Nguyễn Việt Anh Mình hỏi bạn đặt câu hỏi chứ không hỏi bạn ạ. Vô duyên vừa vừa thôi chứ? 

hoc24.vn

Khác số chút thoyy.

C,D là điểm như nào cơ chứ =))))

ĐK : x ≠ 0

\(\dfrac{x+1}{x}=3\)

=> \(3x=x+1\)

=> \(2x=1\)

=> \(x=\dfrac{1}{2}\)

=> \(x^2=\dfrac{1}{4}\)

=> \(\dfrac{1}{x^2}=4\)

Khi đó

\(\dfrac{x^2+1}{x^2}=1+\dfrac{1}{x^2}=1+4=5\)

\(333^{444}=\left(333^4\right)^{111}=\left(111^4.81\right)^{111}\)

\(444^{333}=\left(444^3\right)^{111}=\left(111^3.64\right)^{111}\)

Dễ thấy \(111^4.81>111^3.64\)

\(\Rightarrow333^{444}>444^{333}\)

x đầu ở đa thức A là x^3 chăng?

a/ \(A=x^3-5x^2+8x-4\)

\(=\left(x^3-x^2\right)+\left(-4x^2+4\right)+\left(8x-8\right)\)

\(=x^2\left(x-1\right)-4\left(x-1\right)\left(x+1\right)+8\)

\(=\left(x-1\right)\left(x^2-4x-4\right)=\left(x-1\right)\left(x-2\right)^2\)

b/ \(B=\dfrac{x^5}{30}-\dfrac{x^3}{6}+\dfrac{2x}{15}\)

\(=\dfrac{x^5}{30}-\dfrac{5x^3}{30}+\dfrac{4x}{30}\)

\(=\dfrac{x\left(x^4-5x^2+4\right)}{30}\)

\(=\dfrac{x\left(x^4-x^2-4x^2+4\right)}{30}\)

\(=\dfrac{x\left(x+2\right)\left(x-1\right)\left(x+1\right)\left(x-2\right)}{30}\)

\(\Leftrightarrow\left(x^4-20x^2+100\right)-36=0\)

\(\Leftrightarrow\left(x^2-10\right)^2=36\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-10=6\\x^2-10=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=16\\x^2=4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\pm4\\x=\pm2\end{matrix}\right.\)

Linh tinh thui, chắc sai.

Có

\(x+\dfrac{1}{x}=2\) (x khác 0) 

\(\Rightarrow\dfrac{x^2+1}{x}=2\Rightarrow x^2+1=2x\Rightarrow\left(x-1\right)^2=0\Rightarrow x=1\)(TM)

Thay \(x=1\) vào bt A có \(A=\dfrac{1}{2}\)

\(2A=2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+\dfrac{5}{2^4}+...+\dfrac{100}{2^{99}}\)

=> \(2A-A=A=1+\dfrac{3}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+....+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)

Đặt \(B=\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}\)

=> \(2B=\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{98}}\)

=> \(B=\dfrac{1}{2^2}-\dfrac{1}{2^{99}}\)

=> \(A=1+\dfrac{3}{2^2}+\dfrac{1}{2^2}-\dfrac{100}{2^{100}}-\dfrac{1}{2^{99}}\)

=> \(A=2-\dfrac{102}{2^{100}}< 2\)

a/ Pt \(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2x+8=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)

Vậy \(S=\left\{7;-4\right\}\)

b/ pt \(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\5x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=\dfrac{2}{5}\end{matrix}\right.\)

c/ pt \(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\) (\(x^2+2>0\forall x\))\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)

d/ pt \(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+8=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-8\\x=5\end{matrix}\right.\)