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\(\dfrac{x-3}{10}-\dfrac{x-3}{15}=\dfrac{x-3}{12}-\dfrac{x-3}{18}\)
\(\Rightarrow\dfrac{x-3}{1}-\dfrac{x-3}{15}-\dfrac{x-3}{12}+\dfrac{x-3}{18}=0\)
\(\Rightarrow\left(x-3\right)\left(1-\dfrac{1}{15}-\dfrac{1}{12}+\dfrac{1}{18}\right)=0\)
Mà \(1-\dfrac{1}{15}-\dfrac{1}{12}+\dfrac{1}{18}\ne0\)
\(\Rightarrow x-3=0\Leftrightarrow x=3\)
a) Ta có: \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)
\(\Leftrightarrow\dfrac{x-2}{15}-1+\dfrac{x-3}{14}-1+\dfrac{x-4}{13}-1+\dfrac{x-5}{12}-1=0\)
\(\Leftrightarrow\dfrac{x-17}{15}+\dfrac{x-17}{14}+\dfrac{x-17}{13}+\dfrac{x-17}{12}=0\)
\(\Leftrightarrow\left(x-17\right)\left(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\right)=0\)
mà \(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}>0\)
nên x-17=0
hay x=17
Vậy: x=17
b) Ta có: \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)
\(\Leftrightarrow\dfrac{x+1}{19}+1+\dfrac{x+2}{18}+1+\dfrac{x+3}{17}+1+...+\dfrac{x+18}{2}+1=0\)
\(\Leftrightarrow\dfrac{x+20}{19}+\dfrac{x+20}{18}+\dfrac{x+20}{17}+...+\dfrac{x+20}{2}=0\)
\(\Leftrightarrow\left(x+20\right)\left(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}\right)=0\)
mà \(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}>0\)
nên x+20=0
hay x=-20
Vậy: x=-20
a) Ta có: \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)
\(=\dfrac{-25}{90}+\dfrac{64}{90}-\dfrac{81}{90}\)
\(=\dfrac{-42}{90}=-\dfrac{7}{15}\)
b) Ta có: \(\left(-\dfrac{1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)
\(=\dfrac{-1}{4}+\dfrac{17}{11}-\dfrac{5}{3}+\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{42}{29}\)
\(=\dfrac{-5}{3}+\dfrac{42}{29}\)
\(=\dfrac{-145}{87}+\dfrac{126}{87}=\dfrac{-19}{87}\)
c) Ta có: \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)
\(=\left(1-1\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2-2\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3-3\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+4\)
\(=-1-1-1+4\)
=1
a) Ta có: −518+3245−910−518+3245−910
=−2590+6490−8190=−2590+6490−8190
=−4290=−715=−4290=−715
b) Ta có: (−14+5133−53)−(−1512+611−4229)(−14+5133−53)−(−1512+611−4229)
=−14+1711−53+54−611+4229=−14+1711−53+54−611+4229
=−53+4229=−53+4229
=−14587+12687=−1987=−14587+12687=−1987
c) Ta có: 1−12+2−23+3−34+4−14−3−13−2−12−11−12+2−23+3−34+4−14−3−13−2−12−1
=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4
=−1−1−1+4=−1−1−1+4
=1
\(\dfrac{5}{4}+\left(2x-\dfrac{1}{2}\right)=\dfrac{5}{6}\\ =>2x-\dfrac{1}{2}=\dfrac{5}{6}-\dfrac{5}{4}\\ =>2x-\dfrac{1}{2}=\dfrac{10}{12}-\dfrac{15}{12}\\ =>2x-\dfrac{1}{2}=-\dfrac{5}{12}\\ =>2x=-\dfrac{5}{12}+\dfrac{1}{2}\\ =>2x=-\dfrac{5}{12}+\dfrac{6}{12}\\ =>2x=\dfrac{1}{12}\\ =>x=\dfrac{1}{12}:2\\ =>x=\dfrac{1}{12}\cdot\dfrac{1}{2}\\ =>x=\dfrac{1}{24}\)
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\(\dfrac{3}{2}-\left(x+\dfrac{1}{4}\right)=\dfrac{5}{8}\\ =>x+\dfrac{1}{4}=\dfrac{3}{2}-\dfrac{5}{8}\\ =>x+\dfrac{1}{4}=\dfrac{12}{8}-\dfrac{5}{8}\\ =>x+\dfrac{1}{4}=\dfrac{7}{8}\\ =>x=\dfrac{7}{8}-\dfrac{1}{4}\\ =>x=\dfrac{7}{8}-\dfrac{2}{8}\\ =>x=\dfrac{5}{8}\)
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\(\dfrac{x}{3}=\dfrac{12}{x}\\ =>x^2=3\cdot12\\ =>x^2=36\\ =>x^2=6^2\\ =>x=\pm6\)
Tìm x:
a) \(\dfrac{5}{4}+\left(2x-\dfrac{1}{2}\right)=\dfrac{5}{6}\)
\(=>2x-\dfrac{1}{2}=\dfrac{5}{6}-\dfrac{5}{4}\)
\(=>2x-\dfrac{1}{2}=\dfrac{-5}{12}\)
\(=>2x=\dfrac{-5}{12}+\dfrac{1}{2}\)
\(=>2x=\dfrac{1}{12}\)
\(=>x=\dfrac{1}{12}:2\)
\(=>x=\dfrac{1}{24}\)
b) \(\dfrac{3}{2}-\left(x+\dfrac{1}{4}\right)=\dfrac{5}{8}\)
\(=>x+\dfrac{1}{4}=\dfrac{3}{2}-\dfrac{5}{8}\)
\(=>x+\dfrac{1}{4}=\dfrac{7}{8}\)
\(=>x=\dfrac{7}{8}-\dfrac{1}{4}\)
\(=>x=\dfrac{5}{8}\)
c) \(\dfrac{x}{3}=\dfrac{12}{x}\)
Ta có: \(x.x=3.12\)
\(\Rightarrow x^2=36\)
Vậy x = 6 hoặc x = -6
Chúc bạn học tốt
Giải:
a) \(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{0;\pm5;10\right\}\)
\(\Rightarrow x\in\left\{0;\pm1;2\right\}\)
b) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow-12.\left(x-6\right)=4.18\)
\(\Rightarrow-12x+72=72\)
\(\Rightarrow-12x=72-72\)
\(\Rightarrow-12x=0\)
\(\Rightarrow x=0:-12\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
c) \(\dfrac{x+46}{20}=x.\dfrac{2}{5}\)
\(\dfrac{x+46}{20}=\dfrac{2x}{5}\)
\(\Rightarrow5.\left(x+46\right)=2x.20\)
\(\Rightarrow5x+230=40x\)
\(\Rightarrow5x-40x=-230\)
\(\Rightarrow-35x=-230\)
\(\Rightarrow x=-230:-35\)
\(\Rightarrow x=\dfrac{46}{7}\)
Chúc bạn học tốt!
\(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}\Leftrightarrow\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}\Rightarrow x=\dfrac{1}{10}\)
a: \(\Leftrightarrow-2x=-\dfrac{1}{3}-\dfrac{1}{8}-\dfrac{5}{7}=-\dfrac{197}{168}\)
hay x=197/336
c: \(\Leftrightarrow5x=9+\dfrac{6}{18}-\dfrac{2}{7}=\dfrac{190}{21}\)
hay x=38/21
\(\dfrac{x-2}{5}=\dfrac{1-x}{6}\\ =>\left(x-2\right)\cdot6=\left(1-x\right)\cdot5\\ =>6x-12=5-5x\\ =>6x+5x=5+12\\ =>11x=17\\ x=\dfrac{17}{11}\)
Ta có: \(\dfrac{-4}{15}< \dfrac{5x-1}{18}< \dfrac{5}{12}\)
\(\Leftrightarrow\dfrac{-48}{180}< \dfrac{10\left(5x-1\right)}{180}< \dfrac{75}{180}\)
Suy ra: \(-48< 10\left(5x-1\right)< 75\)
\(\Leftrightarrow10\left(5x-1\right)\in\left\{-40;-30;-20;-10;0;10;20;30;40;50;60;70\right\}\)
\(\Leftrightarrow5x-1\in\left\{-4;-3;-2;-1;0;1;2;3;4;5;6;7\right\}\)
\(\Leftrightarrow5x\in\left\{-3;-2;-1;0;1;2;3;4;5;6;7;8\right\}\)
\(\Leftrightarrow x\in\left\{0;1\right\}\)(Vì x nguyên)