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a) \(\left|x-\dfrac{5}{3}\right|< \dfrac{1}{3}\)
\(\Rightarrow\dfrac{-1}{3}< x-\dfrac{5}{3}< \dfrac{1}{3}\)
\(\Rightarrow\dfrac{-1}{3}+\dfrac{5}{3}< x-\dfrac{5}{3}+\dfrac{5}{3}< \dfrac{1}{3}+\dfrac{5}{3}\)
\(\Rightarrow\dfrac{4}{3}< x< 2\)
b) \(\left|x+\dfrac{11}{2}\right|>\left|-5,5\right|=5,5\)
\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{11}{2}< 5,5\\x+\dfrac{11}{2}>5,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 5,5-\dfrac{11}{2}=0\\x>5,5-\dfrac{11}{2}=0\end{matrix}\right.\)
=> Với x khác 0 thì thõa mãn đề bài
c) \(\dfrac{2}{5}< \left|x-\dfrac{7}{5}\right|< \dfrac{3}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2}{5}< x-\dfrac{7}{5}< \dfrac{3}{5}\\-\dfrac{2}{5}< x-\dfrac{7}{5}< -\dfrac{3}{5}\end{matrix}\right.\)
Ta thấy trường hợp 2 là trường hợp không thể xảy ra
=> Loại
Vậy \(\dfrac{2}{5}< x-\dfrac{7}{5}< \dfrac{3}{5}\)
\(\Rightarrow\dfrac{2}{5}+\dfrac{7}{5}< x< \dfrac{3}{5}+\dfrac{7}{5}\)
\(\Rightarrow\dfrac{9}{5}< x< 2\) (nhận)
p/s : làm đại nha , ko bik đúng sai
a: x>-3/5 nên x+3/5>0
x<1/7 nên x-1/7<0
A=1/7-x-x-3/5+4/5=-2x+12/35
b: B=|x-1/7|+|x+3/5|-1/3
x>-3/5 nên x+3/5>0
x<1/7 nên x-1/7<0
B=1/7-x+3/5+x-1/3=43/105
a: 2x(x-1/7)=0
=>x(x-1/7)=0
=>x=0 hoặc x=1/7
b: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)
nên \(x=\dfrac{-1}{4}:\dfrac{7}{20}=\dfrac{-20}{4\cdot7}=\dfrac{-5}{7}\)
c: \(\Leftrightarrow\dfrac{41}{9}:\dfrac{41}{18}-7< x< \left(3.2:3.2+\dfrac{45}{10}\cdot\dfrac{31}{45}\right):\left(-21.5\right)\)
\(\Leftrightarrow2-7< x< \dfrac{\left(1+3.1\right)}{-21.5}\)
\(\Leftrightarrow-5< x< \dfrac{-41}{215}\)
mà x là số nguyên
nên \(x\in\left\{-4;-3;-2;-1\right\}\)
a, \(\dfrac{3}{5}-4.\left|\dfrac{1}{5}-\dfrac{3}{4}x\right|=\dfrac{1}{3}\)
\(\Rightarrow4\left|\dfrac{1}{5}-\dfrac{3}{4}x\right|=\dfrac{4}{15}\)
\(\Rightarrow\left|\dfrac{1}{5}-\dfrac{3}{4}x\right|=\dfrac{1}{15}\)
\(\Rightarrow\dfrac{1}{5}-\dfrac{3}{4}x\in\left\{-\dfrac{1}{15};\dfrac{1}{15}\right\}\)
\(\Rightarrow\dfrac{3}{4}x\in\left\{\dfrac{4}{15};\dfrac{2}{15}\right\}\Rightarrow x\in\left\{\dfrac{16}{45};\dfrac{8}{45}\right\}\)
b, \(\left|2\dfrac{2}{9}-x\right|=\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\)
\(\Rightarrow\left|2\dfrac{2}{9}-x\right|=\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\)
\(\Rightarrow\left|2\dfrac{2}{9}-x\right|=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{8}-\dfrac{1}{9}\)
(do \(\dfrac{1}{a.\left(a+1\right)}=\dfrac{1}{a}-\dfrac{1}{a+1}\) với mọi \(a\in N\)*)
\(\Rightarrow\left|2\dfrac{2}{9}-x\right|=\dfrac{1}{3}-\dfrac{1}{9}\)
\(\Rightarrow\left|2\dfrac{2}{9}-x\right|=\dfrac{2}{9}\Rightarrow2\dfrac{2}{9}-x\in\left\{-\dfrac{2}{9};\dfrac{2}{9}\right\}\)
\(\Rightarrow x\in\left\{\dfrac{22}{9};2\right\}\)
c,\(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(\Rightarrow\dfrac{1}{3}x+\dfrac{2}{5}x-\dfrac{2}{5}=0\)
\(\Rightarrow\dfrac{11}{15}x=\dfrac{2}{5}\Rightarrow x=\dfrac{6}{11}\)
d, \(60\%x+\dfrac{2}{3}x=\dfrac{1}{3}.6\dfrac{1}{3}\)
\(\Rightarrow\dfrac{3}{5}x+\dfrac{2}{3}x=\dfrac{1}{3}.\dfrac{19}{3}\)
\(\Rightarrow\dfrac{19}{15}x=\dfrac{19}{9}\Rightarrow x=\dfrac{5}{3}\)
Chúc bạn học tốt!!!
a) \(x+\dfrac{3}{10}=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}-\dfrac{3}{10}\)
\(x=\dfrac{-7}{10}\)
b) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\)
\(x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}\)
\(x+\dfrac{5}{6}=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}-\dfrac{5}{6}\)
\(x=\dfrac{7}{30}\)
c) \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)
\(\dfrac{7}{5}x=\dfrac{-43}{35}\)
\(\Rightarrow x=\dfrac{-43}{49}\)
d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)
\(\left[x+\dfrac{3}{4}\right]=0+\dfrac{1}{3}\)
\(\left[x+\dfrac{3}{4}\right]=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}-\dfrac{3}{4}\)
\(x=\dfrac{-5}{12}\)
e) \(\left[x+\dfrac{4}{5}\right]-\left(-3,75\right)=-\left(-2,15\right)\)
\(\left[x+\dfrac{4}{5}\right]+3,75=2,15\)
\(x+\dfrac{4}{5}=2,15-3,75\)
\(x+\dfrac{4}{5}=-\dfrac{8}{5}\)
\(x=\dfrac{-8}{5}-\dfrac{4}{5}\)
\(x=\dfrac{-12}{5}\)
f) \(\left(x-2\right)^2=1\)
\(\Rightarrow x=1\)
Sức chịu đựng có giới hạn -.-
- Mình tiếp tục cho Nguyễn Phương Trâm nhé.
g, \(\left(2x-1\right)^3=-27\)
\(\Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\)
\(\Rightarrow2x-1=-3\)
\(\Rightarrow2x=-2\)
=> \(x=-1\)
- Vậy x = -1
h,\(\dfrac{x-1}{-15}=-\dfrac{60}{x-1}\)
\(\Rightarrow\left(x-1\right)^2=-60.\left(-15\right)\)
\(\Rightarrow\left(x-1\right)^2=900 \)
\(\Rightarrow\left(x-1\right)^2=30^2\Rightarrow x-1=30\)
=> x = 31
i,\(x:\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)
=> \(x:\left(-\dfrac{1}{8}\right)=-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{16}\)
- Vậy x=\(\dfrac{1}{16}\)
j, \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)
\(\Rightarrow \left(\dfrac{3}{4}\right).x=\left(\dfrac{3}{4}\right)^2\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^2:\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
- Vạy x = \(\dfrac{3}{4}\)
k, \(8^x:2^x=4\Rightarrow\left(8:2\right)^x=4\)
=>\(4^x=4\)
=> x = 1
- Vậy x = 1
P = \(\dfrac{-7}{78}x\)
=> Để P > 0 thì x < 0
Để P = 0 thì x = 0
Để P < 0 thì x > 0
1: \(\Leftrightarrow3x+4=2\)
=>3x=-2
=>x=-2/3
2: \(\Leftrightarrow7x-7=6x-30\)
=>x=-23
3: =>\(5x-5=3x+9\)
=>2x=14
=>x=7
4: =>9x+15=14x+7
=>-5x=-8
=>x=8/5
\(1,\)
\(a,\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}+\left(\dfrac{4}{9}-\dfrac{17}{18}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)\)
\(=\dfrac{11}{125}+\left(\dfrac{-1}{2}\right)+\dfrac{1}{2}\)
\(=\dfrac{11}{125}\)
\(b,-1\dfrac{5}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
\(=\dfrac{-12}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
\(=-15.\left[\dfrac{12}{7}+\dfrac{2}{7}+\left(-5\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\right]\)
\(=-15.\left[2+\left(-5\right).\dfrac{1}{105}\right]\)
\(=-15.\left(2-\dfrac{1}{21}\right)\)
\(=-15.\dfrac{41}{21}=\dfrac{-615}{21}\)
\(2,\)
\(a,\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
\(\Leftrightarrow\dfrac{11}{13}-\dfrac{5}{42}+x=\dfrac{-15}{28}+\dfrac{11}{13}\)
\(\Leftrightarrow x=\dfrac{-15}{28}+\dfrac{11}{13}-\dfrac{11}{13}+\dfrac{5}{42}\)
\(\Leftrightarrow x=\left(\dfrac{11}{13}-\dfrac{11}{13}\right)+\left(\dfrac{5}{42}+\dfrac{-15}{28}\right)\)
\(\Leftrightarrow x=\dfrac{5}{12}\)
Vậy \(x=\dfrac{5}{12}\)
\(b,\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|-3,75=-2,15\)
\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2,15+3,75=1,6=\dfrac{16}{10}=\dfrac{8}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=\dfrac{8}{5}\\x+\dfrac{4}{15}=\dfrac{-8}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{5}-\dfrac{4}{15}=\dfrac{4}{3}\\x=\dfrac{-8}{5}-\dfrac{4}{15}=\dfrac{-28}{15}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{4}{3};\dfrac{-28}{15}\right\}\)
\(c,7^{x+2}+2.7^{x-1}=345\)
\(\Leftrightarrow7^{x-1}.\left(7^3+2\right)=345\)
\(\Leftrightarrow7^{x-1}.\left(343+2\right)=345\)
\(\Leftrightarrow7^{x-1}.345=345\)
\(\Leftrightarrow7^{x-1}=345:345=1\)
\(\Leftrightarrow x-1=0\)
\(x=0+1=1\)
Vậy \(x=1\)
b: \(\left|x-\dfrac{3}{5}\right|< \dfrac{1}{3}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{3}{5}>-\dfrac{1}{3}\\x-\dfrac{3}{5}< \dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\dfrac{4}{15}< x< \dfrac{14}{15}\)
c: \(\left|x+\dfrac{11}{2}\right|>-5.5\)
mà \(\left|x+\dfrac{11}{2}\right|\ge0\forall x\)
nên \(x\in R\)