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Giải:
Theo bài ra ta có:
\(\frac{-5}{6}+\frac{8}{3}+\frac{29}{-6}\le x\le\frac{-1}{2}+2+\frac{5}{12}\)
\(\Rightarrow-3\le x\le\frac{23}{12}\)
\(\Rightarrow x\varepsilon\left\{-2;-1;0;1\right\}\)
\(\frac{-5}{6}+\frac{16}{6}+-\frac{29}{6}\le x\le\frac{-6}{12}+\frac{24}{12}+\frac{5}{12}\)
=>-3\(\le\) x\(\le\) 23/12
=> x thuộc{-2-1;0;1}
Ta có: \(\frac{73-x}{98-x}=\frac{1}{6}\)
\(\Rightarrow6\left(73-x\right)=98-x\)
\(\Rightarrow438-6x=98-x\)
\(\Rightarrow438-98=-x+6x\)
\(\Rightarrow340=5x\)
\(\Rightarrow x=340:5=68\)
\(\frac{73-x}{98-x}=\frac{1}{6}\)
=> \(6\left(73-x\right)=98-x\)
=> \(438-6x=98-x\)
\(\Leftrightarrow-6x+x=98-438\)
\(\Leftrightarrow-5x=-340\)
\(\Leftrightarrow x=68\)
Câu hỏi của Nguyễn Thị My Na - Toán lớp 6 - Học toán với OnlineMath bạn tham khảo tại đây nha
\(\frac{x}{6}-\frac{1}{y}=\frac{1}{2}\)
\(\Rightarrow\frac{x}{6}=\frac{1}{y}+\frac{1}{2}\)
\(\Rightarrow\frac{x}{6}=\frac{2+y}{2y}\)
\(6=2y\Rightarrow y=3\)
\(x=2+3\Rightarrow x=5\)
a) \(\frac{x}{3}-\frac{10}{21}=-\frac{1}{7}\)
\(\Rightarrow\frac{x}{3}=-\frac{1}{7}+\frac{10}{21}\)
\(\Rightarrow\frac{x}{3}=\frac{7}{21}\)
\(\Rightarrow\frac{x}{3}=\frac{1}{3}\)
\(\Rightarrow x=1\)
\(x-25\%=\frac{1}{2}\)
\(\Rightarrow x-\frac{1}{4}=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}+\frac{1}{4}\)
\(\Rightarrow x=\frac{3}{4}\)
c) \(-\frac{5}{6}+\frac{8}{3}+-\frac{29}{6}\le x\le-\frac{1}{2}+2+\frac{5}{2}\)
\(\Rightarrow-3\le x\le4\)
\(\Rightarrow x\in\left\{-3;-2;-1;0;1;2;3;4\right\}\)
c)\(\frac{1}{2}x+\frac{1}{8}x=\frac{3}{4}\)
\(\Rightarrow x.\left(\frac{1}{2}-\frac{1}{8}\right)=\frac{3}{4}\)
\(\Rightarrow x.\frac{3}{8}=\frac{3}{4}\)
=>x\(=\frac{3}{4}:\frac{3}{8}\)
=>x=\(2\)
a)\(x+\frac{1}{6}=\frac{-3}{8}\)
=>\(x=\frac{-3}{8}-\frac{1}{6}\)
=>\(x=\frac{-9}{24}-\frac{4}{24}\)
=>\(x=\frac{-13}{24}\)
b)\(2-\left|\frac{3}{4}-x\right|=\frac{7}{12}\)
=>\(\left|\frac{3}{4}-x\right|=2-\frac{7}{12}\)
=>\(\left|\frac{3}{4}-x\right|=\frac{24}{12}-\frac{7}{12}\)
\(\Rightarrow\left|\frac{3}{4}-x\right|=\frac{17}{12}\)
TH1: \(\frac{3}{4}-x=\frac{17}{12}\)
=>x=\(\frac{3}{4}-\frac{17}{12}\)
=>x=\(x=-\frac{2}{3}\)
TH2:\(\frac{3}{4}-x=-\frac{17}{12}\)
=>\(x=\frac{3}{4}-\left(-\frac{17}{12}\right)\)
=>x=\(x=\frac{13}{6}\)
Dzồi nhìu phết
ta có : \(\frac{y}{-78}=\frac{-y}{78};\frac{7}{-6}=\frac{-7}{6}\)
\(\frac{-y}{78}=\frac{-7}{6}\)
=> ( -y ) . 6 = ( -7 ) . 78
=> -y = \(\frac{\left(-7\right).78}{6}=-91\)
=> y = 91
\(\frac{x}{102}=\frac{-91}{78}\)
=> x . 78 = ( -91 ) . 102
=> x = \(\frac{\left(-91\right).102}{78}=-119\)
vậy x = -119 ; y = 91
Theo đề ta có: \(x=\frac{102.7}{-6}=\frac{714}{-6}=-119\)
\(y=\frac{-78.7}{-6}=\frac{-546}{-6}=91\)
Vậy \(x=-119;y=91\)
Ko viết lại đề bài nha
=> (73 - x) . 6 = (98 - x) . 1
438 - 6x = 98 - 1x
-6x + 1x = 98 - 438
-5x = - 340
x = 68
\(\frac{73-x}{98-x}=\frac{1}{6}\)
\(\Rightarrow\left(73-x\right).6=1.\left(98-x\right)\)
\(\Rightarrow\left(73-x\right).6=98-x\)
\(\Rightarrow73+x.6=98-x\)