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a) Ta có:
\(\frac{3}{x+2}=\frac{5}{2x+1}\)
\(\Rightarrow3\left(2x+1\right)=5\left(x+2\right)\)
\(\Rightarrow6x+3=5x+10\)
\(\Rightarrow6x-5x=10-3\)
\(\Rightarrow x=7\)
b)Ta có:
\(\frac{5}{8x-2}=\frac{-4}{7-x}\)
\(\Rightarrow5\left(7-x\right)=-4\left(8x-2\right)\)
\(\Rightarrow35-5x=-32x+8\)
\(\Rightarrow-5x+32x=8-35\)
\(\Rightarrow27x=-27\)
\(\Rightarrow x=-1\)
c) Ta có:
\(\frac{4}{3}=\frac{2x-1}{x}\)
\(\Rightarrow4x=3\left(2x-1\right)\)
\(\Rightarrow4x=6x-3\)
\(\Rightarrow3=6x-4x=2x\)
\(\Rightarrow x=\frac{3}{2}\)
d)Ta có:
\(\frac{2x-1}{3}=\frac{3x+1}{4}\)
\(\Rightarrow4\left(2x-1\right)=3\left(3x+1\right)\)
\(\Rightarrow8x-4=9x+3\)
\(\Rightarrow8x-9x=3+4\)
\(\Rightarrow-x=7\Rightarrow x=-7\)
e)Ta có:
\(\frac{4}{x+2}=\frac{7}{3x+1}\)
\(\Rightarrow4\left(3x+1\right)=7\left(x+2\right)\)
\(\Rightarrow12x+4=7x+14\)
\(\Rightarrow12x-7x=14-4\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
f)Ta có:
\(\frac{-3}{x+1}=\frac{4}{2-2x}\)
\(\Rightarrow-3\left(2-2x\right)=4\left(x+1\right)\)
\(\Rightarrow-6+6x=4x+4\)
\(\Rightarrow6x-4x=4+6\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=5\)
a) \(\frac{5}{6}=\frac{x-1}{x}\)
\(5x=6x-6\)
\(6x-5x=6\)
\(x=6\)
các câu còn lại lm tương tự
hok tốt!!
b) \(\frac{1}{2}=\frac{x+1}{3x}\)
\(\Rightarrow1.3x=2.\left(x+1\right)\)
\(3x=2x+2\)
\(3x-2x=2\)
\(x=2\)
Vậy x=2
các câu khác bạn làm tương tự
a) \(2x-\frac{2}{3}-7x=\frac{3}{2}-1\\ 2x-7x-\frac{2}{3}=\frac{1}{2}\\ -5x=\frac{1}{2}+\frac{2}{3}\\ -5x=\frac{7}{6}\\ x=\frac{7}{6}:\left(-5\right)\\ x=\frac{-7}{30}\)Vậy \(x=\frac{-7}{30}\)
b) \(\frac{3}{2}x-\frac{2}{5}=\frac{1}{3}x-\frac{1}{4}\\ \frac{3}{2}x-\frac{1}{3}x=\frac{2}{5}-\frac{1}{4}\\ \frac{7}{6}x=\frac{3}{20}\\ x=\frac{3}{20}:\frac{7}{6}\\ x=\frac{9}{70}\)Vậy \(x=\frac{9}{70}\)
c) \(\frac{2}{3}-\frac{5}{3}x=\frac{7}{10}x+\frac{5}{6}\\ \frac{2}{3}-\frac{5}{6}=\frac{7}{10}x+\frac{5}{3}x\\ \frac{-1}{6}=\frac{71}{30}x\\ x=\frac{-1}{6}:\frac{71}{30}\\ x=\frac{-5}{71}\)Vậy \(x=\frac{-5}{71}\)
d) \(2x-\frac{1}{4}=\frac{5}{6}-\frac{1}{2}x\\ 2x+\frac{1}{2}x=\frac{5}{6}+\frac{1}{4}\\ \frac{5}{2}x=\frac{13}{12}\\ x=\frac{13}{12}:\frac{5}{2}\\ x=\frac{13}{30}\)Vậy \(x=\frac{13}{30}\)
e) \(3x-\frac{5}{3}=x-\frac{1}{4}\\ 3x-x=\frac{5}{3}-\frac{1}{4}\\ 2x=\frac{17}{12}\\ x=\frac{17}{12}:2\\ x=\frac{17}{24}\)Vậy \(x=\frac{17}{24}\)
Èo, chăm thế? Chăm hơn cả mik cơ, gần 11 h rồi onl thì thấy bài được bạn HISI làm hết rồi :((
g)=>x+1/2=0
x=0-1/2
x=-1/2
hoặc 2/3-2x=0
2x=2/3-0
2x=2/3
x=2/3:2
x=1/3
nhìn @_@ hoa cả mắt đăng từng bài thôi bạn
a) \(\left(4,5-2x\right)\cdot1\frac{4}{7}=\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)\cdot\frac{11}{7}=\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)=\frac{11}{7}\div\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)=1\)
\(2x=\frac{9}{2}-1\)
\(x=\frac{7}{2}\div2\)
\(x=\frac{7}{4}\)
b) \(|\frac{3}{4}\cdot x-\frac{1}{2}|-1=\frac{1}{4}\)
\(|\frac{3}{4}\cdot x-\frac{1}{2}|=\frac{1}{4}+1\)
\(|\frac{3}{4}\cdot x|=\frac{5}{4}+\frac{1}{2}\)
\(x=\frac{7}{4}\div\frac{3}{4}\)
\(x=\frac{7}{3}\)
c) \(\frac{1}{4}-|3-x|=-\frac{3}{4}\)
\(|3-x|=\frac{1}{4}-\left(-\frac{3}{4}\right)\)
\(|3-x|=1\)
\(x=3-1\)
\(\Rightarrow x=2\)
d) \(4\cdot\left(x-\frac{6}{7}\right)-\frac{3}{5}=1,4\)
\(4\cdot\left(x-\frac{6}{7}\right)-\frac{3}{5}=\frac{7}{5}\)
\(4\cdot\left(x-\frac{6}{7}\right)=\frac{7}{5}+\frac{3}{5}\)
\(4\cdot\left(x-\frac{6}{7}\right)=2\)
\(\left(x-\frac{6}{7}\right)=2\div4\)
\(x=\frac{1}{2}+\frac{6}{7}\)
\(x=\frac{19}{14}\)
\(\)
Giải:
a) \(\left(4,5-2x\right).\left(-1\dfrac{4}{7}\right)=\dfrac{11}{14}\)
\(\Leftrightarrow\left(4,5-2x\right).\left(-\dfrac{3}{7}\right)=\dfrac{11}{14}\)
\(\Leftrightarrow4,5-2x=\dfrac{11}{14}:\left(-\dfrac{3}{7}\right)=-\dfrac{11}{6}\)
\(\Leftrightarrow2x=4,5-\left(-\dfrac{11}{6}\right)\)
\(\Leftrightarrow2x=\dfrac{19}{3}\)
\(\Leftrightarrow x=\dfrac{19}{3}:2=\dfrac{19}{6}\)
Vậy ...
b) \(\dfrac{4}{9}x=\dfrac{9}{8}-0,125\)
\(\Leftrightarrow\dfrac{4}{9}x=\dfrac{9}{8}-\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{4}{9}x=1\)
\(\Leftrightarrow x=1:\dfrac{4}{9}=\dfrac{9}{4}\)
Vậy ...
Các câu còn lại làm tương tự.
\(a,\frac{3}{7}+\left|2x-\frac{1}{2}\right|=\frac{4}{5}\)
\(\Rightarrow\left|2x-\frac{1}{2}\right|=\frac{13}{35}\)
\(\Rightarrow2x-\frac{1}{2}=\pm\left(\frac{13}{35}\right)\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=\frac{13}{35}\\2x-\frac{1}{2}=\frac{-13}{35}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=\frac{61}{70}\\2x=\frac{9}{70}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{61}{140}\\x=\frac{9}{140}\end{cases}}\)
~Study well~
#KSJ
\(b,\frac{3}{4}-4\times\left|2x+1\right|=\frac{1}{2}\)
\(\Rightarrow4\times\left|2x+1\right|=\frac{1}{4}\)
\(\Rightarrow\left|2x+1\right|=\frac{1}{16}\)
\(\Rightarrow2x+1=\pm\left(\frac{1}{16}\right)\)
\(\Rightarrow\orbr{\begin{cases}2x+1=\frac{1}{16}\\2x+1=\frac{-1}{16}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=\frac{-15}{16}\\2x=\frac{-17}{16}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-15}{32}\\x=\frac{-17}{32}\end{cases}}\)
~Study well~
#KSJ
\(c,\left(\frac{1}{2}+\frac{3}{4}\right)\times\left|3x-1\right|=\frac{1}{2}\)
\(\Rightarrow\frac{5}{4}\times\left|3x-1\right|=\frac{1}{2}\)
\(\Rightarrow\left|3x-1\right|=\frac{2}{5}\)
\(\Rightarrow3x-1=\pm\left(\frac{2}{5}\right)\)
\(\Rightarrow\orbr{\begin{cases}3x-1=\frac{2}{5}\\3x-1=\frac{-2}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=\frac{7}{5}\\3x=\frac{3}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{15}\\x=\frac{1}{5}\end{cases}}\)
~Study well~
#KSJ
\(\text{a)}\frac{3}{7}+\left|2x-\frac{1}{2}\right|=\frac{4}{5}\)
\(\Rightarrow\left|2x-\frac{1}{2}\right|=\frac{4}{5}-\frac{3}{7}=\frac{13}{35}\)
\(TH1:2x-\frac{1}{2}\ge0\)
\(\Rightarrow2x-\frac{1}{2}=\frac{13}{35}\)
\(\Rightarrow2x=\frac{13}{35}+\frac{1}{2}=\frac{61}{70}\)
\(\Rightarrow x=\frac{61}{70}:2=\frac{61}{140}\)
\(TH2:2x-\frac{1}{2}< 0\)
\(\Rightarrow2x-\frac{1}{2}=\frac{-13}{35}\)
\(\Rightarrow2x=\frac{-13}{35}+\frac{1}{2}=\frac{9}{70}\)
\(\Rightarrow x=\frac{9}{70}:2=\frac{9}{140}\)
Vậy \(x=\left\{\frac{9}{140};\frac{61}{140}\right\}\)
\(\text{b)}\frac{3}{4}-4\left|2x+1\right|=\frac{1}{2}\)
\(\Rightarrow4\left|2x+1\right|=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\)
\(\Rightarrow\left|2x+1\right|=\frac{1}{4}:4=\frac{1}{16}\)
\(TH1:2x+1\ge0\)
\(\Rightarrow2x+1=\frac{1}{16}\)
\(\Rightarrow2x=\frac{1}{16}-1=\frac{-15}{16}\)
\(\Rightarrow x=\frac{-15}{16}:2=\frac{-15}{32}\)
\(TH2:2x+1< 0\)
\(\Rightarrow2x+1=\frac{-1}{16}\)
\(\Rightarrow2x=\frac{-1}{16}+1=\frac{15}{16}\)
\(\Rightarrow x=\frac{15}{16}:2=\frac{15}{32}\)
Vậy \(x=\left\{\frac{15}{32};\frac{-15}{32}\right\}\)
\(\text{c)}\left(\frac{1}{2}+\frac{3}{4}\right).\left|3x-1\right|=\frac{1}{2}\)
\(\Rightarrow\frac{5}{4}\left|3x-1\right|=\frac{1}{2}\)
\(\Rightarrow\left|3x-1\right|=\frac{1}{2}:\frac{5}{4}=\frac{2}{5}\)
\(TH1:3x-1\ge0\)
\(\Rightarrow3x-1=\frac{2}{5}\)
\(\Rightarrow3x=\frac{2}{5}+1=\frac{7}{5}\)
\(\Rightarrow x=\frac{7}{5}:3=\frac{7}{15}\)
\(TH2:3x-1< 0\)
\(\Rightarrow3x-1=\frac{-2}{5}\)
\(\Rightarrow3x=\frac{-2}{5}+1=\frac{3}{5}\)
\(\Rightarrow x=\frac{3}{5}:3=\frac{1}{5}\)
Vậy \(x=\left\{\frac{1}{5};\frac{7}{15}\right\}\)
\(a,\) \(\frac{3}{7}+\left|2x-\frac{1}{2}\right|=\frac{4}{5}\)
\(\left|2x-\frac{1}{2}\right|=\frac{4}{5}-\frac{3}{7}\)
\(\left|2x-\frac{1}{2}\right|=\frac{13}{35}\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=-\frac{13}{35}\\2x-\frac{1}{2}=\frac{13}{35}\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}2x=-\frac{13}{35}+\frac{1}{2}=\frac{9}{70}\\2x=\frac{13}{35}+\frac{1}{2}=\frac{61}{70}\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{70}\text{ }:\text{ }2=\frac{9}{140}\\x=\frac{61}{70}\text{ : }2=\frac{61}{140}\end{cases}}\)
\(b,\) \(\frac{3}{4}-4\left|2x+1\right|=\frac{1}{2}\)
\(4\left|2x+1\right|=\frac{3}{4}-\frac{1}{2}\)
\(4\left|2x+1\right|=\frac{1}{4}\)
\(\left|2x+1\right|=\frac{1}{4}\text{ : }4\)
\(\left|2x+1\right|=\frac{1}{16}\)
\(\Rightarrow\orbr{\begin{cases}2x+1=-\frac{1}{16}\\2x+1=\frac{1}{16}\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}2x=-\frac{1}{16}-1=-\frac{17}{16}\\2x=\frac{1}{16}-1=-\frac{15}{16}\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-\frac{17}{16}\text{ : }2=-\frac{17}{32}\\x=-\frac{15}{16}\text{ : }2=-\frac{15}{32}\end{cases}}\)
\(x\in\left\{-\frac{17}{32}\text{ ; }-\frac{15}{32}\right\}\)
\(c,\) \(\left(\frac{1}{2}+\frac{3}{4}\right)\text{ x }\left|3x-1\right|=\frac{1}{2}\)
\(\frac{5}{4}\text{ x }\left|3x-1\right|=\frac{1}{2}\)
\(\left|3x-1\right|=\frac{1}{2}\text{ : }\frac{5}{4}\)
\(\left|3x-1\right|=\frac{2}{5}\)
\(\Rightarrow\orbr{\begin{cases}3x-1=-\frac{2}{5}\\3x-1=\frac{2}{5}\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}3x=-\frac{2}{5}+1=\frac{3}{5}\\3x=\frac{2}{5}+1=\frac{7}{5}\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=\frac{3}{5}\text{ : }3=\frac{1}{5}\\x=\frac{7}{5}\text{ : }3=\frac{7}{15}\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{\frac{1}{5}\text{ ; }\frac{7}{15}\right\}\)