\(\left|3x-1\right|+\left|1-3x\right|=6\)
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13 tháng 10 2023
a: \(y'< 0\)
=>\(\left(x-3\right)^3\cdot\left(x-1\right)^{22}\cdot\left(-3x-6\right)^7< 0\)
=>\(\left(x-3\right)\left(-3x-6\right)< 0\)
=>\(\left(x+2\right)\left(x-3\right)>0\)
=>\(\left[{}\begin{matrix}x>3\\x< -2\end{matrix}\right.\)
y'>0
=>\(\left(x+2\right)\left(x-3\right)< 0\)
=>\(-2< x< 3\)
y'=0
=>\(\left[{}\begin{matrix}x-3=0\\x-1=0\\-3x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\\x=-2\end{matrix}\right.\)
Ta có bảng xét dấu sau:
| x | \(-\infty\) -2 1 3 +\(\infty\) |
| y' | - 0 + 0 + 0 - |
Vậy: Hàm số đồng biến trên các khoảng \(\left(-2;1\right);\left(1;3\right)\)
Hàm số nghịch biến trên các khoảng \(\left(-\infty;-2\right);\left(3;+\infty\right)\)
b: y'<0
=>\(\left(4x-3\right)^3\cdot\left(x^2-1\right)^{21}\left(3x-9\right)^7< 0\)
=>\(\left(4x-3\right)\left(3x-9\right)\left(x^2-1\right)< 0\)
=>\(\left(4x-3\right)\left(x-3\right)\left(x^2-1\right)< 0\)
TH1: \(\left\{{}\begin{matrix}\left(4x-3\right)\left(x-3\right)>0\\x^2-1< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>3\\x< \dfrac{3}{4}\end{matrix}\right.\\-1< x< 1\end{matrix}\right.\Leftrightarrow-1< x< \dfrac{3}{4}\)
TH2: \(\left\{{}\begin{matrix}\left(4x-3\right)\left(x-3\right)< 0\\x^2-1>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{4}< x< 3\\\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow1< x< 3\)
y'>0
=>\(\left(4x-3\right)\left(x-3\right)\left(x^2-1\right)>0\)
TH1: \(\left\{{}\begin{matrix}\left(4x-3\right)\left(x-3\right)>0\\x^2-1>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>3\\x< \dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -1\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}\left(4x-3\right)\left(x-3\right)< 0\\x^2-1< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{4}< x< 3\\-1< x< 1\end{matrix}\right.\Leftrightarrow\dfrac{3}{4}< x< 1\)
Ta sẽ có bảng xét dấu sau đây:
| x | \(-\infty\) -1 3/4 1 3 +\(\infty\) |
| y' | + 0 - 0 + 0 - 0 + |
Vậy: Hàm số đồng biến trên các khoảng \(\left(-\infty;-1\right);\left(\dfrac{3}{4};1\right);\left(3;+\infty\right)\)
Hàm số nghịch biến trên các khoảng \(\left(-1;\dfrac{3}{4}\right);\left(1;3\right)\)
16 tháng 2 2021
a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)
b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)
\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)
\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)
\(\Leftrightarrow x\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy: S={0;6}
c) Ta có: \(3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)
d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)
\(\Leftrightarrow30-6x=6x-8\)
\(\Leftrightarrow30-6x-6x+8=0\)
\(\Leftrightarrow-12x+38=0\)
\(\Leftrightarrow-12x=-38\)
\(\Leftrightarrow x=\dfrac{19}{6}\)
Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)
e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow6x+4-3x-1=12x+10\)
\(\Leftrightarrow3x+3-12x-10=0\)
\(\Leftrightarrow-9x-7=0\)
\(\Leftrightarrow-9x=7\)
\(\Leftrightarrow x=-\dfrac{7}{9}\)
Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)
2 tháng 10 2018
\(a,\left(x+2\right)^2-4x\left(x-1\right)=-3x\left(x-6\right)\)
\(\Leftrightarrow x^2+4x+4-4x^2+4x=-3x^2+18x\)
\(\Leftrightarrow x^2-4x^2+3x^2+4x+4x-18x=-4\)
\(\Leftrightarrow-10x=-4\)
\(\Leftrightarrow x=\frac{2}{5}\)
Câu b làm tương tự nhé
Học tốt ##
D
2 tháng 10 2018
\(b,\left(x+3\right)^3-\left(x-2\right)^3=\left(5x-1\right)\left(3x+2\right)\)
\(\Rightarrow x^3+6x^2+9x+3x^2+18x+27-x^3+4x^2-4x+2x^2-8x+8=15x^2+10x-3x-2\)
\(\Rightarrow15x^2+15x+35=15x^2+10x-3x-2\)
\(\Rightarrow15x^2+15x+35=15x^2+7x-2\)
\(\Rightarrow15x+35=7x-2\)
\(\Rightarrow15x-7x=-2-35\)
\(\Rightarrow8x=-37\)
\(\Rightarrow x=\frac{-37}{8}\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 4 2019
a/ \(\left(2n^3-5n^2+1\right):\left(2n-1\right)=n^2-2n-1\)
b/ \(x\ne0;\pm2\)
\(\left(\frac{x^2}{x\left(x^2-4\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\right):\left(\frac{x^2-4+10-x^2}{x+2}\right)\)
\(=\left(\frac{x}{x^2-4}-\frac{2\left(x+2\right)}{x^2-4}+\frac{x-2}{x^2-4}\right):\left(\frac{6}{x+2}\right)\)
\(=\left(\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\right).\left(\frac{x+2}{6}\right)\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{\left(x+2\right)}{6}=-\frac{1}{x-2}=\frac{1}{2-x}\)
c/
\(\left(3x-1\right)^2+2\left(3x-1\right)\left(3x+4\right)+\left(3x+4\right)^2\)
\(=\left(3x-1+3x+4\right)^2\)
\(=\left(6x+3\right)^2\)
|3x - 1| + |1 - 3x| = 6
Ta có: 3x - 1 và 1 - 3x là 2 số đối nhau
=> |3x - 1| = |1 - 3x|
Mà |3x -1| + |1 - -3x| = 6
=> 2 . |3x - 1| = 6
=> |3x - 1| = 3
TH1: 3x - 1 = 3 TH2: 3x - 1 = -3
=> 3x = 3 + 1 => 3x = -3 + 1
=> 3x = 4 => 3x = -2
=> x = 4/3 => x = -2/3
Vậy x thuộc {4/3;-2/3}
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