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\(\frac{x-2}{\left(a+3\right)\left(5-a\right)}=\frac{1}{2\left(a+3\right)}+\frac{1}{2\left(5-a\right)}\)
\(\Rightarrow\frac{2\left(x-2\right)}{2\left(a+3\right)\left(5-a\right)}=\frac{5-a+a+3}{2\left(a+3\right)\left(5-a\right)}\)
\(\Rightarrow\) 2x - 4 = 8
\(\Rightarrow\) 2x = 12
\(\Rightarrow\) x = 6
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{a}{b}=\dfrac{2}{3}\Rightarrow\dfrac{a}{2}=\dfrac{b}{3};\dfrac{a}{c}=\dfrac{1}{2}\Rightarrow\dfrac{a}{1}=\dfrac{c}{2}\\ \Rightarrow\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a^3}{8}=\dfrac{b^3}{27}=\dfrac{c^3}{64}\)
Áp dụng tcdtsnb:
\(\dfrac{a^3}{8}=\dfrac{b^3}{27}=\dfrac{c^3}{64}=\dfrac{a^3+b^3+c^3}{8+27+64}=\dfrac{99}{99}=1\\ \Rightarrow\left\{{}\begin{matrix}a^3=8\\b^3=27\\c^3=64\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=2\\b=3\\c=4\end{matrix}\right.\)
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\(\text{1)}\hept{\begin{cases}A=2x-3-3x+1=-x-2\\A=2x-3+3x-1=5x-4\end{cases}}\)
\(\text{2)}\hept{\begin{cases}A=-x-2=4\\A=5x-4=4\end{cases}\Rightarrow\hept{\begin{cases}x=-6\\x=\frac{8}{5}\end{cases}}}\)
\(\text{3)}\hept{\begin{cases}A=-2-2=-4\\A=5\times2-4=6\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a}{b}=\frac{2}{3}\Leftrightarrow\frac{a}{2}=\frac{b}{3}\)
\(\frac{a}{c}=\frac{1}{2}\Leftrightarrow\frac{a}{1}=\frac{c}{2}\Leftrightarrow\frac{a}{2}=\frac{c}{4}\)
=>\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\)
ÁP dụng tc của dãy tỉ số bằng nhau ta có:
\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{a^3+b^3+c^3}{2^3+3^3+4^3}=\frac{99}{99}=1\)
=>\(\begin{cases}a=2\\b=3\\c=4\end{cases}\)
Giải:
Ta có: \(\frac{a}{b}=\frac{2}{3}\Rightarrow\frac{a}{2}=\frac{b}{3}\)
\(\frac{a}{c}=\frac{1}{2}\Rightarrow\frac{a}{1}=\frac{c}{2}\Rightarrow\frac{a}{2}=\frac{c}{4}\)
\(\Rightarrow\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\)
\(\Rightarrow\frac{a^3}{8}=\frac{b^3}{27}=\frac{c^3}{64}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a^3}{8}=\frac{b^3}{27}=\frac{c^3}{64}=\frac{a^3+b^3+c^3}{8+27+64}=\frac{99}{99}=1\)
+) \(\frac{a^3}{8}=1\Rightarrow a^3=8\Rightarrow a=2\)
+) \(\frac{b^3}{27}=1\Rightarrow b=3\)
+) \(\frac{c^3}{64}=1\Rightarrow c=4\)
Vậy bộ số \(\left(a,b,c\right)\) là \(\left(2,3,4\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
M=(2a+9/a+3)+(5a+17/a+3)-(3a/a+3)
M=(2a+9+5a+17-3a)/(a+3)
M=[(2a+5a-3a)+(9+17)]/(a+3)
M=(4a+26)/(a+3)
M=[4.(a+3)+14]/(a+3)=4+14/(a+3)
M nguyên <=>14 chia hết cho a+3
<=>a+3 E Ư(14)={-14;-7;-2;-1;1;2;7;14}
<=>a E {-17;-10;-5;-4;-2;-1;4;11}
Vậy ........
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2.Giải:
Theo bài ra ta có:
\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{d}{5}\) và a + b + c + d = -42
Theo tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{d}{5}=\frac{a+b+c+d}{2+3+4+5}=\frac{-42}{14}=-3\)
+) \(\frac{a}{2}=-3\Rightarrow a=-6\)
+) \(\frac{b}{3}=-3\Rightarrow b=-9\)
+) \(\frac{c}{4}=-3\Rightarrow c=-12\)
+) \(\frac{d}{5}=-3\Rightarrow d=-15\)
Vậy a = -6
b = -9
c = -12
d = -15
Bài 3:
Ta có:\(\frac{a}{2}=\frac{b}{3}\Leftrightarrow\frac{a}{10}=\frac{b}{15}\); \(\frac{b}{5}=\frac{c}{4}\Leftrightarrow\frac{b}{15}=\frac{c}{12}\)
\(\Rightarrow\frac{a}{10}=\frac{b}{15}=\frac{c}{12}\)
Áp dụng tc dãy tỉ:
\(\frac{a}{10}=\frac{b}{15}=\frac{c}{20}=\frac{a+b+c}{10+15+12}=\frac{-49}{37}\)
Với \(\frac{a}{10}=\frac{-49}{37}\Rightarrow a=10\cdot\frac{-49}{37}=\frac{-490}{37}\)
Với \(\frac{b}{15}=\frac{-49}{37}\Rightarrow b=15\cdot\frac{-49}{37}=\frac{-735}{37}\)
Với \(\frac{c}{12}=\frac{-49}{37}\Rightarrow c=12\cdot\frac{-49}{37}=\frac{-588}{37}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x\ne\pm3\)
a
Khi x = 1:
\(A=\dfrac{3.1+2}{1-3}=\dfrac{5}{-2}=-2,5\)
Khi x = 2:
\(A=\dfrac{3.2+2}{2-3}=-8\)
Khi x = \(\dfrac{5}{2}:\)
\(A=\dfrac{3.2,5+2}{2,5-3}=\dfrac{9,5}{-0,5}=-19\)
b
Để A nguyên => \(\dfrac{3x+2}{x-3}\) nguyên
\(\Leftrightarrow3x+2⋮\left(x-3\right)\\3\left(x-3\right)+11⋮\left(x-3\right) \)
Vì \(3\left(x-3\right)⋮\left(x-3\right)\) nên \(11⋮\left(x-3\right)\)
\(\Rightarrow\left(x-3\right)\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\\ \Rightarrow x\left\{4;2;-8;14\right\}\)
c
Để B nguyên => \(\dfrac{x^2+3x-7}{x+3}\) nguyên
\(\Rightarrow x\left(x+3\right)-7⋮\left(x+3\right)\)
\(\Rightarrow-7⋮\left(x+3\right)\\ \Rightarrow x+3\inƯ\left\{\pm1;\pm7\right\}\)
\(\Rightarrow x=\left\{-4;-11;-2;4\right\}\)
d
\(\left\{{}\begin{matrix}A.nguyên.\Leftrightarrow x=\left\{-8;2;4;14\right\}\\B.nguyên\Leftrightarrow x=\left\{-11;-4;-2;4\right\}\end{matrix}\right.\)
=> Để A, B cùng là số nguyên thì x = 4.
xét a<(=)-2:
=>-a-(a+2)=3
=>-a-a-2=3
=>-2a=3+2
=>-2a=5
=>a=-5/2
xét a>-2:
=>a+a+2=3
=>2a=3-2
=>2a=1
=>a=1/2
Vậy a=-5/2;1/2
a thuộc tập hợp ?