqrt{3}+1-2\sqrt{3}-1=-\sqrt{3}\)
c: \(=2-\sqrt{3}+\sqrt{3}-1=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: a>=0; a<>9
\(Q=\dfrac{3}{\sqrt{a}-3}+\dfrac{2}{\sqrt{a}+3}-\dfrac{a-5\sqrt{a}-3}{9-a}\)
\(=\dfrac{3\left(\sqrt{a}+3\right)+2\left(\sqrt{a}-3\right)+a-5\sqrt{a}-3}{a-9}\)
\(=\dfrac{3\sqrt{a}+9+2\sqrt{a}-6+a-5\sqrt{a}-3}{a-9}=\dfrac{a}{a-9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}\right)^2-4\sqrt{5+2^2}}-\sqrt{\left(\sqrt{5}\right)^2+4\sqrt{5}+2^2}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\left|\sqrt{5}-2\right|-\left|\sqrt{5}+2\right|\)
\(=\sqrt{5}-2-\left(\sqrt{5}+2\right)\)
\(=-4\)
\(B=\sqrt[3]{9}.\sqrt[3]{-3}+\left(1+\sqrt{2}\right)^2\)
\(=-\sqrt[3]{27}+3+2\sqrt{2}\)
\(=-3+3+2\sqrt{2}\)
\(=2\sqrt{2}\)