1 tìm a 8: (a-1) + 12: (a-1) = 5
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Tìm số nguyên a
a)4/5<5/a<10/7 b)2/5<a-1/10<8/15(a-1 là tử, 10 là mẫu)
c)12/7<4/a<8/3. d)5<a^2-15<16

a) \(\dfrac{4}{5}< \dfrac{5}{a}< \dfrac{10}{7}\) \(\left(a\inℤ\right)\)
\(\Leftrightarrow\dfrac{7}{10}< \dfrac{a}{5}< \dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{7.5}{10}< a< \dfrac{5}{4}.5\)
\(\Leftrightarrow\dfrac{7}{2}< a< \dfrac{25}{4}\)
\(\Leftrightarrow a\in\left\{4;5;6\right\}\)
b) \(\dfrac{2}{5}< \dfrac{a-1}{10}< \dfrac{8}{15}\)
\(\Leftrightarrow\dfrac{2.10}{5}< a-1< \dfrac{8.10}{15}\)
\(\Leftrightarrow4< a-1< \dfrac{16}{3}\)
\(\Leftrightarrow5< a< \dfrac{19}{3}\)
\(\Leftrightarrow a\in\left\{6\right\}\)
c) \(\dfrac{12}{7}< \dfrac{4}{a}< \dfrac{8}{3}\)
\(\Leftrightarrow\dfrac{3}{8}< \dfrac{a}{4}< \dfrac{7}{12}\)
\(\Leftrightarrow\dfrac{3.4}{8}< a< \dfrac{7.4}{12}\)
\(\Leftrightarrow\dfrac{3}{2}< a< \dfrac{7}{3}\)
\(\Leftrightarrow a\in\left\{2\right\}\)
d) \(5< a^2-15< 16\)
\(\Leftrightarrow10< a^2< 31\)
\(\Leftrightarrow\sqrt[]{10}< a< \sqrt[]{31}\)
\(\Leftrightarrow a\in\left\{4;5\right\}\)

Bài giải:
Câu 1: a, \(\left(-2\right).4.5.38.\left(-25\right)\)
\(=\left[\left(-2\right).5\right].\left[4.\left(-25\right)\right].38\)
\(=\left(-10\right).\left(-100\right).38\)
\(=1000.38=38000\)
b,\(\frac{1}{3}+\frac{3}{8}-\frac{7}{12}\)
\(=\left(\frac{1}{3}+\frac{3}{8}\right)-\frac{7}{12}\)
\(=\frac{17}{24}-\frac{7}{12}=\frac{1}{8}\)
c, \(\frac{-5}{8}.\frac{5}{12}+\frac{-5}{8}.\frac{7}{12}+2\frac{1}{8}\)
\(=\frac{-5}{8}.\left(\frac{5}{12}+\frac{7}{12}\right)+\frac{17}{8}\)
\(=\frac{-5}{8}.1+\frac{17}{8}\)
\(=\frac{3}{2}\)
Câu 2: a, \(x-\frac{2}{5}=0,24\)
\(x-0,4=0,24\)
\(x=0,24+0,4\)
\(\Rightarrow x=0,64\left(\frac{16}{25}\right)\)
b,\(\frac{2}{3}.x+\frac{1}{12}=\frac{1}{10}\)
\(\frac{2}{3}.x=\frac{1}{10}-\frac{1}{12}\)
\(\frac{2}{3}.x=\frac{1}{60}\)
\(x=\frac{1}{60}:\frac{2}{3}\)
\(\Rightarrow x=\frac{1}{40}\)
c, \(\left(3\frac{1}{2}-2x\right).1\frac{1}{3}=7\frac{1}{3}\)
\(\frac{7}{2}-2x=\frac{22}{3}:\frac{4}{3}\)
\(\frac{7}{2}-2x=\frac{11}{2}\)
\(2x=\frac{7}{2}-\frac{11}{2}\)
\(2x=-2\)
\(\Rightarrow x=-2:2\)
\(x=-1\)

\(A=\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{a}\right):\dfrac{4}{5}\) \(\left(a\ne0\right)\)
Tại a = 12 biểu thức \(A=\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{12}\right):\dfrac{4}{5}=\dfrac{5}{4}-\dfrac{11}{24}:\dfrac{4}{5}=\dfrac{5}{4}-\dfrac{11}{24}.\dfrac{5}{4}=\dfrac{5}{4}-\dfrac{55}{96}=\dfrac{65}{96}\)
Để \(A=\dfrac{15}{23}< =>\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{a}\right):\dfrac{4}{5}=\dfrac{15}{23}\)
\(\Leftrightarrow\left(\dfrac{3}{8}+\dfrac{1}{a}\right):\dfrac{4}{5}=\dfrac{55}{92}< =>\dfrac{3}{8}+\dfrac{1}{a}=\dfrac{11}{23}< =>\dfrac{1}{a}=\dfrac{19}{184}< =>a=\dfrac{184}{19}\)
Thay \(a=12\) vào A ta có:
\(A=\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{12}\right):\dfrac{4}{5}=\dfrac{65}{96}\)
Vậy:
____________________
Ta có:
\(A=\dfrac{15}{23}\) khi \(\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{a}\right):\dfrac{4}{5}=\dfrac{15}{23}\)
\(\Rightarrow\left(\dfrac{3}{8}+\dfrac{1}{a}\right)\cdot\dfrac{5}{4}=\dfrac{5}{4}-\dfrac{15}{23}\)
\(\Rightarrow\dfrac{3}{8}+\dfrac{1}{a}=\dfrac{55}{92}:\dfrac{5}{4}\)
\(\Rightarrow\dfrac{3}{8}+\dfrac{1}{a}=\dfrac{11}{23}\)
\(\Rightarrow\dfrac{1}{a}=\dfrac{11}{184}\)
\(\Rightarrow a=\dfrac{1\cdot184}{11}=\dfrac{184}{11}\)

`b)y:12/15=5`
`\qquad y:4/5=5`
`\qquad y=5*4/5=4`
Vậy `y=4`
`a)y-1/8=1/4`
`\qquad y=1/4+1/8`
`\qquad y=3/8`
Vậy `y=3/8`
Giải:
a) \(y-\dfrac{1}{8}=\dfrac{1}{4}\)
\(y=\dfrac{1}{4}+\dfrac{1}{8}\)
\(y=\dfrac{3}{8}\)
b) \(y:\dfrac{12}{15}=5\)
\(y=5\times\dfrac{4}{5}\)
\(y=4\)
Chúc em học tốt!

a) \(\dfrac{2x+5}{2x+1}=\dfrac{2x+1+4}{2x+1}=\dfrac{2x+1}{2x+1}+\dfrac{4}{2x+1}=1+\dfrac{4}{2x+1}\)
Để \(\dfrac{2x+5}{2x+1}\in Z\) thì \(\dfrac{4}{2x+1}\in Z\)
\(\Rightarrow4\) ⋮ \(2x+1\)
\(\Rightarrow2x+1\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
\(\Rightarrow2x\in\left\{0;-2;1;-3;3;-5\right\}\)
\(\Rightarrow x\in\left\{0;-1;\dfrac{1}{2};-\dfrac{3}{2};\dfrac{3}{2};-\dfrac{5}{2}\right\}\)
Mà x nguyên \(\Rightarrow\text{x}\in\left\{0;-1\right\}\)
b) \(\dfrac{3x+5}{x+1}=\dfrac{3x+3+2}{x+1}=\dfrac{3\left(x+1\right)+2}{x+1}=\dfrac{3\left(x+1\right)}{x+1}+\dfrac{2}{x+1}=3+\dfrac{2}{x+1}\)
Để \(\dfrac{3x+5}{x+1}\in Z\) thì \(\dfrac{2}{x+1}\in Z\)
\(\Rightarrow2\) ⋮ \(x+1\)
\(\Rightarrow x+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
\(\Rightarrow x\in\left\{0;-2;1;-3\right\}\)
c) \(\dfrac{3x+8}{x-1}=\dfrac{3x-3+11}{x-1}=\dfrac{3\left(x-1\right)+11}{x-1}=\dfrac{3\left(x-1\right)}{x-1}+\dfrac{11}{x-1}=3+\dfrac{11}{x-1}\)
Để: \(\dfrac{3x+8}{x-1}\in Z\) thì \(\dfrac{11}{x-1}\in Z\)
\(\Rightarrow11\) ⋮ \(x-1\)
\(\Rightarrow x-1\inƯ\left(11\right)=\left\{1;-1;11;-11\right\}\)
\(\Rightarrow x\in\left\{2;0;12;-10\right\}\)
d) \(\dfrac{5x+12}{x-2}=\dfrac{5x-10+22}{x-2}=\dfrac{5\left(x-2\right)+22}{x-2}=\dfrac{5\left(x-2\right)}{x-2}+\dfrac{22}{x-2}=5+\dfrac{22}{x-2}\)
Để: \(\dfrac{5x+12}{x-2}\in Z\) thì \(\dfrac{22}{x-2}\in Z\)
\(\Rightarrow22\) ⋮ \(x-2\)
\(\Rightarrow x-2\inƯ\left(22\right)=\left\{1;-1;2;-2;11;-11;22;-22\right\}\)
\(\Rightarrow x\in\left\{3;1;4;0;13;-9;24;-20\right\}\)
e) \(\dfrac{7x-12}{x+16}=\dfrac{7x+112-124}{x+16}=\dfrac{7\left(x+16\right)-124}{x+16}=\dfrac{7\left(x+16\right)}{x+16}-\dfrac{124}{x+16}=7-\dfrac{124}{x+16}\)
Để \(\dfrac{7x-12}{x+16}\in Z\) thì \(\dfrac{124}{x+16}\in Z\)
\(\Rightarrow124\) ⋮ \(x+16\)
\(\Rightarrow x+16\inƯ\left(124\right)=\left\{1;-1;2;-2;4;-4;31;-31;62;-62;124;-124\right\}\)
\(\Rightarrow x\in\left\{-15;-17;-14;-18;-12;-20;15;-47;46;-78;108;-140\right\}\)

Bài 1:
a) \(\dfrac{9}{20}-\dfrac{8}{15}\times\dfrac{5}{12}\)
\(=\dfrac{9}{20}-\dfrac{2}{9}\)
\(=\dfrac{41}{180}\)
b) \(\dfrac{2}{3}\div\dfrac{4}{5}\div\dfrac{7}{12}\)
\(=\dfrac{2}{3}\times\dfrac{5}{4}\times\dfrac{12}{7}\)
\(=\dfrac{5}{6}\times\dfrac{12}{7}\)
\(=\dfrac{10}{7}\)
c) \(\dfrac{7}{9}\times\dfrac{1}{3}+\dfrac{7}{9}\times\dfrac{2}{3}\)
\(=\dfrac{7}{9}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\dfrac{7}{9}\times1\)
\(=\dfrac{7}{9}\)
Bài 2:
a) \(2\times\left(x-1\right)=4026\)
\(\left(x-1\right)=4026\div2\)
\(x-1=2013\)
\(x=2014\)
Vậy: \(x=2014\)
b) \(x\times3,7+6,3\times x=320\)
\(x\times\left(3,7+6,3\right)=320\)
\(x\times10=320\)
\(x=320\div10\)
\(x=32\)
Vậy: \(x=32\)
c) \(0,25\times3< 3< 1,02\)
\(\Leftrightarrow0,75< 3< 1,02\) ( S )
=> \(0,75< 1,02< 3\)
8 : (a - 1) + 12 : (a - 1) = 5
(8 + 12) : (a - 1) = 5
20 : (a - 1) = 5
a - 1 = 20 : 5
a - 1 = 4
a = 1 + 4
a = 5
vậy a = 5