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a: |x-1/2|=7/2
=>x-1/2=7/2 hoặc x-1/2=-7/2
=>x=4 hoặc x=-3
b: \(x:\dfrac{3}{8}+\dfrac{5}{8}=x\)
=>8/3x-x=-5/8
=>5/3x=-5/8
hay x=-5/8:5/3=-5/8x3/5=-15/40=-3/8
c: \(\dfrac{5}{6}-\left|x-\dfrac{1}{2}\right|=\dfrac{15}{18}=\dfrac{5}{6}\)
=>|x-1/2|=0
=>x-1/2=0
hay x=1/2
e: \(\left(5x-3\right)^2-\dfrac{1}{64}=0\)
=>(5x-3)2=1/64
=>5x-3=1/8 hoặc 5x-3=-1/8
=>5x=25/8 hoặc 5x=23/8
=>x=5/8 hoặc x=23/40
8)\(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)
=\(\frac{4}{9}:\left(-\frac{1}{7}\right)+\frac{59}{9}:\left(-\frac{1}{7}\right)\)
=\(\left(\frac{4}{9}+\frac{59}{9}\right).\left(-7\right)\)
=7.(-7)
=-49
a: \(\Leftrightarrow\dfrac{23}{5}\cdot\dfrac{50}{23}< =x< =\dfrac{-13}{5}:\dfrac{7}{5}\)
=>10<=x<=-13/7
hay \(x\in\varnothing\)
b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< =x< =\dfrac{-2}{3}\cdot\dfrac{-11}{12}\)
=>-13/9<=x<=22/36
hay \(x\in\left\{-1;0\right\}\)
a: \(\Leftrightarrow-\dfrac{2}{3}\cdot\dfrac{4-6-9}{12}\ge x\ge-\dfrac{13}{3}\cdot\dfrac{3-1}{6}\)
\(\Leftrightarrow-\dfrac{2}{3}\cdot\dfrac{-11}{12}\ge x\ge\dfrac{-13}{3}\cdot\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{22}{36}\ge x\ge\dfrac{-13}{9}\)
mà x là số nguyên
nên \(x\in\left\{0;-1\right\}\)
b: \(\Leftrightarrow\dfrac{21}{100}+\dfrac{75}{100}-\dfrac{220}{100}>=2x-1>=-3-\dfrac{1}{2}+3+\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{-124}{100}\ge2x-1\ge\dfrac{-3}{10}\)
\(\Leftrightarrow-\dfrac{124}{100}+1\ge2x>=\dfrac{-3}{10}+1\)
\(\Leftrightarrow\dfrac{-3}{25}\ge2x\ge\dfrac{7}{10}\)(vô lý)
=>x không có giá trị
c: \(\Leftrightarrow43+\dfrac{1}{2}-39-\dfrac{1}{5}\le-3x+4\le9+\dfrac{1}{5}+50+\dfrac{1}{7}\)
\(\Leftrightarrow3+\dfrac{3}{10}\le-3x+4\le59+\dfrac{12}{35}\)
\(\Leftrightarrow\dfrac{33}{10}-4\le-3x\le59+\dfrac{12}{35}-4\)
\(\Leftrightarrow\dfrac{-7}{10}\le-3x\le\dfrac{1937}{35}\)
\(\Leftrightarrow\dfrac{7}{30}\ge x\ge-\dfrac{1937}{105}\)
mà x là số nguyên
nên \(x\in\left\{0;-1;-2;...;-18\right\}\)
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
Giải:
a) \(\dfrac{1}{3}x+\dfrac{1}{5}-\dfrac{1}{2}x=1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{6}x=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{1}{6}x=\dfrac{-21}{20}\)
\(\Leftrightarrow x=\dfrac{-63}{10}\)
Vậy ...
b) \(\dfrac{3}{2}\left(x+\dfrac{1}{2}\right)-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{3}{2}x+\dfrac{3}{4}-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{11}{8}x=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-4}{11}\)
Vậy ...
Các câu sau làm tương tự câu b)
1. Tính:
a. \(\dfrac{\text{−1 }}{\text{4 }}+\dfrac{\text{5 }}{\text{6 }}=\dfrac{-3}{12}+\dfrac{10}{12}=\dfrac{7}{12}\)
b. \(\dfrac{\text{5 }}{\text{12 }}+\dfrac{\text{-7 }}{8}=\dfrac{10}{24}+\dfrac{-21}{24}=\dfrac{-11}{24}\)
c. \(\dfrac{-7}{6}+\dfrac{-3}{10}=\dfrac{-35}{30}+\dfrac{-9}{30}=\dfrac{-44}{30}=\dfrac{-22}{15}\)
d.\(\dfrac{-3}{7}+\dfrac{5}{6}=\dfrac{-18}{42}+\dfrac{35}{42}=\dfrac{17}{42}\)
2. Tính :
a. \(\dfrac{2}{14}-\dfrac{5}{2}=\dfrac{2}{14}-\dfrac{35}{14}=\dfrac{-33}{14}\)
b.\(\dfrac{-13}{12}-\dfrac{5}{18}=\dfrac{-39}{36}-\dfrac{10}{36}=\dfrac{49}{36}\)
c.\(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-2}{5}+\dfrac{3}{11}=\dfrac{-22}{55}+\dfrac{15}{55}=\dfrac{-7}{55}\)
d. \(0,6--1\dfrac{2}{3}=\dfrac{6}{10}--\dfrac{5}{3}=\dfrac{3}{5}+\dfrac{5}{3}=\dfrac{9}{15}+\dfrac{25}{15}=\dfrac{34}{15}\)
3. Tính :
a.\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-7}{156}\)
b.\(\dfrac{-6}{9}-\dfrac{12}{16}=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8}{12}-\dfrac{9}{12}=\dfrac{-17}{12}\)
c. \(\dfrac{-3}{7}-\dfrac{-2}{11}=\dfrac{-3}{7}+\dfrac{2}{11}=\dfrac{-33}{77}+\dfrac{14}{77}=\dfrac{-19}{77}\)
d.\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{1}+\dfrac{1}{10}\)
\(=\dfrac{10}{10}-\dfrac{1}{10}\)
= \(\dfrac{9}{10}\)
Chế Kazuto Kirikaya thử tham khảo thử đi !!!
Mấy câu trên kia dễ rồi mình chữa mình câu \(c\) bài \(3\) thôi nhé Kazuto Kirikaya
d) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
a: \(=\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\cdot\dfrac{18}{5}-\dfrac{6}{5}:\dfrac{-9}{5}+4\)
\(=\dfrac{18}{5}-\dfrac{6}{5}\cdot\dfrac{-5}{9}+4\)
\(=\dfrac{18}{5}+\dfrac{2}{3}+4\)
\(=\dfrac{124}{15}\)
b: \(=\dfrac{9}{25}\cdot\left(\dfrac{3}{5}-\dfrac{1}{5}+\dfrac{1}{2}\right)-\dfrac{3}{8}:\dfrac{9}{8}\)
\(=\dfrac{9}{25}\cdot\dfrac{4}{10}-\dfrac{1}{3}\)
\(=-\dfrac{71}{375}\)
c: \(=\dfrac{7}{10}:\dfrac{4}{5}+\dfrac{2}{9}:\dfrac{5}{9}+\dfrac{1}{8}\)
\(=\dfrac{7}{10}\cdot\dfrac{5}{4}+\dfrac{2}{5}+\dfrac{1}{8}\)
=1+2/5
=7/5
d: \(=\dfrac{3}{7}\left(19+\dfrac{1}{3}-33-\dfrac{1}{3}\right)-\dfrac{2}{7}=\dfrac{3}{7}\cdot\left(-14\right)-\dfrac{2}{7}=-6-\dfrac{2}{7}=\dfrac{-44}{7}\)
e: \(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{-2^{11}\cdot3^{11}-2^{12}\cdot3^{12}}\)
\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{-2^{11}\cdot3^{11}\left(1+2\cdot3\right)}=-\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot7}=\dfrac{-4}{7}\)
1) (1+\( \dfrac{2}{3}\)-\(\dfrac{1}{4}\)). 0,8-\(\dfrac{3}{4}\))2
=( \(\dfrac{17}{12}\).0,8-\(\dfrac{3}{4}\))2
=(\(\dfrac{17}{15}\)-\(\dfrac{3}{4}\))2
=( \(\dfrac{23}{60}\))2
= \(\dfrac{529}{3600}\)