Tìm x thuộc Q bik
a) \(\left|3x-1\right|< 5\)
b) \(\left|15x-1\right|>31\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
TV
2
14 tháng 6 2017
a) \(\left|15x-1\right|>31\)
\(\Rightarrow-31< 15x-1< 31\)
\(\Rightarrow-31+1< 15x-1+1< 31+1\)
\(\Rightarrow-30< 15x< 32\)
\(\Rightarrow-2< x< \frac{32}{15}\)
b) \(\left|2x-4\right|+4\ge25\)
\(\Rightarrow\left|2x-4\right|+4-4\ge25-4\)
\(\Rightarrow\left|2x-4\right|\ge21\)
\(\Rightarrow\hept{\begin{cases}2x-4\le-21\\2x-4\ge21\end{cases}}\Rightarrow\hept{\begin{cases}2x\le-17\\2x\ge25\end{cases}}\Rightarrow\hept{\begin{cases}x\le-\frac{17}{2}\\x\ge\frac{25}{2}\end{cases}}\)
Vậy \(x\le-\frac{17}{2}\) hoặc \(x\ge\frac{25}{2}\)thì thõa mãn đề bài
DL
14 tháng 6 2017
a) \(\left|15x-1\right|>31\)
\(\Rightarrow\left\{x\in N\right\}\left\{x>2\right\}\)
T
29 tháng 6 2017
a) \(\left(3x-5\right)\left(9x^2+15x+25\right)\)
\(=\left(3x\right)^3-5^3\)
\(=27x^3-125\)
b) \(\left(2x+7\right)\left(x^2-14x+49\right)-2x\left(2x-1\right)\left(2x+1\right)\)
\(=2x^3-28x^2+98x+7x^2-98x+343-2x\left(4x^2-1\right)\)
\(=2x^3-28x^2+7x^2+343-8x^3+2x\)
\(=-6x^3-21x^2+343+2x\)
c) \(\left(4x-7\right)\left(16x^2+28x+49\right)\left(3x+1\right)\left(9x^2-3x+1\right)-9x\left(3x^2-1\right)\)
\(=\left(64x^3-343\right)\left(3x+1\right)\left(9x^2-3x+1\right)-27x^3+9x\)
\(=\left(6x^3-343\right)\left(27x^3+1\right)-27x^3+9x\)
\(=1728x^6+64x^3-9261x^3-343-27x^3+9x\)
\(=1728x^6-9224x^3-343+9x\)
4 tháng 12 2017
a)\(\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow6x=36\Leftrightarrow x=6\)
26 tháng 2 2020
a) Ta có: x(x-1)<0
\(\Leftrightarrow\)x; x-1 khác dấu
*Trường hợp 1:
\(\left\{{}\begin{matrix}x>0\\x-1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\x< 1\end{matrix}\right.\Leftrightarrow0< x< 1\)
*Trường hợp 2:
\(\left\{{}\begin{matrix}x< 0\\x-1>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 0\\x>1\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Vậy: 0<x<1
b) Ta có: (2-x)(3x-12)>0
\(\Leftrightarrow\)2-x; 3x-12 cùng dấu
*Trường hợp 1:
\(\left\{{}\begin{matrix}2-x>0\\3x-12>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>2\\3x>12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>2\\x>4\end{matrix}\right.\Leftrightarrow x>4\)
*Trường hợp 2:
\(\left\{{}\begin{matrix}2-x< 0\\3x-12< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2\\3x< 12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2\\x< 4\end{matrix}\right.\Leftrightarrow x< 2\)
Vậy: 2<x<4
c) Ta có: \(\left(x+1\right)^2\cdot\left(5-2x\right)\le0\)
*Trường hợp 1:
\(\left(x+1\right)^2\cdot\left(5-2x\right)< 0\)
\(\Leftrightarrow\)(x+1)2; 5-2x khác dấu
-Trường hợp 1:
\(\left\{{}\begin{matrix}\left(x+1\right)^2< 0\\5-2x>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1< 0\\2x< 5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 1\\x< \frac{5}{2}\end{matrix}\right.\Leftrightarrow x< 1\)
-Trường hợp 2:
\(\left\{{}\begin{matrix}\left(x+1\right)^2>0\\5-2x< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1>0\\2x>5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>1\\x>\frac{5}{2}\end{matrix}\right.\Leftrightarrow x>\frac{5}{2}\)
Vậy: \(1< x< \frac{5}{2}\)
câu d tương tự nhé bạn
HT
18 tháng 8 2017
a) \(\Leftrightarrow\dfrac{15x}{x^2+3x-4}-1=\dfrac{12}{x+4}+\dfrac{4}{x-1}\)
\(\Leftrightarrow\dfrac{15x}{x^2+4x-x-4}-\dfrac{12}{x+4}-\dfrac{4}{x-1}=1\)
\(\Leftrightarrow\dfrac{15x}{\left(x-1\right)\left(x+4\right)}-\dfrac{12}{x+4}-\dfrac{4}{x-1}=1\)
\(\Leftrightarrow\dfrac{15x-12x+12-4x-16}{\left(x-1\right)\left(x+4\right)}=1\)
\(\Leftrightarrow\dfrac{-1}{x-1}=1\)
\(\Leftrightarrow x-1=-1\)
\(\Rightarrow x=0\)
tick cho t vs hik
HT
18 tháng 8 2017
b) \(\Leftrightarrow\left|x-2\right|+3=5\)
\(\Leftrightarrow\left|x-2\right|=5-3\)
\(\Leftrightarrow\left|x-2\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
Giải:
a) \(\left|3x-1\right|< 5\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1< 5\\1-3x< 5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x< 6\\-3x< 4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x< 2\\x>-\dfrac{4}{3}\end{matrix}\right.\)
Vậy ...
b) \(\left|15x-1\right|>31\)
\(\Leftrightarrow\left[{}\begin{matrix}15x-1>31\\1-15x>31\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}15x>32\\-15x>30\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{32}{15}\\x< -2\end{matrix}\right.\)
Vậy ...
a) | 3x - 1 | < 5
<=>\(\left[{}\begin{matrix}3x-1< 5\\3x-1>-5\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}3x>-4\\3x< 6\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x>-\dfrac{4}{3}\left(loại\right)\\x< 2\left(nhân\right)\end{matrix}\right.\)
vậy x < 2
b) | 15x-1 | > 31
<=>\(\left[{}\begin{matrix}15x-1>31\\15x-1< -31\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}15x>32\\15x< -30\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x>\dfrac{32}{15}\left(nhân\right)\\x< -6\left(loại\right)\end{matrix}\right.\)
vậy x > 32/15