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a) \(\dfrac{7}{4}< \dfrac{a}{8}< 3\\ =>\dfrac{7}{4}.8< a< 3.8\\ =>14< a< 24\\ =>a\in\left\{15;16;17;...;23\right\}\)
b) \(\dfrac{2}{3}< \dfrac{a-1}{6}< \dfrac{8}{9}\\ =>\dfrac{2}{3}.6< a-1< \dfrac{8}{9}.6\\ =>4< a-1< \dfrac{16}{3}\\ =>4+1< a< \dfrac{16}{3}+1\\ =>5< a< \dfrac{19}{3}\\ =>a=6\)
b) \(\dfrac{2}{3}< a-\dfrac{1}{6}< \dfrac{8}{9}\\ =>\dfrac{2}{3}+\dfrac{1}{6}< a< \dfrac{8}{9}+\dfrac{1}{6}\\ =>\dfrac{5}{6}< a< \dfrac{19}{18}\\ =>a=1\)
c) \(\dfrac{12}{9}< \dfrac{4}{a}< \dfrac{8}{3}\\ =>\dfrac{24}{18}< \dfrac{24}{6a}< \dfrac{24}{9}\\ =>9< 6a< 18\\ =>\dfrac{9}{6}< a< \dfrac{18}{6}\\ =>1,5< a< 3\\ =>a=2\)
Giup mình với ah.
1- Tính :
A= 5. | x- 5 | - 3x + 1
2 - Tìm các số nguyên x,y ; sao cho :
a) 5/x - y/3 = 1/6 b) 5/x + y/4 = 1/8
3- Tìm giá trị lớn nhất của Q = 27-2x/12-x ( x là số nguyên)
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Bài 1:
a: \(\Leftrightarrow x\cdot\left(-11\right)=121\)
hay x=-11
1) \(A=5.\left|x-5\right|-3x+1\)
\(A=\left[{}\begin{matrix}5.\left(x-5\right)-3x+1\left(x-5\ge0\right)\\5.\left(5-x\right)-3x+1\left(x-5< 0\right)\end{matrix}\right.\)
\(A=\left[{}\begin{matrix}5x-25-3x+1\left(x\ge5\right)\\25-5x-3x+1\left(x< 5\right)\end{matrix}\right.\)
\(A=\left[{}\begin{matrix}2x-24\left(x\ge5\right)\\26-8x\left(x< 5\right)\end{matrix}\right.\)
3:
\(Q=\dfrac{27-2x}{12-x}=\dfrac{2x-27}{x-12}\)
\(\Leftrightarrow Q=\dfrac{2x-24-3}{x-12}=2-\dfrac{3}{x-12}\)
Để Q lớn nhất thì \(2-\dfrac{3}{x-12}\) lớn nhất
=>\(\dfrac{3}{x-12}\) nhỏ nhất
=>x-12 là số nguyên âm lớn nhất
=>x-12=-1
=>x=11
Vậy: \(Q_{min}=2-\dfrac{3}{11-12}=2+3=5\) khi x=11
Bài 2:
a: \(\dfrac{5}{x}-\dfrac{y}{3}=\dfrac{1}{6}\)
=>\(\dfrac{15-xy}{3x}=\dfrac{1}{6}\)
=>\(15-xy=\dfrac{x}{2}\)
=>\(30-2xy=x\)
=>x+2xy=30
=>x(2y+1)=30
mà x,y nguyên
nên \(\left(x;2y+1\right)\in\left\{\left(30;1\right);\left(-30;-1\right);\left(2;15\right);\left(-2;-15\right);\left(10;3\right);\left(-10;-3\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(30;0\right);\left(-30;-1\right);\left(2;7\right);\left(-2;-8\right);\left(10;1\right);\left(-10;-2\right)\right\}\)
b: \(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\)
=>\(\dfrac{20+xy}{4x}=\dfrac{1}{8}\)
=>\(\dfrac{40+2xy}{8x}=\dfrac{x}{8x}\)
=>40+2xy=x
=>x-2xy=40
=>x(1-2y)=40
mà x,y nguyên
nên \(\left(x;1-2y\right)\in\left\{\left(40;1\right);\left(-40;-1\right);\left(8;5\right);\left(-8;-5\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(40;0\right);\left(-40;1\right);\left(8;-2\right);\left(-8;3\right)\right\}\)
\(\text{#040911}\)
Vì \(-\dfrac{5}{12}< 0\)
\(\Rightarrow-\dfrac{5}{12}< \dfrac{a}{5}\text{ }\forall\text{ }a\)
\(\dfrac{a}{5}< \dfrac{1}{4}\)
\(\Rightarrow a=1\)
Vậy, để thỏa mãn \(-\dfrac{5}{12}< \dfrac{a}{5}< \dfrac{1}{4}\) thì \(a=1.\)
\(\frac{-5}{12}< \frac{a}{5}< \frac{1}{4}\)
=>\(\frac{-25}{60}< \frac{12a}{60}< \frac{15}{60}\)
=>-25<12a<15
=>\(12a\in\left\{-24;-23;-22;...;13;14\right\}\)
=> \(a\in\left\{-2;-\frac{23}{12};-\frac{11}{6};...;\frac{13}{2};\frac{7}{6}\right\}\)
Vì a là số nguyên =>\(a\in\left\{-2;-1;0;1\right\}\)
\(\dfrac{-3}{4}< \dfrac{a}{12}< \dfrac{-5}{9}\)
\(\Rightarrow\dfrac{-27}{36}< \dfrac{3a}{36}< \dfrac{-20}{36}\)
\(\Rightarrow-27< 3a< -20\)
\(\Rightarrow a=\left\{-8;-7\right\}\)