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\(\frac{x+1}{2}=\frac{x-2}{3}\)
\(\Rightarrow\left(x+1\right).3=\left(x-2\right).2\)
\(3x+3=2x-4\)
\(\Rightarrow3x-2x=-4-3\)
\(x=-7\)
KL: x= -7
Học tốt nhé bn !!
\(\frac{x+1}{2}=\frac{x-2}{3}\)
=> (x+1).3 = 2(x-2)
=> 3x + 3 = 2x - 2
=> 3 + 2 = 2x - 3x
=> 5 = -x
=> x = -5
Tìm x :
x - 0,27 = \(\frac{73}{100}\)
x = \(\frac{73}{100}+0,27\)
x = 1
Cậu P khó quá mik chưa nghĩ ra cách tính nhanh nhất !
Cậu tự giải nhé !
Hok tốt
\(\frac{4}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}.\left(\frac{-1}{6}+\frac{3}{4}\right)\)
\(\frac{4}{3}.\frac{-1}{3}\le x\le\frac{2}{3}.\frac{7}{12}\)
\(\frac{-4}{9}\le x\le\frac{7}{18}\)
\(\frac{-8}{18}\le x\le\frac{7}{18}\)
\(\Rightarrow\)X \(\in\) {\(\frac{-7}{18};\frac{-6}{18};\frac{-5}{18};\frac{-4}{18};\frac{-3}{18};\frac{-2}{18};\frac{-1}{18};0;\frac{1}{18};\frac{2}{18};\frac{3}{18};\frac{4}{18};\frac{5}{18};\frac{6}{18}\)}
1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)
\(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)
\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)
\(x=\frac{45}{4}+\frac{9}{4}\)
\(x=\frac{27}{2}\)
\(A=3-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}\)
\(A=3-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\)
\(A=3-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)\)
\(A=3-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(A=3-\left(1-\frac{1}{8}\right)\)
\(A=3-\frac{5}{8}\)
\(A=\frac{19}{8}\)
(1-1/3).(1-1/5).(1-1/7).(1-1/9).(1-1/11).(1-1/13).(1-1/2).(1-1/4).(1-1/6).(1-1/8).(1-1/10)
=2/3.4/5.6/7.8/9.10/11.12/13.1/2.3/4.5/6.7/8.9/10
=8/15.48/63.120/143.3/8.35/48.9/10
=384/945.360/1144.315/480
=138240/1081080.315/480
=43545600/518918400=84/1001
a) \(\frac{3x-6}{x+4}=\frac{2\left(x+5\right)+\left(x-3\right)}{x-2}\)
\(\frac{3\left(x-2\right)}{x+4}=\frac{2\left(x+5\right)+x-3}{x-2}\)
\(\frac{3\left(x-4\right)}{x+4}=\frac{3x+7}{x-2}\)
\(3\left(x-2\right)\left(x-2\right)=\left(3x+7\right)\left(x+4\right)\)
\(3\left(x-2\right)^2=\left(3x+7\right)\left(x+4\right)\)
\(3x^2-12x+12=3x^2+12x+7x+28\)
\(3x^2-12x+12=3x^2+19x+28\)
\(-12x+12=19x+28\)
\(12=19x+28+12x\)
\(19x+28+12x=12\) (chuyển vế)
\(31x+28=12\)
\(31x=12-28\)
\(31x=-16\)
\(x=-\frac{16}{31}\)
\(\Rightarrow x=-\frac{16}{31}\)
a/\(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)
=\(\frac{2^3.5^3.7^4}{2^2.5^2.7^4}\)
=2.5
=10
a (\(\frac{9}{2}\)x - \(\frac{16}{3}\)).\(\frac{1}{12}\) + \(\frac{1}{2}\)x=\(\frac{3}{2}\)
\(\frac{9}{2}\)x . \(\frac{1}{12}\) - \(\frac{16}{3}\) . \(\frac{1}{12}\) + \(\frac{1}{2}\)x=\(\frac{3}{2}\)
( \(\frac{9}{2}\)x . \(\frac{1}{12}\) + \(\frac{1}{2}\)x) - \(\frac{16}{3}\) . \(\frac{1}{12}\) = \(\frac{3}{2}\)
x. ( \(\frac{9}{2}\) . \(\frac{1}{12}\) +\(\frac{1}{2}\)) - \(\frac{4}{9}\) = \(\frac{3}{2}\)
x.\(\frac{7}{8}\) = \(\frac{3}{2}\) + \(\frac{4}{9}\) = \(\frac{35}{18}\)
x= \(\frac{35}{18}\) : \(\frac{7}{8}\) = \(\frac{20}{9}\) Vậy x=\(\frac{20}{9}\)
b 60%+2/3x=684
3/5x+2/3x=684
x(3/5+ 2/3) = 684
x. 19/15 = 684
x=540. Vậy x=540
Bạn Kiên giải đúng nhưng chưa rõ nên mình giải lại.
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{202}{201}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{202}{201}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{202}{201}\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{\left(x+1\right)}\right)=\frac{202}{201}\)
\(=2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{202}{201}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{\left(x+1\right)}=\frac{202}{201}:2=\frac{202}{402}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{202}{402}=-\frac{1}{402}=\frac{-1}{402}=\frac{1}{-402}\)
\(\Rightarrow\frac{1}{x+1}=\hept{\begin{cases}\frac{-1}{402}\\\frac{1}{-402}\end{cases}}\Rightarrow x+1=\hept{\begin{cases}402\\-402\end{cases}}\Rightarrow\hept{\begin{cases}x=402-1\\x=\left(-402\right)-1\end{cases}}\Rightarrow x=\hept{\begin{cases}401\\-403\end{cases}}\)
\(\Rightarrow A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}=\frac{202}{201}\)\(\Rightarrow A=2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{202}{201}\)
\(\Rightarrow A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{202}{201}\)
\(\Rightarrow A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{202}{201}\)
\(\Rightarrow A=2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{202}{201}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{202}{402}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{202}{402}=\frac{-1}{402}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{-402}\)
\(\Rightarrow x+1=-402\)
\(\Rightarrow x=-403\)