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1:
a: =7/5(40+1/4-25-1/4)-1/2021
=21-1/2021=42440/2021
b: =5/9*9-1*16/25=5-16/25=109/25
3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
\(\text{a) }\left(x-1\right)\left(x-5\right)>0\\ \text{ Để }\left(x-1\right)\left(x-5\right)>0\text{ thì }\Rightarrow x-1\text{ và }x-5\text{ cùng dấu }\\ \text{+) Xét }x-1\text{ và }x-5\text{ là số nguyên dương }\Rightarrow\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x-5>0\Rightarrow x>5\end{matrix}\right.\Rightarrow x>5\\ \text{+) Xét }x-1\text{ và }x-5\text{ là số nguyên âm }\Rightarrow\left\{{}\begin{matrix}x-1< 0\Rightarrow x< 1\\x-5< 0\Rightarrow x< 5\end{matrix}\right.\Rightarrow x< 1\\ \text{Vậy }\left(x-1\right)\left(x-5\right)>0\text{ khi }x< 1\text{ hoặc }x>5\)
\(\text{b) }\left(x-1\right)\left(x-5\right)< 0\\ \text{ Để }\left(x-1\right)\left(x-5\right)< 0\text{ thì }\Rightarrow x-1\text{ và }x-5\text{ trái dấu }\\ \text{ Mà }x-1>x-5\\ \Rightarrow\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x-5< 0\Rightarrow x< 5\end{matrix}\right.\Rightarrow1< x< 5\\ \text{ Vậy }\left(x-1\right)\left(x-5\right)< 0\text{ khi }1< x< 5\)
\(\text{c) }\dfrac{3}{4}-\dfrac{1}{4}\left|x-\dfrac{1}{7}\right|=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{1}{4}\left|x-\dfrac{1}{7}\right|=\dfrac{1}{2}\\ \Leftrightarrow\left|x-\dfrac{1}{7}\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{7}=-2\\x-\dfrac{1}{7}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{13}{7}\\x=\dfrac{15}{7}\end{matrix}\right.\\ \text{Vậy }x=-\dfrac{13}{7}\text{ hoặc }x=\dfrac{15}{7}\)
\(\text{d) }\left(x-\dfrac{1}{2}\right)^2=\dfrac{1}{16}\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=-\dfrac{1}{4}\\x-\dfrac{1}{2}=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{3}{4}\end{matrix}\right.\\ \text{Vậy }x=\dfrac{1}{4}\text{ hoặc }x=\dfrac{3}{4}\)
\(\text{e) }8\left(x+1\right)-2\left(2x+5\right)=0\\ \Leftrightarrow8x+8-4x+10=0\\ \Leftrightarrow\left(8x-4x\right)+\left(8+10\right)=0\\ \Leftrightarrow4x+18=0\\ \Leftrightarrow4x=-18\\ \Leftrightarrow x=-\dfrac{9}{2}\\ \text{Vậy }x=-\dfrac{9}{2}\)
\(\text{g) }\left(6x-1\right)-\left(x+8\right)=0\\ \Leftrightarrow6x-1-x-8=0\\ \Leftrightarrow\left(6x-x\right)-\left(1+8\right)=0\\ \Leftrightarrow5x-9=0\\ \Leftrightarrow5x=9\\ \Leftrightarrow x=\dfrac{9}{5}\\ \text{Vậy }x=\dfrac{9}{5}\)
\(\text{h) }\left|7x-\dfrac{1}{4}\right|=1\\ \Leftrightarrow\left[{}\begin{matrix}7x-\dfrac{1}{4}=-1\\7x-\dfrac{1}{4}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=-\dfrac{3}{4}\\7x=\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{28}\\x=\dfrac{5}{28}\end{matrix}\right.\\ \text{Vậy }x=-\dfrac{3}{28}\text{ hoặc }x=\dfrac{5}{28}\)
\(\text{q) }-2x-3=-x+7\\ \Leftrightarrow-2x-3-\left(-x+7\right)=0\\ \Leftrightarrow-2x-3+x-7=0\\ \Leftrightarrow\left(-2x+x\right)-\left(3+7\right)=0\\ \Leftrightarrow-x-10=0\\ \Leftrightarrow-x=10\\ \Leftrightarrow x=-10\\ \text{ Vậy }x=-10\)
a/dễ --> tự lm
b/ \(\left(x-\dfrac{4}{7}\right)\left(1\dfrac{3}{5}+2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=0\\1\dfrac{3}{5}+2x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\2x=\dfrac{8}{5}\Rightarrow x=\dfrac{4}{5}\end{matrix}\right.\)
Vậy...............
c/ \(\left(x-\dfrac{4}{7}\right):\left(x+\dfrac{1}{2}\right)>0\)
TH1: \(\left\{{}\begin{matrix}x-\dfrac{4}{7}>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{4}{7}\\x>-\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x>\dfrac{4}{7}\)
TH2: \(\left\{{}\begin{matrix}x-\dfrac{4}{7}< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< \dfrac{4}{7}\\x< -\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x< -\dfrac{1}{2}\)
Vậy \(x>\dfrac{4}{7}\) hoặc \(x< -\dfrac{1}{2}\) thì thỏa mãn đề
d/ \(\left(2x-3\right):\left(x+1\dfrac{3}{4}\right)< 0\)
TH1: \(\left\{{}\begin{matrix}2x-3>0\\x+1\dfrac{3}{4}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>1,5\\x< -\dfrac{7}{4}\end{matrix}\right.\)(vô lý)
TH2: \(\left\{{}\begin{matrix}2x-3< 0\\x+1\dfrac{3}{4}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< 1,5\\x>-\dfrac{7}{4}\end{matrix}\right.\)\(\Rightarrow-\dfrac{7}{4}< x< 1,5\)
Vậy...................
\(\left(x-1\right)\left(x+5\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x+5>0\Rightarrow x>-5\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\Rightarrow x< 1\\x+5< 0\Rightarrow x< -5\end{matrix}\right.\end{matrix}\right.\)
\(\left(x-1\right)\left(x+5\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x+5< 0\Rightarrow x< -5\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\Rightarrow x< 1\\x+5>0\Rightarrow x>-5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-5< x< 1\)
câu dễ tự làm
\(\Rightarrow x>-5;x< -5\)
Bài 1:
a.
$|x+\frac{7}{4}|=\frac{1}{2}$
\(\Leftrightarrow \left[\begin{matrix} x+\frac{7}{4}=\frac{1}{2}\\ x+\frac{7}{4}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-5}{4}\\ x=\frac{-9}{4}\end{matrix}\right.\)
b. $|2x+1|-\frac{2}{5}=\frac{1}{3}$
$|2x+1|=\frac{1}{3}+\frac{2}{5}$
$|2x+1|=\frac{11}{15}$
\(\Leftrightarrow \left[\begin{matrix} 2x+1=\frac{11}{15}\\ 2x+1=\frac{-11}{15}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-2}{15}\\ x=\frac{-13}{15}\end{matrix}\right.\)
c.
$3x(x+\frac{2}{3})=0$
\(\Leftrightarrow \left[\begin{matrix} 3x=0\\ x+\frac{2}{3}=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=\frac{-3}{2}\end{matrix}\right.\)
d.
$x+\frac{1}{3}=\frac{2}{5}-(\frac{-1}{3})=\frac{2}{5}+\frac{1}{3}$
$\Leftrightarrow x=\frac{2}{5}$
Nguyễn Quý Trung:
\(x+\dfrac{1}{3}=\dfrac{2}{5}+\dfrac{1}{3}\)
Bạn bớt 2 vế đi 1/3 thì \(x=\dfrac{2}{5}\)
a, \(\left|2x-3\right|-\dfrac{1}{3}=0\Leftrightarrow\left|2x-3\right|=\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=\dfrac{1}{3}\\2x-3=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)
b, tương tự
c, \(\left|2x-1\right|-\left|x+\dfrac{1}{3}\right|=0\Leftrightarrow\left|2x-1\right|=\left|x+\dfrac{1}{3}\right|\)
TH1 : \(2x-1=x+\dfrac{1}{3}\Leftrightarrow x=\dfrac{4}{3}\)
TH2 : \(2x-1=-x-\dfrac{1}{3}\Leftrightarrow3x=\dfrac{2}{3}\Leftrightarrow x=\dfrac{2}{9}\)
d, \(3x-\left|x+15\right|=\dfrac{5}{4}\Leftrightarrow\left|x+15\right|=3x-\dfrac{5}{4}\)ĐK : x >= 5/12
TH1 : \(x+15=3x-\dfrac{5}{4}\Leftrightarrow-2x=-\dfrac{65}{4}\Leftrightarrow x=\dfrac{65}{8}\)( tm )
TH2 : \(x+15=\dfrac{5}{3}-3x\Leftrightarrow4x=-\dfrac{40}{3}\Leftrightarrow x=-\dfrac{10}{3}\)
a)
\(\left(3x+\dfrac{1}{3}\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+\dfrac{1}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\\x=\dfrac{1}{2}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{3}{2}\right)\left(2x+1\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{3}{2}>0\\2x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{3}{2}< 0\\2x+1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\)
mình làm lại câu b) nha
b) |x-3|=-4
th1: x-3=-4
x=3+(-4)
x=-1
th2: x-3=4
x=3+4
x=7
b) \(\left|x-3\right|=-4\)
t/h1:\(x-3=-4\)
\(x=3-\left(-4\right)\)
\(x=7\)
t/h2:\(x-3=4\)
\(x=3-4\)
\(x=-1\)
\(x\left(x-\dfrac{1}{7}\right)\left(x+\dfrac{1}{9}\right)< 0\)
Vậy phải có 1 số lẻ các số âm
Vậy \(\left\{{}\begin{matrix}x< 0\\x-\dfrac{1}{7}< 0\Rightarrow x< \dfrac{1}{7}\\x+\dfrac{1}{9}< 0\Rightarrow x< -\dfrac{1}{9}\end{matrix}\right.\)
Vậy \(x< \dfrac{1}{7}\)
Vì \(x-\dfrac{1}{7}< x< \dfrac{1}{9}+x\) nên
\(\left\{{}\begin{matrix}x-\dfrac{1}{7}< 0\Rightarrow x< \dfrac{1}{7}\\x>0\\\dfrac{1}{9}+x>0\Rightarrow x>-\dfrac{1}{9}\end{matrix}\right.\)
Vậy \(-\dfrac{1}{9}< x< \dfrac{1}{7}\)
\(\dfrac{4-x}{2x-\dfrac{1}{5}}>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4-x>0\Rightarrow x< 4\\2x-\dfrac{1}{5}>0\Rightarrow x>\dfrac{1}{10}\end{matrix}\right.\\\left\{{}\begin{matrix}4-x< 0\Rightarrow x>4\\2x-\dfrac{1}{5}< 0\Rightarrow x< \dfrac{1}{10}\end{matrix}\right.\end{matrix}\right.\)
Vậy\(\dfrac{1}{10}< x< 4\)