Ta quy đồng:\(\dfrac{3}{5}=\dfrac{18}{30};\dfrac{5}{6}=\dfrac{25}{30}\)
Vậy ta có : \(\dfrac{18}{30}< x< \dfrac{25}{30}\)
=> x có thể = \(\dfrac{19;20;21;22;23;24}{30}\)
Ta quy đồng:\(\dfrac{3}{5}=\dfrac{18}{30};\dfrac{5}{6}=\dfrac{25}{30}\)
Vậy ta có : \(\dfrac{18}{30}< x< \dfrac{25}{30}\)
=> x có thể = \(\dfrac{19;20;21;22;23;24}{30}\)
Tìm x, biết
a)\(\dfrac{1}{2}\)x\(x\)-\(\dfrac{7}{3}\)=\(\dfrac{-5}{6}\)+\(\dfrac{3}{4}\)x\(x\)
b)\(\dfrac{4}{5}\)x\(x\)-\(\dfrac{6}{5}\)=\(\dfrac{1}{2}\)+\(\dfrac{3}{2}\)x\(x\)
c)\(\dfrac{2}{5}\)x(3x\(x\)+\(\dfrac{3}{4}\))=\(1\dfrac{1}{5}\)-\(\dfrac{1}{3}\)x\(x\)
d)2x(3x\(x \)+\(\dfrac{3}{4}\))+\(\dfrac{4}{5}\)=\(\dfrac{1}{2}\)-2x\(x\)
Tìm x :
a, \(\dfrac{11}{12}\) - (\(\dfrac{2}{5}\) +x) = \(\dfrac{2}{3}\) b, x+\(\dfrac{2}{3}\) =\(\dfrac{3}{5}\) - \(\dfrac{-1}{6}\) c, x-[\(\dfrac{17}{2}\) -(\(\dfrac{-3}{7}\) + \(\dfrac{5}{3}\) )]=\(\dfrac{-1}{3}\)
d,\(\dfrac{9}{2}\) - [\(\dfrac{2}{3}\) - (x+\(\dfrac{7}{4}\) )] =\(\dfrac{-5}{4}\)
Mọi người giúp mình nha
\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\)
\(\dfrac{x+1}{22}=\dfrac{6}{x}\)
\(\dfrac{2x-1}{2}=\dfrac{5}{x}\)
\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\)
\(\dfrac{2x+1}{9}=\dfrac{5}{x+1}\)
Tìm x
Tìm x:
a) \(\dfrac{11}{12}\) - (\(\dfrac{2}{5}\) + \(\dfrac{3}{4}\)x) \(\dfrac{2}{3}\)
b) \(\dfrac{-2}{5}\) + \(\dfrac{5}{3}\) . (\(\dfrac{3}{2}\) - \(\dfrac{4}{15}\)x) = \(\dfrac{-7}{6}\)
c) \(\dfrac{1}{2}\) + \(\dfrac{3}{4}\)x = \(\dfrac{1}{4}\)
Tìm số nguyên x, biết:
a) \(-4\dfrac{3}{5}\). \(2\dfrac{4}{3}\) < x < \(-2\dfrac{3}{5}\) : \(1\dfrac{6}{15}\)
b) \(-4\dfrac{1}{3}\).(\(\dfrac{1}{2}\)-\(\dfrac{1}{6}\)) < x < - \(\dfrac{2}{3}\).(\(\dfrac{1}{3}\) - \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\))
\(b,\dfrac{x}{2}-\left(\dfrac{3x}{5}-\dfrac{13}{5}\right)=-\left(\dfrac{7}{5}+\dfrac{7}{10}.x\right)\)
\(c,\dfrac{2x-3}{3}+\dfrac{-3}{2}=\dfrac{5-3x}{6}-\dfrac{1}{3}\)
\(d,\dfrac{2}{3x}-\dfrac{3}{12}=\dfrac{4}{5}-\left(\dfrac{7}{x}-2\right)\)
\(e,2\left(x-1\right)=\left(x-1\right)^2\)
Bài 1: Tìm x; y ϵ \(ℤ\)
a) 2x - y\(\sqrt{6}\) = 5 + (x + 1)\(\sqrt{6}\)
b) 5x + y - (2x -1)\(\sqrt{7}\) = y\(\sqrt{7}\) + 2
Bài 2: So sánh M và N
M = \(\dfrac{\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{6}{4}+\dfrac{6}{5}+\dfrac{6}{7}-\dfrac{6}{11}}\)
N = \(\dfrac{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}{\dfrac{6}{2}+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}\)
Bài 3: Chứng minh:
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)
\(\dfrac{x-5}{3}=\dfrac{-12}{5-x}\)
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2023}{2024}\)
tính hợp lí
a,\(\dfrac{27}{2}\) x 7,5 + \(\dfrac{27}{2}\) x 2,5 -150
b,3\(^3\)x \(\dfrac{18}{5}\) - 3\(^3\) x 2\(\dfrac{2}{5}\) - 3\(^3\) x \(\dfrac{6}{5}\)
Tìm x
\(a,3-x=x+1,8\)
\(b,2x-5=7x+35\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(k,7x^2-11=6x^2-2\)
\(m,5\left(x+3.2^3\right)=10^2\)
\(n,\dfrac{4}{9}-(\dfrac{1}{6^2})=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)