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a) \(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30.75\right).x-8=\left(\dfrac{3}{5}+0.415\right)\)
\(=\left(\dfrac{1}{12}+3\dfrac{1}{6}-\dfrac{123}{4}\right).x-8=\left(\dfrac{3}{5}+\dfrac{83}{200}\right)\)
\(=\dfrac{-55}{2}.x-8=\dfrac{203}{200}\)\(=\dfrac{-55}{2}.x=\dfrac{203}{200}+8=\dfrac{1803}{200}\)
\(x=\dfrac{1803}{200}:\dfrac{-55}{2}=\dfrac{-1803}{5500}\)
a, \(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,75\right).x-8=\dfrac{3}{5}+0,415\)
\(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,75\right).x-8=\dfrac{203}{200}\)
\(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,75\right).x=\dfrac{203}{200}+8\)
\(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,75\right).x=\dfrac{1803}{200}\)
\(\left(\dfrac{13}{4}-30,75\right).x=\dfrac{1803}{200}\)
\(\dfrac{-55}{2}.x=\dfrac{1803}{200}\)
\(x=\dfrac{1803}{200}:\dfrac{-55}{2}\)
\(x=\dfrac{-1803}{5500}\)
Nếu là tìm số nguyên thì hình như đề sai rồi bạn
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b, \(4\dfrac{1}{3}.\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\le x\le\dfrac{2}{3}.\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)
Cho \(A=4\dfrac{1}{3}.\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\)
\(A=4\dfrac{1}{3}.\dfrac{-1}{3}\)
\(A=\dfrac{13}{3}.\dfrac{-1}{3}\)
\(A=\dfrac{-13}{9}\)
Cho \(B=\dfrac{2}{3}.\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)
\(B=\dfrac{2}{3}.\left(\dfrac{-1}{6}-\dfrac{3}{4}\right)\)
\(B=\dfrac{2}{3}.\dfrac{-11}{12}\)
\(B=\dfrac{-11}{18}\)
Ta có: \(A\le x\le B\)
\(\dfrac{-13}{9}\le x\le\dfrac{-11}{18}\)
\(\Rightarrow x=-1\)
Bài 4:
=>(x-5)*3/10=1/5x+5
=>3/10x-3/2=1/5x+5
=>1/10x=5+3/2=6,5
=>0,1x=6,5
=>x=65
a, (\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)).10 - x = 0
<=> \(\dfrac{5}{6}.10-x=0\)
<=> \(\dfrac{25}{3}-x=0\)
<=> x = \(\dfrac{25}{3}\) (thỏa mãn)
@Hoàng Mạnh Quân
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
a: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
b: 
c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
Tìm x biết:
\(\left(4,5-2x\right).1\dfrac{4}{7}=\dfrac{11}{4}\)
\(\Leftrightarrow\left(4,5-2x\right)=\dfrac{11}{4}:1\dfrac{4}{7}\)
\(\Leftrightarrow4,5-2x=\dfrac{7}{4}\)
\(\Leftrightarrow2x=4,5-\dfrac{7}{4}\)
\(\Leftrightarrow2x=\dfrac{11}{4}\)
Vậy \(x=\dfrac{11}{8}\)
Tìm số nguyên x biết:
Theo đề bài, ta có:
\(4\dfrac{1}{3}\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\le x\le\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)
\(\Leftrightarrow-\dfrac{13}{9}\le x\le-\dfrac{11}{18}\) hay \(-\dfrac{26}{18}\le x\le-\dfrac{11}{18}\)
\(\Leftrightarrow-1,\left(4\right)\le x\le-0,6\left(1\right)\)
Mà \(x\in Z\) nên x=-1
Vậy x = -1
a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)
<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)
<=>\(4x-17=0\)
<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)
Câu 1:
a, \(\left|-5\right|=5\)
b, \(\left|10\right|=10\)
c, \(\left|-5\right|-\left|10\right|=5-10=-5\)
d, -15.30= -450
Câu 2:
a, Ta có: \(\dfrac{10}{21}.\dfrac{14}{25}=\dfrac{10.14}{21.25}=\dfrac{5.2.7.2}{3.7.5.5}=\dfrac{2.2}{3.5}=\dfrac{4}{15}\)
c, Ta có: \(-\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{-5.2+3.3}{12}=\dfrac{-10+9}{12}=\dfrac{-1}{12}\)
d, \(\dfrac{11}{17}.\dfrac{3}{2017}+\dfrac{11}{17}.\dfrac{2014}{2017}-1\dfrac{11}{17}=\dfrac{11}{17}\left(\dfrac{3}{2017}+\dfrac{2014}{2017}\right)-1\dfrac{11}{17}\)
\(=\dfrac{11}{17}.\dfrac{2017}{2017}-1\dfrac{11}{17}=\dfrac{11}{17}-1-\dfrac{11}{17}=-1\)
Câu 7: a, Để A có nghĩa khi \(x+2\ne0\) \(\Leftrightarrow x=-2\)
b, Ta có: \(A=2\)
<=> \(\dfrac{x-1}{x+2}=2\)
<=> \(\dfrac{x-1}{x+2}-2=0\)
<=> \(\dfrac{x-1}{x+2}-\dfrac{2x+4}{x+2}=0\)
<=> \(\dfrac{x-1-2x-4}{x+2}=0\)
<=> \(\dfrac{-x-5}{x+2}=0\)
<=> -x-5=0
<=> -x=5
<=> x= -5