Cmr
nếu a=\(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\) thì
a^3+3a=4
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Lập phương cả hai vế ta được
\(a^3=4-3a\)
\(\Rightarrow a^3+3a\Leftrightarrow4-3a+3a=4\left(đpcm\right)\)
a: Sửa đề: căn 6+2căn 5-căn 5
\(a=\dfrac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{\sqrt{5}+1-\sqrt{5}}=\dfrac{2}{1}=2\)
b: \(a^3=2-\sqrt{3}+2+\sqrt{3}+3a\)
=>a^3-3a-4=0
=>a^3-3a=4
\(\dfrac{64}{\left(a^2-3\right)^3}-3a=\left(\dfrac{4}{a^2-3}\right)^3-3a\)
\(=\left(\dfrac{a^3-3a}{a^2-3}\right)^3-3a=a^3-3a\)
=4
\(1) \sqrt{9a^2.b^2}\)=3ab
\(2) \sqrt{3a}.\sqrt{27a}=\sqrt{3a}.3\sqrt{3a}=9a\)
\(3) \sqrt{3a^5}.12a=12\sqrt{3a^7}\)
\(4) \sqrt{5a}.\sqrt{45a}-3a=15a-3a=12a\)
\(5) \sqrt{3+\sqrt{a}}.\sqrt{3-\sqrt{a}}=\sqrt{(3+\sqrt{a}).(3-\sqrt{a})} =\sqrt{9-a} \)
\(6) \sqrt{3+\sqrt{5}}.\sqrt{3\sqrt{5}} =\sqrt{\sqrt{3\sqrt{5}}.(3+\sqrt{5})} =\sqrt{9+\sqrt{15}}\)
1) \(\sqrt{9a^2b^2}=3ab\)
2) \(\sqrt{3a}\cdot\sqrt{27a}=9a\)
4) \(\sqrt{5a}\cdot\sqrt{45a}-3a=15a-3a=12a\)
Bình phương a ta được
\(a^2=3+3+\sqrt{5+2\sqrt{3}}-\sqrt{5+2\sqrt{3}}+2\sqrt{\left(3+\sqrt{5+2\sqrt{3}}\right)\left(3-\sqrt{5+2\sqrt{3}}\right)}\)
\(a^2=6+2\sqrt{9-3\sqrt{5+2\sqrt{3}}+3\sqrt{5+2\sqrt{3}}-5-2\sqrt{3}}\)
\(a^2=6+2\sqrt{9-5-2\sqrt{3}}\Rightarrow a^2=6+2\sqrt{4-2\sqrt{3}}\Rightarrow a^2=6+2\sqrt{3+1-2.1.\sqrt{3}}\)\(a^2=6+2\sqrt{\left(\sqrt{3}-1\right)^2}\Rightarrow a^2=6+2\sqrt{3}-2=4+2\sqrt{3}=3+1+2.1.\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow a=\sqrt{3}+1\)
Rồi bạn tự thay vào tính típ nha
Chúc bạn học tốt
T I C K ủng hộ nha
Đặt \(m=\sqrt[3]{x^2}\)và \(n=\sqrt[3]{y^2}\)
=> m3 = x2 và n3 = y2
Ta có :\(\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^4}}=a\)
=> \(\sqrt{m^3+\sqrt[3]{m^6n^3}}+\sqrt{n^3+\sqrt[3]{m^3n^6}}=a\)
=> \(\sqrt{m^3+m^2n}+\sqrt{n^3+mn^2}=a\)
=> \(\sqrt{m^2\left(m+n\right)}+\sqrt{n^2\left(m+n\right)}=a\)
=> \(\sqrt{m+n}\left(m+n\right)=a\)
=> \(\left(\sqrt{m+n}\right)^3=\left(\sqrt[3]{a}\right)^3\)
=>\(\sqrt{m+n}=\sqrt[3]{a}\)
=> \(m+n=\left(\sqrt[3]{a}\right)^2\)
=> \(\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}\)
1) \(ĐK:3-2a>0\Leftrightarrow a< \dfrac{3}{2}\)
2) \(ĐK:2x-5< 0\Leftrightarrow x< \dfrac{5}{2}\)
3) \(ĐK:3-5a< 0\Leftrightarrow a>\dfrac{3}{5}\)
4) \(ĐK:a< 0\)
5) \(ĐK:-a\ge0\Leftrightarrow a\le0\)
a) \(=5\left|a\right|+3a=5a+3a=8a\)
b) \(=3\left|a^2\right|+3a^2=3a^2+3a^2=6a^2\)
c) \(=5.2\left|a^3\right|-3a^3=-10a^3-3a^3=-13a^3\)
\(a^3=\sqrt{5}+2-\sqrt{5}+2-3a\)
=>a^3=4-3a
=>a^3+3a=4