Tìm x:
|3x - 1| \(\le\)11
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: =>|3x-5|=|x+2|
=>3x-5=x+2 hoặc 3x-5=-x-2
=>2x=7 hoặc 4x=3
=>x=7/2 hoặc x=3/4
b: \(\Leftrightarrow\left\{{}\begin{matrix}3x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
c: \(\Leftrightarrow\left|3x-5\right|=x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(3x-5-x+2\right)\left(3x-5+x-2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(2x-3\right)\left(4x-7\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
d: \(\dfrac{11}{2}\le\left|x\right|< \dfrac{17}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{11}{2}< =x< \dfrac{17}{2}\\-\dfrac{17}{2}< x< =-\dfrac{11}{2}\end{matrix}\right.\)
\(a,\dfrac{x^2+x+2}{\sqrt{x^2+x+1}}=\dfrac{x^2+x+1+1}{\sqrt{x^2+x+1}}=\sqrt{x^2+x+1}+\dfrac{1}{\sqrt{x^2+x+1}}\left(1\right)\)
Áp dụng BĐT cosi: \(\left(1\right)\ge2\sqrt{\sqrt{x^2+x+1}\cdot\dfrac{1}{\sqrt{x^2+x+1}}}=2\)
Dấu \("="\Leftrightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Bài giải:
a, \(11.xx-66=4.x+11\)
\(11x^2-66=4.x+11\)
\(11x^2-66-4.x-11=0\)
\(11x^2-77-4x=0\)
\(11x^2-4x-77=0\)
\(x=\frac{-\left(-4\right)+\sqrt{\left(-4\right)^2-4.11.\left(-77\right)}}{2.11}\)
\(x=\frac{4+\sqrt{16}+3388}{22}\)
\(x=\frac{4+\sqrt{3404}}{22}\)
\(x=\frac{4+2\sqrt{851}}{22}\)
\(x=\frac{2-\sqrt{851}}{11}\)
\(\Rightarrow\)Có hai trường hợp: \(x_1=\frac{2-\sqrt{851}}{11};x_2=\frac{2+\sqrt{851}}{11}\)
Tớ bận rồi, cậu coi câu trên đã nhé ! Tớ xin lỗi, khi nào tớ sẽ làm tiếp =))
\(Q=x^2\left(4-3x\right)=\dfrac{4}{9}.\dfrac{3}{2}x.\dfrac{3}{2}x\left(4-3x\right)\)
\(Q\le\dfrac{1}{27}.\dfrac{4}{9}.\left(\dfrac{3x}{2}+\dfrac{3x}{2}+4-3x\right)^3=\dfrac{256}{243}\)
\(Q_{maxx}=\dfrac{256}{243}\) khi \(\dfrac{3x}{2}=4-3x\Leftrightarrow x=\dfrac{8}{9}\)
\(A=\dfrac{x^2+x-2+x^2-x-2-4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}=\dfrac{2\left(x-2\right)\left(x+2\right)\left(x-3\right)}{2\left(x-2\right)\left(x+2\right)^2}=\dfrac{x-3}{x+2}\\ A\le0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\x+2< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\x+2>0\end{matrix}\right.\end{matrix}\right.\Rightarrow-2< x< 3;x\ne0\left(ĐKXD\right)\)
\(\left(3x+11\right)⋮\left(x+1\right)\)
\(\Leftrightarrow3\left(x+1\right)+8⋮\left(x+1\right)\)
\(\Leftrightarrow\left(x+1\right)\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)
\(\Leftrightarrow x\in\left\{-9;-5;-3;-2;0;1;3;7\right\}\)
=> -11\(\le\)3x-1 \(\le\)11
-10 \(\le\)3x \(\le\)12
-10/3\(\le\)x\(\le\)4