Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
cái chữ này cj nhìn khó hiểu quá nên em thông cảm nếu muốn bt đáp án thì viết rõ ra đc chứ^^
\(\frac{3}{4}x-\frac{1}{2}=2\left(x-4\right)+\frac{1}{4}x\)
\(\Leftrightarrow\frac{3}{4}x-\frac{1}{2}=2\text{x}-8+\frac{1}{4}x\)
\(\Leftrightarrow\frac{3}{4}x-2\text{x}-\frac{1}{4}x=-8+\frac{1}{2}\)
\(\Leftrightarrow\frac{3-8-1}{4}x=\frac{-15}{2}\)
\(\Leftrightarrow-\frac{3}{2}x=-\frac{15}{2}\Leftrightarrow x=\frac{-15}{-3}=5\)
Vậy x = 5
\(\frac{x-1}{12}+\frac{x-1}{20}+\frac{x-1}{30}+\frac{x-1}{42}+\frac{x-1}{56}+\frac{x-1}{72}=\frac{16}{9}\)
\(\Rightarrow\left(x-1\right)\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)
\(\Rightarrow\left(x-1\right)\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=\frac{16}{9}\)
\(\Rightarrow\left(x-1\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\Rightarrow\left(x-1\right)\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\Rightarrow\left(x-1\right)\cdot\frac{2}{9}=\frac{16}{9}\)
\(\Rightarrow\left(x-1\right)=\frac{16}{9}\div\frac{2}{9}\)
\(\Rightarrow\left(x-1\right)=\frac{16}{9}\cdot\frac{9}{2}\)
\(\Rightarrow x-1=8\Rightarrow x=9\)
Vậy x = 9
\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=\frac{4008}{2005}\)
\(\Rightarrow\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{4008}{2005}\)
\(\Rightarrow\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{4008}{2005}\)
\(\Rightarrow2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{4008}{2005}\)
\(\Rightarrow\left(1-\frac{1}{x+1}\right)=\frac{4008}{2005}\div2\)
\(\Rightarrow\frac{x}{x+1}=\frac{2004}{2005}\)
\(\Rightarrow2005\text{x}=2004\left(x+1\right)\)
\(\Rightarrow2005\text{x}=2004\text{x}+2004\)
\(\Rightarrow2005\text{x}-2004\text{x}=2004\)
\(\Rightarrow x=2004\)
Vậy x = 2004
a. 3/5 . 15/7 - 15/7 . 8/5
= 15/7(3/5-8/5)
=15/7. -\(\frac{1}{1}\)
=22/7
b. 4/5 . 1 3/7 + 4/5 . 4/7
=4/5(13/7+4/7)
=4/5.17/7
= 68/35
a) \(\left(x-5\right)-\frac{1}{3}=\frac{2}{5}\)
\(\Rightarrow\left(x-5\right)=\frac{2}{5}+\frac{1}{3}\)
\(\Rightarrow\left(x-5\right)=\frac{11}{15}\)
\(\Rightarrow x-5=\frac{11}{15}\)
\(\Rightarrow x=\frac{11}{15}+5\)
\(\Rightarrow x=\frac{86}{15}\)
b) \(\frac{2}{3}\cdot x-\frac{3}{2}\cdot x=\frac{5}{12}\)
\(\Rightarrow x\cdot\left(\frac{2}{3}-\frac{3}{2}\right)=\frac{5}{12}\)
\(\Rightarrow x\cdot\left(-\frac{5}{6}\right)=\frac{5}{12}\)
\(\Rightarrow x=\frac{5}{12}:\left(-\frac{5}{6}\right)\)
\(\Rightarrow x=-\frac{1}{2}\)
c) \(-\frac{2}{3}\cdot x+\frac{1}{5}=\frac{3}{10}\)
\(\Rightarrow-\frac{2}{3}\cdot x=\frac{3}{10}-\frac{1}{5}\)
\(\Rightarrow-\frac{2}{3}\cdot x=\frac{1}{10}\)
\(\Rightarrow x=\frac{1}{10}:\left(-\frac{2}{3}\right)\)
\(\Rightarrow x=-\frac{3}{20}\)
d) \(4-\left(\frac{1}{2}\cdot x+\frac{3}{4}\right)=-\frac{1}{5}\)
\(\Rightarrow\left(\frac{1}{2}\cdot x+\frac{3}{4}\right)=4-\left(-\frac{1}{5}\right)\)
\(\Rightarrow\)\(\frac{1}{2}\cdot x+\frac{3}{4}=\frac{21}{5}\)
\(\Rightarrow\)\(\frac{1}{2}\cdot x=\frac{21}{5}-\frac{3}{4}\)
\(\Rightarrow\)\(\frac{1}{2}\cdot x=\frac{69}{20}\)
\(\Rightarrow\)\(x=\frac{69}{20}:\frac{1}{2}\)
\(\Rightarrow\)\(x=\frac{69}{10}\)
c: =>2/3x=1/10+1/2=1/10+5/10=6/10=3/5
hay \(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
d: \(\Leftrightarrow\dfrac{4}{9}:x=\dfrac{2}{3}-\dfrac{3}{5}=\dfrac{1}{15}\)
hay \(x=\dfrac{4}{9}:\dfrac{1}{15}=\dfrac{4}{9}\cdot15=\dfrac{20}{3}\)
f: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3