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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/ \(\left(x+1\right)\left(x-2\right)< 0\)
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, \(\left(x-3\right)\left(2x+5\right)>0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3>0\\2x+5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\\2x+5< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>3\\x>-\dfrac{5}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 3\\x< -\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -\dfrac{5}{2}\end{matrix}\right.\)
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, \(\dfrac{-3}{x+2}< 0\Leftrightarrow x+2>0\Leftrightarrow x>-2\)
1, a/ \(\left|x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy .............
b/ \(\left|x\right|=3,12\Leftrightarrow\left[{}\begin{matrix}x=3,12\\x=-3,12\end{matrix}\right.\)
Vậy ...........
c/ \(\left|x\right|=0\Leftrightarrow x=0\)
Vậy ..........
d/ \(\left|x\right|=2\dfrac{1}{7}\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\dfrac{1}{7}\\x=-2\dfrac{1}{7}\end{matrix}\right.\)
Vậy ..............
2, a/ \(\left|x\right|=2,1\Leftrightarrow\left[{}\begin{matrix}x=2,1\\x=-2,1\end{matrix}\right.\)
Vậy ...........
b/ \(\left|x\right|=\dfrac{17}{9}\) ; \(x< 0\)
\(\Leftrightarrow x=-\dfrac{17}{9}\)
Vậy ..........
c/ \(\left|x\right|=1\dfrac{2}{5}\Leftrightarrow\left[{}\begin{matrix}x=1\dfrac{2}{5}\\x=-1\dfrac{2}{5}\end{matrix}\right.\)
Vậy ...........
d/ \(\left|x\right|=0,35\) ; \(x>0\Leftrightarrow x=0,35\)
3, a/ \(\left|x-1,7\right|=2,3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1,7=2,3\\x-1,7=-2,3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-0,6\end{matrix}\right.\)
Vậy ...........
b/ \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{3}=0\)
\(\Leftrightarrow\left|x+\dfrac{3}{4}\right|=\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{3}\\x+\dfrac{3}{4}=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{12}\\x=-\dfrac{13}{12}\end{matrix}\right.\)
Vậy ...........
a: \(\left(2x+3\right)\left(3x-5\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5\ge0\\2x+3\le0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>=\dfrac{5}{3}\\x< =-\dfrac{3}{2}\end{matrix}\right.\)
b: \(\dfrac{x}{3-x}>-1\)
\(\Leftrightarrow\dfrac{x}{3-x}+1>0\)
\(\Leftrightarrow\dfrac{x+3-x}{3-x}>0\)
=>3-x>0
hay x<3
c: \(\dfrac{x-1}{x+5}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x-1}{x+5}-\dfrac{3}{2}\ge0\)
\(\Leftrightarrow\dfrac{2x-2-3x-15}{2\left(x+5\right)}>=0\)
\(\Leftrightarrow\dfrac{x+17}{2\left(x+5\right)}< =0\)
=>-17<=x<-5
d: \(\dfrac{7}{4x^2-1}\ge0\)
=>4x2-1>0
=>(2x-1)(2x+1)>0
=>x>1/2 hoặc x<-1/2
b: \(ab\cdot bc\cdot ac=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow\left(abc\right)^2=\dfrac{1}{4}\)
Trường hợp 1: abc=1/2
\(\Leftrightarrow\left\{{}\begin{matrix}c=\dfrac{1}{2}:\dfrac{1}{2}=1\\a=\dfrac{1}{2}:\dfrac{2}{3}=\dfrac{3}{4}\\b=\dfrac{1}{2}:\dfrac{3}{4}=\dfrac{1}{2}\cdot\dfrac{4}{3}=\dfrac{2}{3}\end{matrix}\right.\)
Trường hợp 2: abc=-1/2
\(\Leftrightarrow\left\{{}\begin{matrix}c=-1\\a=-\dfrac{3}{4}\\b=-\dfrac{2}{3}\end{matrix}\right.\)
c: Theo đề, ta có: \(\left\{{}\begin{matrix}\dfrac{x-1}{2}=\dfrac{y-2}{1}\\\dfrac{y-2}{3}=\dfrac{z-3}{4}\end{matrix}\right.\Leftrightarrow\dfrac{x-1}{6}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-1}{6}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{2\cdot6-3\cdot6+3\cdot4}=\dfrac{45}{6}=\dfrac{15}{2}\)
Do đó: x-1=45; y-2=45/2; z-3=30
=>x=46; y=49/2; z=33
Bài 1:
a) \(x^2+5x+6=x^2+2x+3x+6=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
b) \(2x^2+5x+3=2x^2+2x+3x+3=2x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(2x+3\right)\)
c) \(x^2-10x+16=x^2-2x-8x+16=x\left(x-2\right)-8\left(x-2\right)=\left(x-2\right)\left(x-8\right)\)
d) \(4x^2+9x+5=4x^2+4x+5x+5=4x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(4x+5\right)\)
Bài 2:
không rõ đề --> k lm
2, \(\Rightarrow\left\{{}\begin{matrix}\\\dfrac{5}{4}x-\dfrac{7}{2}=0\\\dfrac{5}{8}x+\dfrac{3}{5}=0\\\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{5}\\\\x=\dfrac{-24}{25}\\\end{matrix}\right.\)
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 ...