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1, \(x^3=\left(7+\sqrt{\frac{49}{8}}\right)+\left(7-\sqrt{\frac{49}{8}}\right)+3x\sqrt[3]{\left(7+\sqrt{\frac{49}{8}}\right)\left(7-\sqrt{\frac{49}{8}}\right)}\)
\(=14+3x\cdot\frac{7}{2}=14+\frac{21x}{2}\)
\(\Leftrightarrow x^3-\frac{21}{2}x-14=0\)
Ta có: \(f\left(x\right)=\left(2x^3-21-29\right)^{2019}=\left[2\left(x^3-\frac{21}{2}x-14\right)-1\right]^{2019}=\left(-1\right)^{2019}=-1\)
2, ta có: \(1^3+2^3+...+n^3=\left(1+2+...+n\right)^2=\left[\frac{n\left(n+1\right)}{2}\right]^2\) (bạn tự cm)
Áp dụng công thức trên ta được n=2016
3, \(x=\frac{\sqrt[3]{17\sqrt{5}-38}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}=\frac{\sqrt[3]{\left(\sqrt{5}\right)^3-3.\left(\sqrt{5}\right)^2.2+3\sqrt{5}.2^2-2^3}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{9-2.3\sqrt{5}+5}}\)
\(=\frac{\sqrt[3]{\left(\sqrt{5}-2\right)^3}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}=\frac{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}{\sqrt{5}+3-\sqrt{5}}=\frac{5-4}{3}=\frac{1}{3}\)
Thay x=1/3 vào A ta được;
\(A=3x^3+8x^2+2=3.\left(\frac{1}{3}\right)^3+8.\left(\frac{1}{3}\right)^2+2=3\)
Bài 2:Áp dụng BĐT AM-GM ta có:
\(\frac{1}{x}+\frac{1}{y}\ge2\sqrt{\frac{1}{xy}}\)
\(\frac{1}{y}+\frac{1}{z}\ge2\sqrt{\frac{1}{yz}}\)
\(\frac{1}{x}+\frac{1}{z}\ge2\sqrt{\frac{1}{xz}}\)
CỘng theo vế 3 BĐT trên có:
\(2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge2\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\right)\)
Khi x=y=z
Ta có: \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{100}}\)
\(..........................\)
\(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\)
Cộng theo vế ta có:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}=\frac{100}{10}=10\)
\(B1,1,S_{3n}+3S_n=\left(2-\sqrt{3}\right)^{3n}+\left(2+\sqrt{3}\right)^{3n}+3\left[\left(2-\sqrt{3}\right)^n+\left(2+\sqrt{3}\right)^n\right]\)
\(=\left[\left(2-\sqrt{3}\right)^n\right]^3+\left[\left(2+\sqrt{3}\right)^n\right]^3\)
\(+3\left[\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)\right]^n\left[\left(2-\sqrt{3}\right)^n+\left(2+\sqrt{3}\right)^n\right]\)
Ta có hằng đẳng thức \(a^3+b^3+3ab\left(a+b\right)=\left(a+b\right)^3\)
Ở đây với \(a=\left(2-\sqrt{3}\right)^n\)và \(b=\left(2+\sqrt{3}\right)^n\)
Nên \(S_{3n}+3S_n=\left[\left(2-\sqrt{3}\right)^n+\left(2+\sqrt{3}\right)^n\right]^3=S_n^3\)
\(2,S_3=\left(2-\sqrt{3}\right)^3+\left(2+\sqrt{3}\right)^3\)
\(=\left(2-\sqrt{3}+2+\sqrt{3}\right)\left(2-\sqrt{3}-\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)+2+\sqrt{3}\right)\)
\(=4\left[4-\left(4-3\right)\right]\)
\(=12\)
Ta có \(S_4=\left(2-\sqrt{3}\right)^4+\left(2+\sqrt{3}\right)^4\)
\(=\left[\left(2-\sqrt{3}\right)^2\right]^2+\left[\left(2+\sqrt{3}\right)^2\right]^2\)
\(=\left(7-4\sqrt{3}\right)^2+\left(7+4\sqrt{3}\right)^2\)
\(=97-56\sqrt{3}+97+56\sqrt{3}\)
\(=194\)
\(B2,F=x^4+6x^3+13x^2+12x+12\)(Bài này cẩn thận dấu "=")
\(=\left(x^4+6x^3+9x^2\right)+4x^2+12x+12\)
\(=\left(x^2+3x\right)^2+4\left(x^2+3x\right)+4+8\)
\(=\left(x^2+3x+2\right)^2+8\ge8\)
Dấu "=" tại \(x^2+3x+2=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)