cú t ccau oiii ét ô ét giải giúp t bài này zới
4,7 : 0,25 + 5,3 x 4
3x ( a - 2) +150 = 240
1/9 + a + 7/12 = 17/18
( 1/2 x 1/3 + 1/3 x 1/4 + 1/4 x 1/5 + 1/5 x 1/6 + 1/6 x 1/7 + 1/7 x 1/8 ) x a = 9/16
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22 tháng 6 2016
\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
= \(\frac{1}{4}+\frac{1}{2}\)
= \(\frac{3}{4}\)
b)\(-\frac{7}{3}.\frac{5}{9}+\frac{4}{9}.\left(-\frac{3}{7}\right)+\frac{17}{7}\)
=\(-\frac{35}{27}+\left(-\frac{4}{21}\right)+\frac{17}{7}\)
= \(-\frac{35}{27}+\frac{47}{21}\)
= \(\frac{178}{189}\)
c) \(\frac{117}{13}-\left(\frac{2}{5}+\frac{57}{13}\right)\)
= \(\frac{117}{13}-\frac{311}{65}\)
= \(\frac{274}{65}\)
d) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{3}+\frac{5}{2}\)
= \(\frac{1}{3}+\frac{5}{2}\)
= \(\frac{17}{6}\)
7 tháng 2 2022
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 \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 9
Bài 1:
a; \(\dfrac{7}{8}\) + \(x\) = \(\dfrac{4}{7}\)
\(x\) = \(\dfrac{4}{7}\) - \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{32}{56}\) - \(\dfrac{49}{56}\)
\(x=-\) \(\dfrac{49}{56}\)
Vậy \(x=-\dfrac{49}{56}\)
b; 6 - \(x\) = - \(\dfrac{3}{4}\)
\(x\) = 6 + \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{24}{4}+\dfrac{3}{4}\)
\(x=\dfrac{27}{4}\)
Vậy \(x=\dfrac{27}{4}\)
c; \(\dfrac{1}{-5}\) + \(x\) = \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{3}{4}\) + \(\dfrac{1}{5}\)
\(x=\dfrac{15}{20}\) + \(\dfrac{4}{20}\)
\(x=\dfrac{19}{20}\)
Vậy \(x=\dfrac{19}{20}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 9
Bài 1:
d; - 6 - \(x\) = - \(\dfrac{3}{5}\)
\(x\) = - 6 + \(\dfrac{3}{5}\)
\(x=-\dfrac{30}{5}\) + \(\dfrac{3}{5}\)
\(x=-\dfrac{27}{5}\)
Vậy \(x=-\dfrac{27}{5}\)
e; - \(\dfrac{2}{6}\) + \(x\) = \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{5}{7}\) + \(\dfrac{2}{6}\)
\(x\) = \(\dfrac{15}{21}\) + \(\dfrac{1}{3}\)
\(x=\dfrac{15}{21}\) + \(\dfrac{7}{21}\)
\(x=\dfrac{22}{21}\)
Vậy \(x=\dfrac{22}{21}\)
f; - 8 - \(x\) = - \(\dfrac{5}{3}\)
\(x\) = \(-\dfrac{5}{3}\) + 8
\(x\) = \(\dfrac{-5}{3}\) + \(\dfrac{24}{3}\)
\(x\) = \(\dfrac{-19}{3}\)
Vậy \(x=-\dfrac{19}{3}\)
11 tháng 7 2017
1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
11 tháng 7 2017
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)
MH
7 tháng 5 2022
+) \(\dfrac{1}{3}x=-\dfrac{4}{3}-\dfrac{1}{2}=-\dfrac{11}{6}\)
\(x=-\dfrac{11}{6}:\dfrac{1}{3}=-\dfrac{11}{2}\)
+) \(\dfrac{4}{3}x=-\dfrac{2}{3}+\dfrac{1}{2}=-\dfrac{1}{6}\)
\(x=-\dfrac{1}{6}:\dfrac{4}{3}=-\dfrac{1}{8}\)
+) \(2\left(x-1\right)=\dfrac{5}{2}+\dfrac{2}{3}=\dfrac{19}{6}\)
\(x-1=\dfrac{19}{12}\)
\(x=\dfrac{31}{12}\)
O
7 tháng 5 2022
\(\dfrac{1}{3}x+\dfrac{1}{2}=-\dfrac{4}{3}\)
\(\dfrac{1}{3}x=\left(-\dfrac{4}{3}\right)-\dfrac{1}{2}\)
\(\dfrac{1}{3}x=-\dfrac{11}{6}\)
\(x=\left(-\dfrac{11}{6}\right):\dfrac{1}{3}\)
\(x=-\dfrac{11}{2}\)
\(-\dfrac{2}{3}-\dfrac{4}{3}x=-\dfrac{1}{2}\)
\(\dfrac{4}{3}x=\left(-\dfrac{2}{3}\right)-\dfrac{-1}{2}\)
\(\dfrac{4}{3}x=-\dfrac{1}{6}\)
\(x=\left(-\dfrac{1}{6}\right):\dfrac{4}{3}\)
\(x=-\dfrac{1}{8}\)
\(\dfrac{5}{2}-2\left(x-1\right)=-\dfrac{2}{3}\)
\(2\left(x-1\right)=\dfrac{5}{2}-\left(-\dfrac{2}{3}\right)\)
\(2\left(x-1\right)=\dfrac{19}{6}\)
\(\left(x-1\right)=\dfrac{19}{6}:2\)
\(x-1=\dfrac{19}{12}\)
\(x=\dfrac{19}{12}+1\)
\(x=\dfrac{31}{12}\)
\(4,7\div0,25+5,3\times4\)
\(=18,8+21,2\)
\(=40\)
\(3\times\left(a-2\right)+150=240\)
\(3\times\left(a-2\right)=90\)
\(a-2=30\)
\(a=32\)
\(\dfrac{1}{9}+a+\dfrac{7}{12}=\dfrac{17}{18}\)
\(\dfrac{1}{9}+a=\dfrac{13}{36}\)
\(a=\dfrac{1}{4}\)
\(\left(\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+\dfrac{1}{4}\times\dfrac{1}{5}+\dfrac{1}{5}\times\dfrac{1}{6}+\dfrac{1}{6}\times\dfrac{1}{7}+\dfrac{1}{7}\times\dfrac{1}{8}\right)\times a=\dfrac{9}{16}\)
\(\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}\right)\times a=\dfrac{9}{16}\)
\(\left(\dfrac{1}{2}-\dfrac{1}{8}\right)\times a=\dfrac{9}{16}\)
\(\dfrac{3}{8}\times a=\dfrac{9}{16}\)
\(a=\dfrac{3}{2}\)