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\(-2\dfrac{1}{5}=-\dfrac{11}{5}\)
\(\dfrac{110}{-50}=-\dfrac{11}{5}\)
\(\Rightarrow-2\dfrac{1}{5}=\dfrac{110}{-50}\)
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\(\dfrac{x}{7}=\dfrac{y}{9}=\dfrac{z}{11}\)
Áp dụng tính chất của dãy tỉ sô bằng nhau ta có :
\(\dfrac{x}{7}=\dfrac{y}{9}=\dfrac{z}{11}=\dfrac{y-x}{9-7}=\dfrac{-1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{7}=\dfrac{-1}{2}\\\dfrac{y}{9}=\dfrac{-1}{2}\\\dfrac{z}{11}=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-7}{2}\\y=\dfrac{-9}{2}\\z=\dfrac{-11}{2}\end{matrix}\right.\)
Ta có : \(\dfrac{x}{7}=\dfrac{y}{9}=\dfrac{z}{11}\)
Áp dụng tính chất dãy tỉ số bằng nhâu , ta có:
\(\dfrac{x}{7}=\dfrac{y}{9}=\dfrac{z}{11}=\dfrac{y-x}{9-7}=-\dfrac{1}{2}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}.7=-\dfrac{7}{2}\\y=-\dfrac{1}{2}.9=-\dfrac{9}{2}\\z=-\dfrac{1}{2}.11=-\dfrac{11}{2}\end{matrix}\right.\)
Vậy \(x=-\dfrac{7}{2}\); \(y=-\dfrac{9}{2}\); \(z=-\dfrac{11}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{7}=\dfrac{y}{8}=\dfrac{y-x}{8-7}=\dfrac{4}{1}=4\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{7}=4\Rightarrow x=7.4=28\\\dfrac{y}{8}=4\Rightarrow y=8.4=32\end{matrix}\right.\)
Vậy..............
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a)\(\left|-0.75\right|+\dfrac{1}{4}-2\dfrac{1}{2}\)
=0.75+0.25-2.5
=1-2.5=-1.5
b)\(15.\dfrac{1}{5}:\left(\dfrac{-5}{7}\right)-2\dfrac{1}{5}.\left(\dfrac{-7}{5}\right)\)
=3.(-1.4)+3.08
=-4.2+3.08=-1.12
c)\(\dfrac{5}{17}+\dfrac{2}{3}-\dfrac{20}{12}+\dfrac{7}{9}+\dfrac{12}{17}\)
=\(\dfrac{49}{51}-\dfrac{5}{3}+\dfrac{7}{9}+\dfrac{12}{17}\)
=\(\dfrac{-12}{17}+\dfrac{7}{9}+\dfrac{12}{17}\)
=\(\dfrac{11}{153}+\dfrac{12}{17}\)
=\(\dfrac{7}{9}\)
d)\(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}\)
=\(\dfrac{67}{75}-\dfrac{4}{3}+\dfrac{2}{7}+\dfrac{11}{25}\)
=-0.44+\(\dfrac{127}{175}\)
=\(\dfrac{2}{7}\)
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\(\dfrac{17}{40}=\dfrac{7}{10}-x+\dfrac{13}{20}\)
\(\Rightarrow\dfrac{17}{40}=\dfrac{28}{40}-x+\dfrac{26}{40}\)
\(\Rightarrow\dfrac{17}{40}=\dfrac{44}{40}-x\)
\(\Rightarrow x=\dfrac{44}{40}-\dfrac{17}{40}\)
\(\Rightarrow x=\dfrac{27}{40}\)
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a) 1/20 - (x - 8/5) = 1/10
x - 8/5 = 1/20 - 1/10
x - 8/5 = -1/20
x = -1/20 + 8/5
x = 31/20
b) 7/4 - (x + 5/3) = -12/5
x + 5/3 = 7/4 + 12/5
x + 5/3 = 83/20
x = 83/20 - 5/3
x = 149/60
c) x - [17/2 - (-3/7 + 5/3)] = -1/3
x - (17/2 - 26/21) = -1/3
x - 305/42 = -1/3
x = -1/3 + 305/42
x = 97/14
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=\(\dfrac{11}{31}\)*\(\left(\dfrac{-2}{17}-\dfrac{-9}{17}\right)\)+\(\dfrac{7}{31}\)*\(\dfrac{-20}{17}\)
=\(\dfrac{11}{31}\)*\(\dfrac{-7}{17}\)+\(\dfrac{7}{31}\)*\(\dfrac{-20}{17}\)
=\(\dfrac{-77}{512}\)+\(\dfrac{-140}{512}\)
=\(\dfrac{-217}{512}\)
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a) Ta có: \(\dfrac{x}{-15}=\dfrac{-60}{x}\)
\(\Leftrightarrow x^2=\left(-15\right)\cdot\left(-60\right)=900\)
hay \(x\in\left\{30;-30\right\}\)
Vậy: \(x\in\left\{30;-30\right\}\)
b) Ta có: \(\left|x\right|+0.573=2\)
\(\Leftrightarrow\left|x\right|=1.427\)
hay \(x\in\left\{1.427;-1.427\right\}\)
Vậy: \(x\in\left\{1.427;-1.427\right\}\)
c) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=-1\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{8}{3};-\dfrac{10}{3}\right\}\)
d) Ta có: \(0.01:2.5=\left(0.75x\right):0.75\)
\(\Leftrightarrow\dfrac{0.75\cdot x}{0.75}=\dfrac{0.01}{2.5}\)
\(\Leftrightarrow x=\dfrac{1}{250}\)
Vậy: \(x=\dfrac{1}{250}\)
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a) \(-\dfrac{3}{5}-x=-0,75\)
\(\Rightarrow-\dfrac{3}{5}-x=-\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}-\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{15}{20}-\dfrac{12}{20}=\dfrac{8}{20}=\dfrac{2}{5}\)
b) \(1\dfrac{4}{5}=-0,15-x\)
\(\Rightarrow\dfrac{9}{5}=-\dfrac{3}{20}-x\)
\(\Rightarrow x=-\dfrac{3}{20}-\dfrac{9}{5}\)
\(\Rightarrow x=-\dfrac{3}{20}-\dfrac{36}{20}=-\dfrac{39}{20}\)
c) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\left(-\dfrac{1}{3}\right)\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{2}{5}+\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}+\dfrac{1}{3}-\dfrac{1}{3}=\dfrac{2}{5}\)
a) \(-\dfrac{3}{5}-x=-0,75\)
\(x=-\dfrac{3}{5}+0,75=\dfrac{3}{5}+\dfrac{3}{4}\)
\(x=\dfrac{27}{20}\)
________
b) \(1\dfrac{4}{5}=-0,15-x\)
\(=>-0,15-x=\dfrac{9}{5}\)
\(x=\dfrac{-3}{20}-\dfrac{9}{5}=\dfrac{-3}{20}-\dfrac{36}{20}\)
\(x=\dfrac{-39}{20}\)
c) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\left(-\dfrac{1}{3}\right)=\dfrac{6}{15}+\dfrac{5}{15}\)
\(x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(x=\dfrac{11}{15}-\dfrac{1}{3}=\dfrac{11}{15}-\dfrac{5}{15}\)
\(x=\dfrac{6}{15}=\dfrac{2}{5}\)
Ta có: \(y=0,75=\dfrac{3}{4}=\dfrac{15}{20}\Rightarrow\dfrac{17}{20}>\dfrac{15}{20}\)
Vậy x > y