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dễ mk bn cho mình hỏi nhé câu 4 là \(\frac{1}{2\cdot3}\)hay là\(\frac{1}{2}\cdot3\)
Câu 1:
A = \(\dfrac{-7}{12}+\dfrac{11}{8}-\dfrac{5}{9}=\dfrac{-42}{72}+\dfrac{99}{72}-\dfrac{40}{72}=\dfrac{-42+99-40}{72}=\dfrac{17}{72}\)
\(B=\dfrac{1}{7}-\dfrac{8}{7}:8-3:\dfrac{3}{4}.2^2=\dfrac{1}{7}-\dfrac{8}{7}.\dfrac{1}{8}-3.\dfrac{4}{3}.4=\dfrac{1}{7}-\dfrac{1}{7}-16=0-16=-16\)
\(C=1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}=\dfrac{7}{5}.\dfrac{15}{49}-\dfrac{22}{15}.\dfrac{5}{11}=\dfrac{3}{7}-\dfrac{2}{3}=\dfrac{9-14}{21}=-\dfrac{5}{21}\)
Vậy A=\(\dfrac{17}{72};B=-16;C=\dfrac{-5}{21}\)
Câu 2:
a. \(\dfrac{-11x}{12}+\dfrac{3}{4}=-\dfrac{1}{6}\)
\(\Rightarrow\dfrac{-11x}{12}=-\dfrac{1}{6}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{-11x}{12}=\dfrac{-2}{12}-\dfrac{9}{12}\)
\(\Rightarrow\dfrac{-11x}{12}=\dfrac{-11}{12}\)
\(\Rightarrow-11x=\dfrac{-11.12}{12}\)
\(\Rightarrow-11x=-11\Rightarrow x=1\)
Vậy x=1
b. \(3-(\dfrac{1}{6}-x).\dfrac{2}{3}=\dfrac{2}{3}\Rightarrow3-\left(\dfrac{1}{6}-x\right)=1\)
\(\Rightarrow-(\dfrac{1}{6}-x)=1-3\Rightarrow\dfrac{1}{6}+x=-2\)
\(\Rightarrow x=2-\dfrac{1}{6}\Rightarrow x=\dfrac{11}{6}\)
Vậy x = \(\dfrac{11}{6}\)
Câu 4:
Ta có: \(\dfrac{1}{2.3}=\dfrac{1}{6}\)
\(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{3}{6}-\dfrac{2}{6}=\dfrac{1}{6}\)
\(\Rightarrow\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)
các bn lm đến đâu cx dc miễn là lm hộ mk cái ạ, ai đang lm vào nhắn tin vs mk để mk bít nha
a; \(-\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{71}{15}< x< -\dfrac{13}{7}+\dfrac{19}{14}-\dfrac{7}{2}\)
-\(\dfrac{19}{15}\) - \(\dfrac{71}{15}\) < \(x\) < -\(\dfrac{1}{2}\) - \(\dfrac{7}{2}\)
-6 < \(x\) < -4
vì \(x\) \(\in\) Z nên \(x\) = -5
\(-\frac{1}{10}< =x< =\frac{3}{5}\)
\(\frac{-4}{9}< x< =\frac{2}{3}\)
Câu 1 :
a, 8.( -5 ).( -4 ).2
= [ 8.2 ].[( -5 ).(-4 ]
= 16.20
= 320
b, \(1\frac{3}{7}+\frac{-1}{3}+2\frac{4}{7}\)
\(=\frac{10}{7}+\frac{-1}{3}+\frac{18}{7}\)
\(=\frac{11}{3}\)
c, \(\frac{8}{5}.\frac{2}{3}+\frac{-5.5}{3.5}\)
\(=\frac{8}{3}+\frac{-5}{3}\)
\(=\frac{3}{3}=1\)
d, \(\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}.\left(-2\right)^2\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3}{16}.4\)
\(=\frac{55}{56}-\frac{3}{4}\)
\(=\frac{13}{56}\)
Câu 2 :
a, 2x + 10 = 16
2x = 16 + 10
2x = 26
x = 26 : 2
x = 13
b, \(x-\frac{1}{3}=\frac{5}{4}\)
\(x=\frac{5}{4}+\frac{1}{3}\)
\(x=\frac{19}{12}\)
c, \(2x+3\frac{1}{3}=7\frac{1}{3}\)
\(2x+\frac{10}{3}=\frac{22}{3}\)
\(2x=\frac{22}{3}-\frac{10}{3}\)
\(2x=4\)
\(x=4:2\)
\(x=2\)
d, \(\left(\frac{2}{11}+\frac{1}{3}\right)x=\left(\frac{1}{7}-\frac{1}{8}\right).56\)
\(\frac{17}{33}x=1\)
\(x=1-\frac{17}{33}\)
\(x=\frac{16}{33}\)
Câu 1:a) \(\left(\frac{-5}{12}+\frac{6}{11}\right)+\left(\frac{7}{17}+\frac{5}{11}+\frac{5}{12}\right)\)
\(=\left(\frac{-5}{12}+\frac{5}{12}\right)+\left(\frac{6}{11}+\frac{5}{11}\right)+\frac{7}{17}\)
\(=0+1+\frac{7}{17}\)
\(=\frac{17}{17}+\frac{7}{17}\)
\(=\frac{24}{17}\)
b) \(\frac{7}{12}-\left(\frac{5}{12}-\frac{5}{6}\right)\)
\(=\frac{7}{12}-\frac{5}{12}+\frac{5}{6}\)
\(=\frac{7}{12}-\frac{5}{12}+\frac{10}{12}\)
\(=\frac{7-5+10}{12}\)
\(=1\)
c) \(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{12}+\frac{1}{30}\)
\(=\frac{5}{60}+\frac{2}{60}\)
\(=\frac{7}{60}\)
Câu 2:a) \(\frac{x}{8}=2+\frac{-3}{2}\)
\(\Leftrightarrow\frac{x}{8}=\frac{4-3}{2}\)
\(\Leftrightarrow\frac{x}{8}=\frac{1}{2}\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=\frac{8}{2}\)
\(\Leftrightarrow x=4\)
b) \(\frac{-5}{6}+\frac{8}{3}+\frac{29}{-6}\le x\le\frac{-1}{2}+2+\frac{5}{2}\)
\(\Leftrightarrow\frac{-18}{6}\le x\le4\)
\(\Leftrightarrow-3\le x\le4\)
\(\Leftrightarrow x\in\left\{-3;-2;-1;0;1;2;3;4\right\}\)
\(\dfrac{-7}{12}+\dfrac{11}{8}-\dfrac{5}{9}=\dfrac{-7}{12}+\dfrac{59}{72}=\dfrac{17}{72}\) \(\dfrac{1}{7}-\dfrac{8}{7}:8-3:\dfrac{3}{4}.\left(-2\right)^2=\dfrac{1}{7}-\dfrac{1}{7}-4.4=\left(-16\right)\) \(\dfrac{1}{4}.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}=\dfrac{15}{49}-\dfrac{22}{15}:\dfrac{11}{5}=\dfrac{15}{49}-\dfrac{2}{3}=\dfrac{-53}{147}\)
\(\dfrac{-11x}{12}+\dfrac{3}{4}=\dfrac{1}{6}\Leftrightarrow\dfrac{-11x}{12}=\dfrac{1}{6}-\dfrac{3}{4}\Leftrightarrow\dfrac{-11x}{12}=\dfrac{-7}{12}\Leftrightarrow-11x=-7\Leftrightarrow x=\dfrac{7}{11}\)