-215/21<21/215 

am lúc nào chả nhỏ hơn dương

\(\dfrac{x+50}{49}+\dfrac{x+2}{48}+\dfrac{x+50}{47}=-1\)

\(\dfrac{x+50}{49}+\left(\dfrac{x+2}{48}+1\right)+\dfrac{x+50}{47}=0\)

\(\dfrac{x+50}{49}+\dfrac{x+50}{48}+\dfrac{x+50}{47}=0\)

\(\left(x+50\right)\left(\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}\right)=0\)

\(\Leftrightarrow x=-50\)  Vì \(\left(\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}\right)>0\)

\(\dfrac{-188}{2017}< 0;\dfrac{1}{3678}>0\\ \Rightarrow\dfrac{-188}{2017}< \dfrac{1}{3678}\)

 Dấu <

HT

vì\(\dfrac{-188}{2017}\)bé hơn 0 nên \(\dfrac{-188}{2017}\)<\(\dfrac{1}{3678}\)

Bài 3:

theo đề bài ta có:

\(\left\{{}\begin{matrix}2a-3b=0\\5b-7c=0\\3a-7b+5c=30\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=42\\b=28\\c=20\end{matrix}\right.\)

Bài 4: 

Đặt \(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{z}{6}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=4k\\y=5k\\z=6k\end{matrix}\right.\)

Ta có: \(x^2-2y^2+z^2=18\)

\(\Leftrightarrow16k^2-50k^2+36k^2=18\)

\(\Leftrightarrow k^2=9\)

Trường hợp 1: k=3

\(\Leftrightarrow\left\{{}\begin{matrix}x=4k=4\cdot3=12\\y=5k=5\cdot3=15\\z=6k=6\cdot3=18\end{matrix}\right.\)

Trường hợp 2: k=-3

\(\Leftrightarrow\left\{{}\begin{matrix}x=4k=-3\cdot4=-12\\y=5k=-3\cdot5=-15\\z=6k=-3\cdot6=-18\end{matrix}\right.\)

Ta thấy x=0 không là nghiệm của pt

Chia cả 2 vế cho \(x^2\ne0\) ta được:

\(x^4+\text{ax}^3+bx^2+cx+1=0\)

\(\Leftrightarrow x^2+\dfrac{1}{x^2}+\text{ax}+b+\dfrac{c}{x}=0\)

\(\Leftrightarrow x^2+\dfrac{1}{x^2}=-\text{ax}-b-\dfrac{c}{x}\)

\(\Rightarrow\left(x^2+\dfrac{1}{x^2}\right)^2=\left(\text{ax}+\dfrac{c}{x}+b\right)^2\le\left(a^2+b^2+c^2\right)\left(x^2+\dfrac{1}{x^2}+1\right)\)

( theo BĐT Bunhiacopxki)

\(\Rightarrow\left(a^2+b^2+c^2\right)\ge\dfrac{\left(x^2+\dfrac{1}{x^2}\right)^2}{x^2+\dfrac{1}{x^2}+1}\ge\dfrac{4}{3}\)( theo bánh Cosi)

Dấu '=' xảy ra khi \(x^2=\dfrac{1}{x^2}\Leftrightarrow x=\pm1\)

==> Chọn A

\(\dfrac{4}{x-2}=\dfrac{x-2}{9}\left(x\ne2\right)\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-4\end{matrix}\right.\)

a: \(A=-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6+5\le5\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{10}\)

b: \(B=-\left(x-\dfrac{1}{6}\right)^4-\left(2y+1\right)^2\le0\forall x,y\)

Dấu '=' xảy ra khi \(\left(x,y\right)=\left(\dfrac{1}{6};-\dfrac{1}{2}\right)\)