\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{\frac{-2}{12}-\frac{10}{24}+\frac{14}{39}}\) (x=0,5;y=3)
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a) \(\frac{x+1}{3}=\frac{x-2}{4}\)
=> (x+1).4 = (x - 2) . 3
=> 4x + 4 = 3x - 6
=> 4x - 3x = - 6 - 4
=> x = - 10
b) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
\(\Rightarrow\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}\) = 0
\(\Rightarrow\left(x+1\right).\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)\)
Vì \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\) nên x + 1 =0
=> x = -1
c) Xem lại đề
a)<=>2/7:(13/24+5/24) b)<=>15/24-9/24-4/24 c)=7.5.39/13.14.15
<=>2/7:3/4 <=>2/24=1/12 =3/2.3
<=>2/7x4/3 =1/2
<=>8/21
a. \(\frac{2}{7}:\frac{13}{24}+\frac{2}{7}:\frac{5}{24}\)
= \(\frac{2}{7}:\left(\frac{13}{24}+\frac{5}{24}\right)\)
= \(\frac{2}{7}:\frac{3}{4}\)
= \(\frac{8}{21}\)
b. \(\frac{15}{24}-\frac{3}{8}-\frac{1}{6}\)
= \(\frac{15}{24}-\frac{9}{24}-\frac{4}{24}\)
= \(\frac{2}{24}=\frac{1}{12}\)
c. \(\frac{7}{13}.\frac{5}{14}.\frac{39}{15}\)
= \(\frac{7.5.3.13}{13.2.7.3.5}\)
= \(\frac{1}{2}\)
1
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Ez lắm =)
Bài 1:
Với mọi gt \(x,y\in Q\) ta luôn có:
\(x\le\left|x\right|\) và \(-x\le\left|x\right|\)
\(y\le\left|y\right|\) và \(-y\le\left|y\right|\Rightarrow x+y\le\left|x\right|+\left|y\right|\) và \(-x-y\le\left|x\right|+\left|y\right|\)
Hay: \(x+y\ge-\left(\left|x\right|+\left|y\right|\right)\)
Do đó: \(-\left(\left|x\right|+\left|y\right|\right)\le x+y\le\left|x\right|+\left|y\right|\)
Vậy: \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)
Dấu "=" xảy ra khi: \(xy\ge0\)
\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\frac{2}{12}-\frac{10}{24}+\frac{14}{39}}\)
\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\left(\frac{2}{12}+\frac{10}{24}-\frac{14}{39}\right)}\)
\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\frac{2}{3}\left(\frac{1}{4}+\frac{5}{8}-\frac{7}{13}\right)}\)
\(B=\frac{x}{y}+\frac{1}{-\frac{2}{3}}\)
\(B=\frac{x}{y}-\frac{3}{2}\)
Thế x = 0, 5 = 1/2 ; y = 3 ta được :
\(B=\frac{\frac{1}{2}}{3}-\frac{3}{2}=\frac{1}{6}-\frac{9}{6}=-\frac{8}{6}=-\frac{4}{3}\)
Ta có:\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{\frac{-2}{12}-\frac{10}{24}+\frac{14}{39}}\)
\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\left(\frac{2}{12}+\frac{10}{24}-\frac{14}{39}\right)}\)
\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\frac{2}{3}\left(\frac{1}{4}+\frac{5}{8}-\frac{7}{13}\right)}\)
\(B=\frac{x}{y}+\frac{1}{-\frac{2}{3}}\)(Do\(\frac{1}{4}+\frac{5}{8}-\frac{7}{13}\ne0\))
\(B=\frac{x}{y}-\frac{3}{2}\)
Thay x = 0,5; y = 3 vào B ta được:
\(B=\frac{0,5}{3}-\frac{3}{2}\)
\(B=\frac{1}{6}-\frac{3}{2}\)
\(B=\frac{1}{6}-\frac{9}{6}\)
\(B=-\frac{4}{3}\)
Vậy\(B=-\frac{4}{3}\)tại x = 0,5; y = 3
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