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\(\left|2x-\frac{1}{2}\right|+\frac{3}{7}=5\frac{3}{7}\)
\(\left|2x-\frac{1}{2}\right|=\frac{38}{7}-\frac{3}{7}\)
\(\left|2x-\frac{1}{2}\right|=5\)
Ta xét hai trường hợp:
TH1: \(2x-\frac{1}{2}=5\)
2x = 5 + 1/2
2x = 11/2
x = 11/2 : 2
x = 11/4 (loại vì x < 0)
TH2: 2x - 1/2 = -5
2x = -5 + 1/2
2x = -9/2
x = -9/2:2
x = -9/4 (chọn)
Vậy x = -9/4
\(\left|2x-\frac{1}{2}\right|+\frac{3}{7}=5\frac{3}{7}\)
\(\left|2x-\frac{1}{2}\right|=5\frac{3}{7}-\frac{3}{7}=5\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=5\\2x-\frac{1}{2}=-5\end{cases}}\Rightarrow\orbr{\begin{cases}2x=\frac{11}{2}\\2x=\frac{-9}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=-9\end{cases}}}\)
\(x=-9\)
\(\left|2x-\dfrac{1}{2}\right|+\dfrac{3}{7}=5\dfrac{3}{7}\)
<=> \(\left|2x-\dfrac{1}{2}\right|=5\)
<=> \(\left\{{}\begin{matrix}2x-\dfrac{1}{2}=5\\2x-\dfrac{1}{2}=-5\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}2x=\dfrac{11}{2}\\2x=-\dfrac{9}{2}\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=\dfrac{11}{4}\\x=-\dfrac{9}{4}\end{matrix}\right.\)
Do x < 0 => x = \(-\dfrac{9}{4}\)
Giải:
\(\left|2x-\dfrac{1}{2}\right|+\dfrac{3}{7}=5\dfrac{3}{7}\)
\(\Leftrightarrow\left|2x-\dfrac{1}{2}\right|=5\dfrac{3}{7}-\dfrac{3}{7}\)
\(\Leftrightarrow\left|2x-\dfrac{1}{2}\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{2}=4\\2x-\dfrac{1}{2}=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{9}{2}\\2x=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=-\dfrac{7}{4}\end{matrix}\right.\)
Mà x < 0
Nên \(x=-\dfrac{7}{4}\)
Vậy \(x=-\dfrac{7}{4}\).
Chúc bạn học tốt!
a) \(\dfrac{3}{4}x-1>\dfrac{1}{2}x+5\)
\(\Rightarrow\dfrac{3}{4}x-\dfrac{1}{2}x>1+5\)
\(\Rightarrow\dfrac{1}{4}x>6\)
\(\Rightarrow x>24\)
b) Đơn giản.
c) \(\left(x+1\right)\left(x-2\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}x+1>0;x-2>0\\x+1< 0;x-2< 0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>-1;x>2\\x< -1;x< 2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>2\\x< -1\end{matrix}\right.\)
Vậy ....
a: =>5x=3x-6
=>2x=-6
hay x=-3
b: \(\Leftrightarrow\left(x-3\right)^2=4\cdot5^2=100\)
=>x-3=10 hoặc x-3=-10
=>x=13 hoặc x=-7
c: \(\left|x^3+1\right|+2\ge2\forall x\)
Dấu '=' xảy ra khi x=-1
b) \(\dfrac{x-12}{4}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{x-12}{4}=\dfrac{2}{4}\)
\(\Rightarrow x-12=2\)
\(\Leftrightarrow x=2+12\)
\(\Leftrightarrow x=14\)
c) \(1< \dfrac{x}{3}< 2\)
\(\Rightarrow\dfrac{3}{3}< \dfrac{x}{3}< \dfrac{6}{3}\)
\(\Rightarrow x\in\left\{4;5\right\}\)
Vậy \(x\in\left\{4;5\right\}\)
1. a, \(\dfrac{x}{7}=\dfrac{9}{y}\Leftrightarrow xy=9.7\)
<=> xy = 63
=> x; y \(\inƯ\left(63\right)\)
Lại có x > y nên ta có bảng :
| x | 63 | -1 | 21 | -3 | 9 | -7 |
| y | 1 | -63 | 3 | -21 | 7 | -9 |
@Đặng Hoài An
1. b, \(\dfrac{-2}{x}=\dfrac{y}{5}\Leftrightarrow-2.5=xy\)
<=> -10 = xy
=> x; y \(\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
Lại có : x < 0 < y
=> x = -1; -2; -5; -10
Tương ứng y = 10; 5; 2; 1
@Đặng Hoài An