Đề là \(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-3\right)\sqrt{4-\sqrt{15}}\)

Hay \(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\) bạn?

Như bạn ghi thì ko có gì đặc biệt để tính ra kết quả đẹp đâu

a, \(\sqrt{\left(2x+3\right)^2}=x+1\)

\(\Leftrightarrow\left|2x+3\right|=x+1\)

TH1: \(\left\{{}\begin{matrix}2x+3=x+1\\2x+3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x\ge-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\) vô nghiệm.

Vậy phương trình vô nghiệm.

TH2: \(\left\{{}\begin{matrix}-2x-3=x+1\\2x+3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{4}{3}\\x< -\dfrac{3}{2}\end{matrix}\right.\Rightarrow\) vô nghiệm.

b, 

a, \(\sqrt{\left(2x-1\right)^2}=x+1\)

\(\Leftrightarrow\left|2x-1\right|=x+1\)

TH1: \(\left\{{}\begin{matrix}2x-1=x+1\\2x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x\ge\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x=2\)

TH2: \(\left\{{}\begin{matrix}-2x+1=x+1\\2x-1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x< \dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x=0\)

tìm x, biết

\(a=\sqrt{2}+\sqrt{7-2\sqrt{5}-1}+1\)

\(=\sqrt{2}+\sqrt{5}-1+1=\sqrt{2}+\sqrt{5}\)

f(x)=x^4(x+2)-14x^2(x+2)+9(x+2)+1

=(x+2)(x^4-14x^2+9)+1

\(=\left(\sqrt{2}+\sqrt{5}+2\right)\left[\left(7+2\sqrt{10}\right)^2-14\left(7+2\sqrt{10}\right)+1\right]\)+1

\(=\left(\sqrt{2}+\sqrt{5}+2\right)\left(89+28\sqrt{10}-84-28\sqrt{10}+1\right)\)+1

=6(căn 2+căn 5+1)+1

Yêu cầu?

\(a^3=38+17\sqrt{5}+38-17\sqrt{5}+3\cdot a\cdot\sqrt[3]{\left(38\right)^2-\left(17\sqrt{5}\right)^2}\)

=>a^3=76-3a

=>a^3+3a-76=0

=>a=4

f(x)=(4^3+3*4+1940)^2016=2016^2016

a = \(\sqrt{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}\left(\sqrt{10}-\sqrt{6}\right)=\sqrt{16-15}\left(\sqrt{10}-\sqrt{6}\right)=\sqrt{10}-\sqrt{6}\)

f(a)= \(\frac{4\left(\sqrt{10}-\sqrt{6}\right)\left(\sqrt{10}-\sqrt{6}+1\right)}{\sqrt{10}-\sqrt{6}-1}\)

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 4

ÁP DỤNG BĐT CAUCHY 

là ra

\(\frac{1}{\sqrt{1^3+2^3}}+\frac{1}{\sqrt{1^3+2^3+3^3}}+...+\frac{1}{\sqrt{1^3+2^3+3^3+...+n^3}}=\frac{2015}{2017}\) (1) 

Cần CM: \(1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2\) quy nạp nhé bn, trên mạng có nhìu 

(1) \(\Leftrightarrow\)\(\frac{1}{\sqrt{\left(1+2\right)^2}}+\frac{1}{\sqrt{\left(1+2+3\right)^2}}+...+\frac{1}{\sqrt{\left(1+2+3+...+n\right)^2}}=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+n}=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(\frac{1}{\frac{2\left(2+1\right)}{2}}+\frac{1}{\frac{3\left(3+1\right)}{2}}+...+\frac{1}{\frac{n\left(n+1\right)}{2}}=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\right)=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(n=2016\)