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\(a,\frac{2}{3}.\left(3-x\right)+\frac{1}{2}=\frac{3}{4}.\left(2.x+1\right)
\)
\(2-\frac{2}{3}x+\frac{1}{2}=\frac{3}{2}.\frac{3}{4}x+\frac{3}{4}
\)
\(\frac{2}{3}x+2-\frac{1}{2}=\frac{9}{8}x+\frac{3}{4}\)
\(\frac{2}{3}x+\frac{3}{2}=\frac{9}{8}x+\frac{3}{4}\)
\(\frac{3}{2}-\frac{3}{4}=\frac{9}{8}x-\frac{2}{3}x\)
\(\frac{6}{4}-\frac{3}{4}=\frac{27}{24}x-\frac{16}{24}x\)
\(\frac{11}{24}x=\frac{3}{4}\)
\(x=\frac{3}{4}:\frac{11}{24}\)
\(x=\frac{3}{4}.\frac{24}{11}\)
\(x=\frac{18}{11}\)
\(Vậy
x=\frac{18}{11}\)
\(b,\frac{5-x}{3}=\frac{2x+1}{5}\)
\(\frac{\left(5-x\right).5}{15}=\frac{\left(2x+1\right).3}{15}\)
\(\Rightarrow\left(5-x\right).5=\left(2x+1\right).3\)
\(25-5x=6x+3\)
\(25-3=6x+5x\)
\(\Rightarrow11x=22\)
\(\Rightarrow x=22:11\)
\(\Rightarrow x=2\)
\(Vậy
x=2\)
Ta có :
\(\left|2x-3\right|\le2\)
\(\Leftrightarrow\)\(-2\le2x-3\le2\)
\(\Leftrightarrow\)\(-2+3\le2x-3+3\le2+3\)
\(\Leftrightarrow\)\(1\le2x\le5\)
\(\Leftrightarrow\)\(\frac{1}{2}\le\frac{2x}{2}\le\frac{5}{2}\)
\(\Leftrightarrow\)\(\frac{1}{2}\le x\le\frac{5}{2}\)
Vậy \(\frac{1}{2}\le x\le\frac{5}{2}\)
Chúc bạn học tốt ~
1a) \(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}=>\left(\frac{2}{3}-\frac{3}{2}\right)x=\frac{5}{12}\)
\(\left(\frac{4}{6}-\frac{9}{6}\right)x=\frac{5}{12}=>\frac{-5}{6}x=\frac{5}{12}\)
\(x=\frac{5}{12}:\frac{-5}{6}=>x=\frac{5}{12}.\frac{-6}{5}=>x=\frac{-1}{2}\)
b)\(\frac{-2}{3}x+\frac{1}{5}=\frac{3}{10}=>\frac{-2}{3}x=\frac{3}{10}-\frac{1}{5}\)
\(\frac{-2}{3}x=\frac{3}{10}-\frac{2}{10}=>\frac{-2}{3}x=\frac{1}{10}\)
\(x=\frac{1}{10}:\frac{-2}{3}=>x=\frac{1}{10}.\frac{-3}{2}=>x=\frac{-3}{20}\)
2a)\(\frac{1}{3}.\frac{5}{7}-\frac{7}{27}.\frac{36}{14}=\frac{1}{3}.\frac{5}{7}-\frac{7}{27}.\frac{18}{7}=\frac{1.5}{3.7}-\frac{7.18}{27.7}=\frac{5}{21}-\frac{1.2}{3.1}=\frac{5}{21}-\frac{2}{3}\)
\(\frac{5}{21}-\frac{14}{21}=\frac{-9}{21}=\frac{-3}{7}\)
b) \(\frac{-3}{7}+\frac{15}{26}-\left(\frac{2}{13}-\frac{3}{7}\right)=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}=\left(\frac{-3}{7}+\frac{3}{7}\right)+\left(\frac{15}{26}-\frac{2}{13}\right)\)
\(0+\left(\frac{15}{26}-\frac{4}{26}\right)=0+\frac{11}{26}=\frac{11}{26}\)
\(\left(x+1\right)\left(y-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\y-2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0-1\\y=0+2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\y=2\end{cases}}\)
Vậy x = - 1 ; y = 2
\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{9}{1}+\frac{8}{2}+...+\frac{1}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10-1}{1}+\frac{10-2}{2}+...+\frac{10-9}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10}{1}-1+...+\frac{10}{9}-1\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10-9+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}\)= \(\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}\)
=>\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
=> \(x=10\)
b) Tương tự câu a
b)
+)x>=2 được
2(x-2)-x=1
=>2x-4-x=1
=>x=5
+)x<2 được
2(2-x)-x=1
=>4-2x-x=1
=>4-3x=1
=>3x=3
=>x=1
Vậy có 2 giá trị là 5;1
\(\left|2x-3\right|+5=x\)
\(\left|2x-3\right|=x-5\)
Xét hai trường hợp :
TH1: \(x\le\frac{3}{2}\)=> không có giá trị thỏa mãn
th2: \(x\ge\frac{3}{2}\)=> không có giá trị thỏa mãn
b) Tương tự