tính x^3 bạn ơi

Ta có:

\(x=1+\sqrt[3]{5}+\sqrt[3]{25}\)

\(\Rightarrow x^3=\left(1+\sqrt[3]{5}+\sqrt[3]{25}\right)^3=61+33\sqrt[3]{5}+21\sqrt[3]{25}\)

\(=\left(33+21\sqrt[3]{5}+9\sqrt[3]{25}\right)+\left(12+12\sqrt[3]{5}+12\sqrt[3]{25}\right)+16=3x^2+12x+16\)

\(\Rightarrow P=\left(x^3-3x^2-12x-15\right)^{10}+2018\)

\(=\left(3x^2+12x+16-3x^2-12x-15\right)^{10}+2018=2019\)

chắc bạn chép sai đề rồi , hai căn đầu phải 1 cộng 1 trừ chứ

Đặt

\(x=\dfrac{y+1}{2}\Rightarrow y=2x-1\)

\(\Rightarrow y=\sqrt[3]{4+\sqrt{15}}+\sqrt[3]{4-\sqrt{15}}\)

\(y^3=8+3\sqrt[3]{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}=8+3y\)

\(\Rightarrow y^3-3y-8=0\\ \)

\(\Leftrightarrow8x^3-12x^2-6=0\)

\(\Rightarrow4x^3-6x^2-3=0\)

thay p vào ta có

\(P=12x^5-18x^4+4x^3-15x^2-21\)

\(=12x^5-18x^4-9x^2-4x^3-6x^2-21\)

\(=3x^2\left(4x^2-6x^2-3\right)+4x^3-6x^2-3\\ =3x^2.0+0-18\\ =-18\)

ai nay dung kinh nghiem la chinh

cau a)

ta thay \(10+6\sqrt{3}=\left(1+\sqrt{3}\right)^3\)

\(6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)

khi do \(x=\frac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{5}}\)

\(x=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{1+\sqrt{5}-\sqrt{5}}\)

\(x=\frac{3-1}{1}=2\)

suy ra 

x^3-4x+1=1

A=1^2018

A=1

b)

ta thay

\(7+5\sqrt{2}=\left(1+\sqrt{2}\right)^3\)

khi do 

\(x=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\frac{1}{\sqrt[3]{\left(1+\sqrt{2}\right)^3}}\)

\(x=1+\sqrt{2}-\frac{1}{1+\sqrt{2}}=\frac{\left(1+\sqrt{2}\right)^2-1}{1+\sqrt{2}}=\frac{2+2\sqrt{2}}{1+\sqrt{2}}\)

x=2

thay vao

x^3+3x-14=0

B=0^2018

B=0

1.

\(\lim\limits_{x\to +\infty}(x^3+3x^2+2)=+\infty\)

2. 

\(\lim\limits_{x\to -\infty}\sqrt{4x^2-x+5}=\lim\limits_{x\to -\infty}-x.\sqrt{4+\frac{1}{x}+\frac{5}{x^2}}=+\infty\) do $-x\to +\infty$ và $\lim\limits_{x\to -\infty}\sqrt{4+\frac{1}{x}+\frac{5}{x^2}}=4>0$

3.

\(\lim\limits_{x\to +\infty}(\sqrt{x^2-2x-1}-\sqrt{x^2-7x+3})=\lim\limits_{x\to +\infty}\frac{x^2-2x-1-(x^2-7x+3)}{\sqrt{x^2-2x-1}+\sqrt{x^2-7x+3}}\)

\(=\lim\limits_{x\to +\infty}\frac{5x-4}{\sqrt{x^2-2x-1}+\sqrt{x^2-7x+3}}=\lim\limits_{x\to +\infty}\frac{5-\frac{4}{x}}{\sqrt{1-\frac{2}{x}-\frac{1}{x^2}}+\sqrt{1-\frac{7}{x}+\frac{3}{x^2}}}\)

\(=\frac{5}{1+1}=\frac{5}{2}\)

a) ĐKXĐ : \(3\le x\le7\)

Ta có \(A=1.\sqrt{x-3}+1.\sqrt{7-x}\)

\(\le\sqrt{\left(1+1\right)\left(x-3+7-x\right)}=\sqrt{8}\)(BĐT Bunyacovski)

Dấu "=" xảy ra <=> \(\dfrac{1}{\sqrt{x-3}}=\dfrac{1}{\sqrt{7-x}}\Leftrightarrow x=5\)

\(1,\\ a,A\le\sqrt{\left(x-3+7-x\right)\left(1+1\right)}=\sqrt{8}=2\sqrt{2}\\ A^2=4+2\sqrt{\left(x-3\right)\left(7-x\right)}\ge4\Leftrightarrow A\ge2\\ \Leftrightarrow2\le A\le2\sqrt{2}\\ \left\{{}\begin{matrix}A_{min}\Leftrightarrow\left(x-3\right)\left(7-x\right)=0\Leftrightarrow...\\A_{max}\Leftrightarrow x-3=7-x\Leftrightarrow x=5\end{matrix}\right.\)

\(B=\dfrac{\dfrac{5}{2}\left(4x^4+4x^2+1\right)+2\left(x^4-x^2+\dfrac{1}{4}\right)}{\left(2x^2+1\right)^2}\\ B=\dfrac{\dfrac{5}{2}\left(2x^2+1\right)^2+2\left(x^2-\dfrac{1}{2}\right)^2}{\left(2x^2+1\right)^2}=\dfrac{5}{2}+\dfrac{2\left(x^2-\dfrac{1}{2}\right)^2}{\left(2x^2+1\right)^2}\ge\dfrac{5}{2}\)

\(B=\dfrac{3\left(4x^4+4x^2+1\right)-4x^2}{\left(1+2x^2\right)^2}=\dfrac{3\left(1+2x^2\right)^2-4x^2}{\left(1+2x^2\right)^2}=3-\dfrac{4x^2}{\left(1+2x^2\right)^2}\)

Vì \(-\dfrac{4x^2}{\left(1+2x^2\right)^2}\le0\Leftrightarrow B\le3\)

\(\Leftrightarrow\left\{{}\begin{matrix}B_{min}\Leftrightarrow x^2=\dfrac{1}{2}\Leftrightarrow x=\pm\dfrac{1}{\sqrt{2}}\\B_{max}\Leftrightarrow x=0\end{matrix}\right.\)

\(2,\)

Ta có \(\left(y-2x\right)^2=\left(-2x+y\right)^2=\left[\dfrac{1}{3}\left(-6x\right)+\dfrac{1}{4}\left(4y\right)\right]^2\)

\(\Leftrightarrow\left(y-2x\right)^2\le\left[\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{4}\right)^2\right]\left[\left(-6x\right)^2+\left(4y\right)^2\right]=\dfrac{5^2}{3^2\cdot4^2}\left(36x^2+16y^2\right)=\dfrac{5^2}{4^2}\\ \Leftrightarrow\left|y-2x\right|\le\dfrac{5}{4}\\ \Leftrightarrow-\dfrac{5}{4}\le y-2x\le\dfrac{5}{4}\\ \Leftrightarrow\dfrac{15}{4}\le y-2x+5\le\dfrac{25}{4}\)

\(Max\Leftrightarrow\left\{{}\begin{matrix}-18x=16y\\y-2x=\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{5}\\y=\dfrac{9}{20}\end{matrix}\right.\\ Min\Leftrightarrow\left\{{}\begin{matrix}-18x=16y\\y-2x=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-\dfrac{9}{20}\end{matrix}\right.\)