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Bài 1:
a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=\dfrac{8}{5}\)
=>x+4/15=8/5 hoặc x+4/15=-8/5
=>x=4/3 hoặc x=-28/15
b: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=-\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{6}:\dfrac{5}{3}=\dfrac{-3}{30}=\dfrac{-1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)
c: \(\Leftrightarrow\left|x-1\right|-1=1\)
=>|x-1|=2
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
Bài 2:
b: \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)
Bài 3:
a: \(A=\left|x+\dfrac{15}{19}\right|-1>=-1\)
Dấu '=' xảy ra khi x=-15/19
b: \(\left|x-\dfrac{4}{7}\right|+\dfrac{1}{2}>=\dfrac{1}{2}\)
Dấu '=' xảy ra khi x=4/7
Bài 1:
\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right):\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)
\(=\left(-\dfrac{9}{5}-\dfrac{12}{5}-\dfrac{7}{3}\right):\left(\dfrac{9}{20}-\dfrac{5}{12}+-3\right)\)
\(=\left(-\dfrac{27}{15}-\dfrac{36}{15}-\dfrac{21}{15}\right):\left(\dfrac{27}{60}-\dfrac{25}{60}+-3\right)\)
\(=\left(-\dfrac{28}{5}\right):\left(-\dfrac{89}{30}\right)\)
\(=\left(-\dfrac{28}{5}\right).\left(-\dfrac{30}{89}\right)\)
\(=\dfrac{168}{89}\)
K chép lại đề, lm luôn nhé:
*\(\Rightarrow\) \(\left(\dfrac{7}{2}+2x\right)\cdot\dfrac{8}{3}=\dfrac{16}{3}\)
\(\Rightarrow\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)
\(\Rightarrow2x=2-\dfrac{7}{2}=-\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{3}{4}\)
* \(\Rightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{\dfrac{3}{4}-2}{2}=-\dfrac{5}{8}\)
=> K có gt x nào t/m đề
* Đề sai
* \(\Rightarrow\left[{}\begin{matrix}3x-1=0\\-\dfrac{1}{2}x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)
*\(\Rightarrow\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=-\dfrac{21}{4}\)
\(\Rightarrow2x-1=\dfrac{1}{3}:\left(-\dfrac{21}{4}\right)=-\dfrac{4}{63}\)
\(\Rightarrow2x=-\dfrac{4}{63}+1=\dfrac{59}{63}\)
\(\Rightarrow x=\dfrac{59}{63}:2=\dfrac{59}{126}\)
* \(\Rightarrow\left(2x+\dfrac{3}{5}\right)^2=\dfrac{9}{25}\)
\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{3}{5}=\dfrac{3}{5}\\2x+\dfrac{3}{5}=-\dfrac{3}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=0\Rightarrow x=0\\2x=-\dfrac{6}{5}\Rightarrow x=-\dfrac{3}{5}\end{matrix}\right.\)
* \(\Rightarrow-5x-1-\dfrac{1}{2}x+\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(\Rightarrow-5x-\dfrac{1}{2}x-\dfrac{3}{2}x=-\dfrac{5}{6}+1-\dfrac{1}{3}\)
\(\Rightarrow-7x=-\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{1}{6}:7=-\dfrac{1}{42}\)
a)\(\left(3\dfrac{1}{2}+2x\right).2\dfrac{2}{3}=5\dfrac{1}{3}\)
\(\left(\dfrac{7}{2}+2x\right).\dfrac{8}{3}=\dfrac{16}{3}\)
\(\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)
\(2x=2-\dfrac{7}{2}=\dfrac{-3}{2}\Rightarrow x=\dfrac{-3}{4}\)
b)\(\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)
\(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2=\dfrac{-1}{4}\)
\(\Rightarrow\left|2x-3\right|=\dfrac{-1}{8}\)
\(\Rightarrow x\in\varnothing\)
c) Đề sai,bạn có viết chữ x đâu,đó là phép tính mà.
d)\(\left(3x-1\right)\left(\dfrac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{-1}{2}x+5=0\Rightarrow x=10\)
e)\(\dfrac{1}{4}+\dfrac{1}{3}:\left(2x-1\right)=-5\)
\(\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=\dfrac{-21}{4}\)
\(2x-1=\dfrac{1}{3}:\dfrac{-21}{4}=\dfrac{-4}{63}\)
\(\Rightarrow2x=\dfrac{59}{63}\Rightarrow x=\dfrac{59}{126}\)
g)\(\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)
\(\left(2x+\dfrac{3}{5}\right)^2=0+\dfrac{9}{25}=\dfrac{9}{25}\)
\(\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2=\left(\dfrac{-3}{5}\right)^2\)
\(th1:x=0\)
\(th2:x=\dfrac{-3}{5}\)
h)\(-5\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(-5x+-1-\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(\Leftrightarrow-5x+-1+\dfrac{5}{6}-\dfrac{1}{3}=2x\)
\(-5x+\dfrac{-1}{2}=2x\)
\(\dfrac{-1}{2}=2x+5x\)
\(\dfrac{-1}{2}=7x\Rightarrow x=\dfrac{-1}{14}\)
a: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
b: 
c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)
<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)
<=>\(4x-17=0\)
<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)
\(B=\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)\left(1+\dfrac{1}{24}\right).....\left(1+\dfrac{1}{440}\right)\left(1+\dfrac{1}{483}\right)\)
\(B=\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}.....\dfrac{441}{440}.\dfrac{484}{483}\)
\(B=\dfrac{9.16.25.....441.484}{8.15.24.....440.483}\)
\(B=\dfrac{3.3.4.4.5.5.....21.21.22.22}{2.4.3.5.4.6.....20.22.21.23}\)
\(B=\dfrac{3.4.5.....21.22}{2.3.4.....20.21}.\dfrac{3.4.5.....21.22}{4.5.6.....22.23}\)
\(B=11.\dfrac{3}{23}=\dfrac{33}{23}\)
B = \(\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}...\dfrac{121}{120}.\dfrac{144}{143}\)
B = \(\dfrac{4.9.16.25...121.144}{3.8.15.24....120.143}\)
B = \(\dfrac{2.2.3.3.4.4.5.5...11.11.12.12}{1.3.2.4.3.5.4.6...10.12.11.13}\)
B = \(\dfrac{2.3.4.5...11.12}{1.2.3.4.5...10.11}.\dfrac{2.3.4.5...11.12}{3.4.5.6.7...12.13}\)
B = 12 . \(\dfrac{2}{13}\)
B = \(\dfrac{24}{13}\)
Bài 3: A=2018-|x+2019|. Vì |x+2019|\(\ge\)0 nên -|x+2019|\(\le\)0=>2018-|x+2019|\(\le\) 2. Vậy A có GTLN = 2 khi x+2019=0 hay x=-2019. B=-10-\(\left|2x-\dfrac{1}{1009}\right|\). Vì \(\left|2x-\dfrac{1}{1009}\right|\ge0\Rightarrow-\left|2x-\dfrac{1}{1009}\right|\le0\Rightarrow-10-\left|2x-\dfrac{1}{1009}\right|\le-10\). Vậy B có GTLN = -10 khi 2x-\(\dfrac{1}{1009}=0\) => \(2x=\dfrac{1}{1009}\Rightarrow x=\dfrac{1}{1009}:2=\dfrac{1}{2018}\)
Bài 2: A=\(\left|5x+1\right|-\dfrac{3}{8}\). Vì \(\left|5x+1\right|\ge0\Rightarrow\left|5x+1\right|-\dfrac{3}{8}\ge\dfrac{-3}{8}\). Vậy A có GTNN = \(\dfrac{-3}{8}\) khi 5x+1= 0=> 5x= -1=> x = \(\dfrac{-1}{5}\). B=\(\left|2-\dfrac{1}{6}x\right|+0,25\) , vì \(\left|2-\dfrac{1}{6}x\right|\ge0\Rightarrow\left|2-\dfrac{1}{6}x\right|+0,25\ge0,25\) . Vậy B có GTNN = 0,25 khi \(2-\dfrac{1}{6}x=0\Rightarrow\dfrac{x}{6}=2\Rightarrow x=2.6=12\)
(1 - \(\dfrac{1}{2}\)) . (1 - \(\dfrac{1}{3}\)) . (1 - \(\dfrac{1}{4}\)) ... (1 - \(\dfrac{1}{x}\)) = \(\dfrac{1}{2014}\)
=> \(\dfrac{1}{2}\) . \(\dfrac{2}{3}\) . \(\dfrac{3}{4}\) ... \(\dfrac{x-1}{x}\) =\(\dfrac{1}{2014}\)
=> \(\dfrac{1.2.3...x-1}{2.3.4...x}\) = \(\dfrac{1}{2014}\)
=:> \(\dfrac{1}{x}\) = \(\dfrac{1}{2014}\)
Vậy x = 2014
(Mình nghĩa là vậy, có gì sai mong bạn bỏ qua nha)
1. a, 3x + |x - 2| = 8
<=> |x - 2| = 8 - 3x
Xét 2 TH :
TH1: x - 2 = 8 - 3x
<=> x + 3x = 8 + 2
<=> 4x = 10
<=> x = \(\dfrac{5}{2}\) (thỏa mãn)
TH2: x - 2 = -(8 - 3x)
<=> x - 2 = -8 + 3x
<=> -2 + 8 = 3x - x
<=> 6 = 2x
<=> x = 3 (thỏa mãn)
b, 5 - |x - 1| = 4
<=> |x - 1| = 1
<=> \(\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\) (thỏa mãn)
@Nguyễn Hoàng Vũ
2. 5.(x - 2) - 4.(1 - 3x) = |3 - 7| + 2.(1 + 2x)
<=> 5x - 10 - 4 + 12x = 4 + 2 + 4x
<=> 17x - 14 = 6 + 4x
<=> 17x - 4x = 6 + 14
<=> 13x = 20
<=> x = \(\dfrac{20}{13}\) (thỏa mãn)
@Nguyễn Hoàng Vũ
a)Ta thấy:
\(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{x+a}{x\left(x+a\right)}-\dfrac{x}{x\left(x+a\right)}\)
\(=\dfrac{\left(x+a\right)-x}{x\left(x+a\right)}\)
\(=\dfrac{a}{x\left(x+a\right)}\)
\(\Rightarrowđpcm\)
b)Ta thấy:
\(\dfrac{1}{x\left(x+1\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)^2\left(x+2\right)}-\dfrac{x\left(x+1\right)}{x\left(x+1\right)^2\left(x+2\right)}\)
\(=\dfrac{x+2}{x\left(x+1\right)\left(x+2\right)}-\dfrac{x}{x\left(x+1\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)-x}{x\left(x+1\right)\left(x+2\right)}=\dfrac{2}{x\left(x+1\right)\left(x+2\right)}\Rightarrowđpcm\)
c)Ta thấy:
\(\dfrac{1}{x\left(x+1\right)\left(x+2\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+3\right)}{x\left(x+1\right)^2\left(x+2\right)^2\left(x+3\right)}-\dfrac{x\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)^2\left(x+2\right)^2\left(x+3\right)}=\dfrac{x+3}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}-\dfrac{x}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\dfrac{x+3-x}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\dfrac{3}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\Rightarrowđpcm\)
a/ \(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{a}{x\left(x+a\right)}\)
Ta có: \(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{x+a}{x\left(x+a\right)}-\dfrac{x}{x\left(x+a\right)}\)
\(=\dfrac{\left(x-x\right)+a}{x\left(x+a\right)}\) hay \(\dfrac{a}{x\left(x+a\right)}\)
\(\Rightarrow\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{a}{x\left(x+a\right)}\left(đpcm\right)\)