giải chi tiết hộ mk nhá

\(\sqrt[3]{16-8\sqrt{5}}\)=\(\sqrt[3]{1-3\sqrt{5}+15-5\sqrt{5}}\)=\(\sqrt[3]{1-3\sqrt{5}+3\left(\sqrt{5}\right)^2-\left(\sqrt{5}\right)^3}\)=\(\sqrt[3]{\left(1-\sqrt{5}\right)^3}\)=\(1-\sqrt{5}\)

làm tương tự: \(\sqrt[3]{16+8\sqrt{5}}\)=\(1+\sqrt{5}\)

suy ra: a = 2

`c)root{3}{4}.root{3}{1-sqrt3}.root{6}{(sqrt3+1)^2}`

`=root{3}{4(1-sqrt3)}.root{3}{1+sqrt3}`

`=root{3}{4(1-sqrt3)(1+sqrt3)}`

`=root{3}{4(1-3)}=-2`

`d)2/(root{3}{3}-1)-4/(root{9}-root{3}{3}+1)`

`=(2(root{3}{9}+root{3}{3}+1))/(3-1)-(4(root{3}{3}+1))/(3+1)`

`=root{3}{9}+root{3}{3}+1-root{3}{3}-1`

`=root{3}{9}`

`a)root{3}{8sqrt5-16}.root{3}{8sqrt5+16}`

`=root{3}{(8sqrt5-16)(8sqrt5+16)}`

`=root{3}{320-256}`

`=root{3}{64}=4`

`b)root{3}{7-5sqrt2}-root{6}{8}`

`=root{3}{1-3.sqrt{2}+3.2.1-2sqrt2}-root{6}{(2)^3}`

`=root{3}{(1-sqrt2)^3}-sqrt2`

`=1-sqrt2-sqrt2=1-2sqrt2`

Đặt \(a=\sqrt[3]{16-8\sqrt{5}};b=\sqrt[3]{16+8\sqrt{5}}\)

Ta có \(a^3+b^3=32\)

\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)=32\left(^∗\right)\)

\(a^3.b^3=\left(16-8\sqrt{5}\right)\left(16+8\sqrt{5}\right)=16^2-\left(8\sqrt{5}\right)^2=-64\)

\(\Rightarrow ab=-4\)

Kết hợp với \(\left(^∗\right)\) \(\Rightarrow\left(a+b\right)^3+12\left(a+b\right)=32\)

\(\Rightarrow a+b=2=x\)

Thay \(x=2\)vào \(f\left(x\right)\)ta được :

\(F\left(2\right)=\left(2^3+12.2-31\right)^{2016}^{^{2017}}=1^{2016^{2017}}=1\)

Chúc bạn học tốt !!!

\(a=\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\)

\(\Leftrightarrow a^3=16-8\sqrt{5}+16+8\sqrt{5}+3\sqrt[3]{\left(16-8\sqrt{5}\right)\left(16+8\sqrt{5}\right)}\cdot a\)

\(\Leftrightarrow a^3=32+3\sqrt[3]{256-320}\cdot a\)

\(\Leftrightarrow a^3=32-12a\)

Giải pt được \(a=2\).

Khi đó : \(P\left(a\right)=\left(2^2+12\cdot2-31\right)=-3\)

Vậy...

Đặt \(a=\sqrt[3]{16-8\sqrt{5}};b=\sqrt[3]{16+8\sqrt{5}}\)

Ta có: a³ + b³ = 32

=> (a + b)³ - 3ab(a + b) = 32 (*)

a³.b³ = \(\left(16-8\sqrt{5}\right)\left(16+8\sqrt{5}\right)=16^2-\left(8\sqrt{5}\right)^2=-64\)

=> ab = -4

Kết hợp với (*) => (a + b)³ + 12(a + b) = 32

=> a + b = 2 = x

Thay x = 2 vào f(x) ta được:

\(F\left(2\right)=\left(2^3+12.2-31\right)^{2016^{2017}}=1^{2016^{2017}}=1\)

\(x=\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16-8\sqrt{5}}\)

\(\Rightarrow x^3=32+3\sqrt[3]{16^2-8^2.5}\left(\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\right)\)

\(\Rightarrow x^3=32-12x\)

\(\Rightarrow x^3+12x-32=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+2x+16\right)=0\)

\(\Rightarrow x=2\)

Vậy \(\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}=2\)

$x=\sqrt[3]{316-8\sqrt{5}}+\sqrt[3]{316-8\sqrt{5}}$

$\Rightarrow x^3=32+3\sqrt[3]{3{16}^2-8^2.5}\left(\sqrt[3]{316-8\sqrt{5}}+\sqrt[3]{316+8\sqrt{5}}\right)$

$\Rightarrow x^3=32-12x$

$\Rightarrow x^3+12x-32=0$

$\Rightarrow \left(x-2\right)\left(x^2+2x+16\right)=0$

$\Rightarrow x=2$

Vậy $\sqrt[3]{316-8\sqrt{5}}+\sqrt[3]{316+8\sqrt{5}}=2$

M bằng gì bạn

\(a^3=16-8\sqrt{5}+16+8\sqrt{5}+96\sqrt[3]{\left(16-8\sqrt{5}\right)\left(16+8\sqrt{5}\right)}\)

\(a^3=32+96\sqrt[3]{-64}=32+96.\left(-4\right)=-352\)

đến đây dễ r 

\(a^3=32+3\sqrt[3]{\left(16-8\sqrt{5}\right)\left(16+8\sqrt{5}\right)}\left(\sqrt[3]{16+8\sqrt{5}}+\sqrt[3]{16-8\sqrt{5}}\right)\)

Từ \(a=\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\)

\(\Rightarrow a^3=32+3\sqrt[3]{\left(16-8\sqrt{5}\right)\left(16+8\sqrt{5}\right)}\)\(\left[\sqrt[3]{16+8\sqrt{5}}+\sqrt[3]{18-8\sqrt{5}}\right]\)

\(=32+3\sqrt[3]{-64}a=32-12a\)

\(\Rightarrow a^3+12a=32\)

\(\Rightarrow f\left(a\right)=\left(a^3+12a-31\right)^{2012}=\left(32-31\right)^{2012}=1\)

Vậy f(a) = 1