Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(-\dfrac{1}{4}-\dfrac{3}{4}:x=-\dfrac{11}{36}\)
\(\Rightarrow\dfrac{3}{4}:x=-\dfrac{1}{4}-\left(-\dfrac{11}{36}\right)=\dfrac{1}{18}\)
\(\Rightarrow x=\dfrac{3}{4}:\dfrac{1}{18}=\dfrac{27}{2}\)
b, \(70:\dfrac{4x+720}{x}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{4x+720}{x}=140\)
\(\Rightarrow4x+720=140x\Rightarrow140x-4x=720\)
\(\Rightarrow136x=720\Rightarrow x=\dfrac{90}{17}\)
Chúc bạn học tốt!!!
a)\(\dfrac{-1}{4}-\dfrac{3}{4}:x=\dfrac{-11}{36}\)
\(\dfrac{3}{4}:x=\dfrac{-1}{4}-\left(\dfrac{-11}{36}\right)=\dfrac{1}{18}\)
\(\Rightarrow x=\dfrac{3}{4}:\dfrac{1}{18}=\dfrac{27}{2}\)
b)\(70:\dfrac{4x+720}{x}=\dfrac{1}{2}\)
\(\dfrac{4x+720}{x}=70:\dfrac{1}{2}=140\)
\(\Rightarrow4x+720=140x\)
\(\Rightarrow140x-4x=720\)
\(\Rightarrow136x=720\)
\(\Rightarrow x=\dfrac{90}{17}\)
mấy cái này đơn dãng vô cùng nhưng có đều bn ra đề dài quá nha
a) \(3x+4\ge7\Leftrightarrow3x\ge7-4\Leftrightarrow3x\ge3\Leftrightarrow x\ge1\) vậy \(x\ge1\)
b) \(-5x+1< 11\Leftrightarrow-5x< 11-1\Leftrightarrow-5x< 10\Leftrightarrow x>\dfrac{10}{-5}\)
\(\Leftrightarrow x>-2\) vậy \(x>-2\)
c) \(\dfrac{5}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow x< 3\) vậy \(x< 3\)
d) \(\dfrac{-7}{2-x}\ge0\Leftrightarrow2-x\le0\Leftrightarrow x\ge2\) vậy \(x\ge2\)
e) \(x^2+4x>0\Leftrightarrow x\left(x+4\right)>0\) \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>0\\x+4>0\end{matrix}\right.\\\left[{}\begin{matrix}x< 0\\x+4< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>0\\x>-4\end{matrix}\right.\\\left[{}\begin{matrix}x< 0\\x< -4\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x>0\\x< -4\end{matrix}\right.\) vậy \(x>0\) hoặc \(x< -4\)
f) \(\dfrac{x-2}{x-6}< 0\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-2>0\\x-6>0\end{matrix}\right.\\\left[{}\begin{matrix}x-2< 0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>2\\x>6\end{matrix}\right.\\\left[{}\begin{matrix}x< 2\\x< 6\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>6\\x< 2\end{matrix}\right.\)
vậy \(x>6\) hoặc \(x< 2\)
g) \(\left(x-1\right)\left(x+2\right)\left(3-x\right)< 0\Leftrightarrow-\left[\left(x-1\right)\left(x+2\right)\left(x-3\right)\right]< 0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x-3\right)>0\)
th1: 3 số hạng đều dương : \(\Leftrightarrow\left[{}\begin{matrix}x-1>0\\x+2>0\\x-3>0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>1\\x>-2\\x>3\end{matrix}\right.\) \(\Rightarrow x>3\)
th2: 2 âm 1 dương : (vì trong 3 số hạng ta có : \(\left(x+2\right)\) lớn nhất \(\Rightarrow\left(x+2\right)\) dương)
\(\Leftrightarrow\left[{}\begin{matrix}x-1< 0\\x+2>0\\x-3< 0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x< 1\\x>-2\\x< 3\end{matrix}\right.\) \(\Rightarrow-2< x< 1\)
vậy \(x>3\) hoặc \(-2< x< 1\)
h) \(\dfrac{x^2-1}{x}>0\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x^2-1>0\\x>0\end{matrix}\right.\\\left[{}\begin{matrix}x^2-1< 0\\x< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x^2>1\\x>0\end{matrix}\right.\\\left[{}\begin{matrix}x^2< 1\\x< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\\x>0\end{matrix}\right.\\\left[{}\begin{matrix}-1< x< 1\\x< 0\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>1\\-1< x< 0\end{matrix}\right.\) vậy \(x>1\) hoặc \(-1< x< 0\)
i) \(x^2+x-2< 0\Leftrightarrow x^2+x+\dfrac{1}{4}-\dfrac{9}{4}< 0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2-\dfrac{9}{4}< 0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2< \dfrac{9}{4}\Leftrightarrow\dfrac{-3}{2}< \left(x+\dfrac{1}{2}\right)< \dfrac{3}{2}\Leftrightarrow-2< x< 1\)
vậy \(-2< x< 1\)
Mysterious Person, Đoàn Đức Hiếu, Nguyễn Đình Dũng , ... giúp mình!
a, \(\left(x-3\right)\left(2x+5\right)>0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3>0\\2x+5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\\2x+5< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>3\\x>-\dfrac{5}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 3\\x< -\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -\dfrac{5}{2}\end{matrix}\right.\)
b,\(\left(1-4x\right)\left(x-2\right)< 0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}1-4x>0\\x-2< 0\end{matrix}\right.\\\left\{{}\begin{matrix}1-4x< 0\\x-2>0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{1}{4}\\x< 2\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{1}{4}\\x>2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< 2\\x>2\end{matrix}\right.\)
c, \(\dfrac{-3}{x+2}< 0\Leftrightarrow x+2>0\Leftrightarrow x>-2\)
a: 5x+2>3x-1
=>5x-3x>-1-2
=>2x>-3
hay x>-3/2
b: \(\dfrac{3}{4}x-\dfrac{1}{2}>\dfrac{1}{2}x+\dfrac{3}{4}\)
=>3/4x-1/2x>3/4+1/2
=>1/2x>5/4
hay x>5/4:1/2=5/2
c: (x-2)(x-3)>0
=>x-3>0 hoặc x-2<0
=>x>3 hoặc x<2
d: (2x+4)(x-5)<0
=>(x+2)(x-5)<0
=>-2<x<5
a/ \(\left(x+1\right)\left(x-2\right)< 0\)
TH1:\(\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\) (vô lý)
TH2:\(\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\)\(\Rightarrow-1< x< 2\)
Vậy.........
b/ \(\left(x-3\right)\left(x-4\right)>0\)
TH1:\(\left\{{}\begin{matrix}x-3>0\\x-4>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>3\\x>4\end{matrix}\right.\)\(\Rightarrow x>4\)
TH2:\(\left\{{}\begin{matrix}x-3< 0\\x-4< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< 3\\x< 4\end{matrix}\right.\)\(\Rightarrow x< 3\)
Vậy...............
c/ \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{7}{12}< x< \dfrac{1}{48}-\dfrac{1}{8}\)
\(\Rightarrow\dfrac{-1}{12}< x< -\dfrac{5}{48}\)
Vậy...............
Để ( x + 1 ) ( x - 2 ) < 0
=> x + 1 và x - 2 phải khác dấu mà x + 1 > x + 2
=> x + 1 dương x + 2 âm
Tức là x + 1 > 0 => x > - 1 và x - 2 < 0 => x < 2
a) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\)
\(20x^2-16x-34=10x^2+3x-34\)
\(10x^2-19x=0\)
\(x\left(10x-19\right)=0\)
\(\Leftrightarrow x=0\)
hoặc \(10x-19=0\)
\(\Leftrightarrow x=\dfrac{19}{10}\)
Vạy ..............
b) \(\dfrac{x-1}{x+5}=\dfrac{6}{7}\)
\(\Leftrightarrow1-\dfrac{x-1}{x+5}=1-\dfrac{6}{7}\)
\(\Leftrightarrow\dfrac{x+5}{x+5}-\dfrac{x-1}{x+5}=\dfrac{7}{7}-\dfrac{6}{7}\)
\(\Leftrightarrow\dfrac{\left(x+5\right)-\left(x-1\right)}{x+5}=\dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{x+5-x+1}{x+5}=\dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{\left(x-x\right)+\left(5+1\right)}{x+5}=\dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{6}{x+5}=\dfrac{1}{7}\)
\(\Leftrightarrow x+5=42\)
\(\Leftrightarrow x=37\)
a) Ta có:
(x - 1)5 = - 243
=> (x - 1)5 = (-3)5
=> x - 1 = - 3
=> x = -3 + 1
=> x = -2
Vậy x = -2
b) Ta có:
\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Rightarrow\left(x+2\right).\dfrac{1}{11}+\left(x+2\right).\dfrac{1}{12}+\left(x+2\right).\dfrac{1}{13}=\left(x+2\right).\dfrac{1}{14}+\left(x+2\right).\dfrac{1}{15}\)
=> \(\left(x+2\right).\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}\right)=\left(x+2\right).\left(\dfrac{1}{14}+\dfrac{1}{15}\right)\)
=> \(\left(x+2\right).\dfrac{431}{1716}=\left(x+2\right).\dfrac{29}{210}\)
=> \(\left(x+2\right).\dfrac{431}{1716}-\left(x+2\right).\dfrac{29}{210}=0\)
=> (x + 2).(\(\dfrac{431}{1716}-\dfrac{29}{210}\)) = 0
mà \(\dfrac{431}{1716}-\dfrac{29}{210}\) \(\ne\) 0
=> x + 2 = 0
=> x = -2
Vậy x = -2
c) Ta có :
\(\left|3x-2\right|+5x=4x-10\)
=> \(\left|3x-2\right|=4x-5x-10\)
=> \(\left|3x-2\right|=-x-10\)
=> 3x - 2 = -x - 10
hoặc 3x - 2 = -(-x -10)
*) Nếu 3x - 2 = -x - 10
=> 3x + x = -10 + 2
=> 4x = -8
=> x = -2
*) Nếu 3x - 2 = -(-x -10)
=> 3x - 2 = x +10
=> 3x - x = 10 + 2
=> 2x = 12
=> x = 6
Vậy x = -2 hoặc x = 6
a: \(\Leftrightarrow\dfrac{x+1}{2x+1}=\dfrac{x+4}{2x+6}\)
=>(x+1)(2x+6)=(2x+1)(x+4)
\(\Leftrightarrow2x^2+6x+2x+6=2x^2+8x+x+4\)
=>9x+4=8x+6
=>x=2
b: \(x^2+5x=0\)
=>x(x+5)=0
=>x=0 hoặc x=-5
Giải:
a) \(70.\dfrac{4x+720}{x}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{280x+50400}{x}=\dfrac{1}{2}\)
\(\Leftrightarrow2\left(280x+50400\right)=x\)
\(\Leftrightarrow560x+100800=x\)
\(\Leftrightarrow560x-x=-100800\)
\(\Leftrightarrow549x=-100800\)
\(\Leftrightarrow x=-\dfrac{11200}{61}\)
Vậy ...
b) \(x^2+5x< 0\)
\(\Leftrightarrow x\left(x+5\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x+5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x+5< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x>-5\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x< -5\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0>x>-5\\x\in\varnothing\end{matrix}\right.\)
Vậy ...