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a) \(\frac{x-3}{3}-1=\frac{x}{-4}\)
\(\Leftrightarrow\frac{x-3}{3}-\frac{3}{3}=\frac{x}{-4}\)
\(\Leftrightarrow\frac{x-6}{3}=\frac{x}{-4}\)
\(\Leftrightarrow-4\left(x-6\right)=3x\)
\(\Leftrightarrow-4x+24=3x\)
\(\Leftrightarrow24=3x+4x\)
\(\Leftrightarrow7x=24\)
\(\Leftrightarrow x=\frac{24}{7}\)
b) \(\frac{5}{8}-\left(x-\frac{1}{2}\right)=\frac{-3}{4}\)
\(\Leftrightarrow\frac{5}{8}-x+\frac{1}{2}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{5}{8}+\frac{4}{8}-x=\frac{-3}{4}\)
\(\Leftrightarrow\frac{9}{8}-x=\frac{-3}{4}\)
\(\Leftrightarrow x=\frac{9}{8}+\frac{3}{4}\)
\(\Leftrightarrow x=\frac{15}{8}\)
Bài 2 :
1) \(x-70=-45\) 2) \(\frac{4}{7}:x=\frac{12}{28}\)
\(\Rightarrow\) \(x=-45+70\) \(\Rightarrow x=\frac{4}{7}:\frac{12}{28}\)
\(\Rightarrow\) \(x=25\) \(\Rightarrow x=\frac{4}{3}\)
Vậy \(x=25\) Vậy \(x=\frac{4}{3}\)
3) Giống câu c) ở bài 1
4) \(x-50=-35\) 5) \(\frac{4}{7}.x=\frac{11}{18}\)
\(\Rightarrow x=-35+50\) \(\Rightarrow x=\frac{11}{28}:\frac{4}{7}\)
\(\Rightarrow x=15\) \(\Rightarrow x=\frac{77}{72}\)
Vậy \(x=15\) Vậy \(x=\frac{77}{72}\)
6) \(\left(\frac{2}{3}x+2,5\right):2\frac{2}{6}=6\)
\(\Rightarrow\)\(\left(\frac{2}{3}x+2,5\right):\frac{14}{6}=6\)
\(\Rightarrow\) \(\frac{2}{3}x+2,5=6.\frac{14}{6}\)
\(\Rightarrow\frac{2}{3}x+2,5=14\)
\(\Rightarrow\frac{2}{3}x=\frac{23}{2}\)
\(\Rightarrow x=\frac{23}{2}:\frac{2}{3}\)
\(\Rightarrow x=\frac{69}{4}\)
Vậy \(x=\frac{69}{4}\)
Bài 1:
1) \(\frac{7}{5}+\frac{-8}{5}=-\frac{1}{5}\)
2) \(-\frac{6}{5}.\frac{15}{24}=-\frac{3}{4}\)
3) \(\left(\frac{2}{3}+1,5\right)-3,5:7\frac{1}{2}=\)\(\frac{13}{6}-\frac{7}{15}=\frac{17}{10}\)
4) \(\frac{5}{8}-\frac{-7}{9}=\frac{5}{8}+\frac{7}{9}=\frac{101}{72}\)
5)\(\frac{-7}{3}.\frac{12}{28}=-1\)
a) \(x+\frac{1}{6}=-\frac{3}{8}\)
\(x=-\frac{3}{8}-\frac{1}{6}\)
\(x=-\frac{13}{24}\)
~ Thiên mã ~
b) \(\frac{1}{2}.x+\frac{1}{8}.x=\frac{3}{4}\)
\(x.\left(\frac{1}{2}+\frac{1}{8}\right)=\frac{3}{4}\)
\(\frac{5}{8}.x=\frac{3}{4}\)
\(x=\frac{6}{5}\)
~ Thiên Mã ~
\(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{4}{9}\)
\(\frac{3}{2}x=\frac{4}{9}+\frac{2}{3}\)
\(\frac{3}{2}x=\frac{10}{9}\)
\(x=\frac{10}{9}:\frac{3}{2}\)
\(x=\frac{20}{27}\)
Vậy x=\(\frac{20}{27}\)
\(\left(\frac{9}{11}-x\right):\frac{-10}{11}=1-\frac{4}{5}\)
\(\left(\frac{9}{11}-x\right):\frac{-10}{11}=\frac{1}{5}\)
\(\frac{9}{11}-x=\frac{1}{5}\cdot\frac{-10}{11}\)
\(\frac{9}{11}-x=\frac{-2}{11}\)
\(x=\frac{9}{11}-\frac{-2}{11}\)
\(x=1\)
Vậy x=1
\(\frac{-11}{12}\cdot x+\frac{3}{4}=\frac{-1}{6}\)
\(\frac{-11}{12}\cdot x=\frac{-1}{6}-\frac{3}{4}\)
\(\frac{-11}{12}\cdot x=\frac{21}{12}\)
\(x=\frac{-21}{11}\)
Vậy x=\(\frac{-21}{11}\)
\(\frac{-5}{4}-\left(1\frac{1}{2}+x\right)=4,5\)
\(\frac{3}{2}+x=\frac{-5}{4}-\frac{9}{2}\)
\(\frac{3}{2}+x=\frac{23}{4}\)
\(x=\frac{17}{4}\)
Vậy x=\(\frac{17}{4}\)
\(\left(\frac{3}{4}-x:\frac{2}{15}\right)\cdot\frac{1}{5}=-2,6\)
\(\frac{3}{4}-x:\frac{2}{15}=\frac{-13}{5}:\frac{1}{5}\)
\(\frac{3}{4}-x:\frac{2}{15}=-13\)
\(x:\frac{2}{15}=\frac{3}{4}-\left(-13\right)\)
\(x:\frac{2}{15}=\frac{45}{4}\)
\(x=\frac{3}{2}\)
Vậy x=\(\frac{3}{2}\)
\(3-\left(\frac{1}{6}-x\right)\cdot\frac{2}{3}=\frac{2}{3}\)
\(3-\left(\frac{1}{6}-x\right)=\frac{2}{3}:\frac{2}{3}\)
\(3-\left(\frac{1}{6}-x\right)=1\)
\(\frac{1}{6}-x=2\)
\(x=\frac{1}{6}-2\)
\(x=\frac{-11}{6}\)
Vậy x=\(\frac{-11}{6}\)
\(\left(1-2x\right)\cdot\frac{4}{5}=\left(-2\right)^3\)
\(1-2x=\frac{-1}{10}\)
\(2x=1-\frac{-1}{10}\)
\(2x=\frac{11}{10}\)
\(x=\frac{11}{20}\)
Vậy x=\(\frac{11}{20}\)
\(\frac{1}{6}-\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{1}{8}\)
\(\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{7}{12}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{7}{12}\) \(\frac{1}{2}x-\frac{1}{3}=\frac{-7}{12}\)
\(\frac{1}{2}x=\frac{11}{12}\) \(\frac{1}{2}x=\frac{-1}{4}\)
\(x=\frac{11}{6}\) \(x=\frac{-1}{2}\)
Vậy \(x\in\left\{\frac{11}{6};\frac{-1}{2}\right\}\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)
\(\frac{3}{2}x=\frac{4}{9}+\frac{6}{9}\)
\(\frac{3}{2}x=\frac{10}{9}\)
\(x=\frac{10}{9}:\frac{3}{2}\)
\(x=\frac{20}{27}\)
tk mình đi mình làm nốt cho hjhj ^^
\(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x=0+\frac{2}{5}\)
\(\Leftrightarrow x\left(\frac{1}{3}+\frac{2}{5}\right)=\frac{2}{5}\)
\(\Leftrightarrow x\left(\frac{5}{15}+\frac{6}{15}\right)=\frac{2}{5}\)
\(\Leftrightarrow\frac{11}{15}x=\frac{2}{5}\)
\(\Leftrightarrow x=\frac{2}{5}\div\frac{11}{15}\)
\(\Leftrightarrow x=\frac{6}{11}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{49}{50}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{49}{50}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{49}{50}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{49}{50}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{49}{50}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{49}{50}\div2\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{49}{50}\times\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{49}{100}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{49}{100}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{50}{100}-\frac{49}{100}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{100}\)
\(\Leftrightarrow x+1=100\)
\(\Leftrightarrow x=100-1\)
\(\Leftrightarrow x=99\)
\(a,\left[x+\frac{1}{3}\right]^3=-\frac{8}{27}\)
\(\Leftrightarrow\left[x+\frac{1}{3}\right]^3=\left[-\frac{2}{3}\right]^3\)
\(\Leftrightarrow x+\frac{1}{3}=-\frac{2}{3}\)
\(\Leftrightarrow x=-\frac{2}{3}-\frac{1}{3}=-1\)
Vậy x = -1
a) (x+1/3)3 = -8/27
<=> (x+1/3)3 = (-2/3)3
<=> x + 1/3 = -2/3
<=> x = -1
b) \(\frac{1}{4}x-\frac{8}{12}=1-\frac{3}{2}x\)
\(\Leftrightarrow\frac{7}{4}x=\frac{5}{3}\)
<=> x = 20/21