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a, \(\Leftrightarrow2x^2=72\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow x=\pm6\)
Vậy ...
\(b,\Leftrightarrow\dfrac{3}{5}x-0,75=2\dfrac{4}{5}.\dfrac{3}{7}=\dfrac{6}{5}\)
\(\Leftrightarrow\dfrac{3}{5}x=\dfrac{6}{5}+0,75=\dfrac{39}{20}\)
\(\Leftrightarrow x=\dfrac{39}{20}:\dfrac{3}{5}=\dfrac{13}{4}\)
Vậy ...
\(c,\Leftrightarrow2x=1\dfrac{5}{6}.\dfrac{6}{11}-\dfrac{3}{10}=\dfrac{7}{10}\)
\(\Leftrightarrow x=\dfrac{7}{10}:2=\dfrac{7}{20}\)
Vậy ...
\(d,\Leftrightarrow\dfrac{1}{x-7\dfrac{1}{3}}=1.5:2\dfrac{1}{4}=\dfrac{2}{3}\)
\(\Leftrightarrow x-7\dfrac{1}{3}=\dfrac{3}{2}\)
\(\Leftrightarrow x=\dfrac{3}{2}+7\dfrac{1}{3}=\dfrac{53}{6}\)
Vậy ...
a) 2x2 - 72 = 0
\(\Rightarrow\) 2x2 = 72
\(\Rightarrow\) x2 = 36 = 62 = (- 6)2
\(\Rightarrow\) x = 6 hoặc x = - 6
Vậy x = 6 hoặc x = - 6
b) (\(\dfrac{3}{5}\)x - 0,75) : \(\dfrac{3}{7}\) = \(2\dfrac{4}{5}\)
\(\Rightarrow\) (\(\dfrac{3}{5}\)x - 0,75) : \(\dfrac{3}{7}\) = \(\dfrac{14}{5}\)
\(\Rightarrow\) \(\dfrac{3}{5}\)x - \(\dfrac{3}{4}\) = \(\dfrac{6}{5}\)
\(\Rightarrow\) \(\dfrac{3}{5}\)x = \(\dfrac{39}{20}\)
\(\Rightarrow\) x = \(\dfrac{13}{4}\)
Vậy x = \(\dfrac{13}{4}\)
a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
tìm x biết:
(3x-1) [- 1/2x+5]=0
1/4+1/3:(2x-1)=-5
[2x+3/5]2 - 9/25=0
-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6
[x+1/2]x [2/3-2x]=0
17/2-|2x-3/4|=-7/4
2/3x-1/2x =5/12
(x+1/5)2+17/25=26/25
[x.44/7+3/7].11/5-3/7=-2
3[3x-1/2]+1/9=0
Toán lớp 6Tìm x
Trả lời Câu hỏi tương tự
Chưa có ai trả lời câu hỏi này,bạn hãy là người đâu tiên giúp nguyenvanhoang giải bài toán này !
Bài 1:
a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)
b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)
c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)
d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)
hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)
e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)
hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)
Bài 2:
a: =>3/4x=-3/5-1/2=-11/10
\(\Leftrightarrow x=\dfrac{-11}{10}:\dfrac{3}{4}=\dfrac{-11}{10}\cdot\dfrac{4}{3}=-\dfrac{44}{30}=-\dfrac{22}{15}\)
b: \(\Leftrightarrow x+\dfrac{3}{4}x=\dfrac{1}{3}+\dfrac{5}{4}=\dfrac{19}{12}\)
=>7/4x=19/12
=>x=19/21
c: \(\Leftrightarrow-\dfrac{2}{3}x+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\)
=>-4/3x=-1/3-1/6=-1/2
=>x=1/2:4/3=1/2x3/4=3/8
Bài 1
\(\dfrac{1}{7}:\dfrac{5}{17}-\dfrac{3}{2}.\left(\dfrac{1}{6}-\dfrac{7}{12}\right)\)
\(\dfrac{1}{7}.\dfrac{17}{5}-\dfrac{3}{2}.\left(-\dfrac{5}{12}\right)\)
\(\dfrac{17}{35}-\left(-\dfrac{5}{8}\right)\)
\(\dfrac{17}{35}+\dfrac{5}{8}\)
\(\dfrac{311}{280}\)
\(\frac{2}{3}.x-\frac{3}{2}.\left(x-\frac{1}{2}\right)=\frac{5}{12}\)
\(\Leftrightarrow\frac{2}{3}.x-\frac{3}{2}.x+\frac{3}{4}=\frac{5}{12}\)
\(\Leftrightarrow x.\left(\frac{2}{3}-\frac{3}{2}\right)=\frac{5}{12}-\frac{3}{4}\)
\(\Leftrightarrow x.\frac{-5}{6}=\frac{-4}{12}\)
\(\Leftrightarrow x=\frac{-4}{12}:\frac{-5}{12}\)
\(\Leftrightarrow x=\frac{4}{5}\)
Vậy : \(x=\frac{4}{5}\)
Đề câu 2 bạn viết lại hộ mình nhé ! vũ thị diệu hương