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a) \(x\in\left\{1;-1;5;-5;7;-7;35;-35\right\}\)
b) \(x\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)
c) \(x\in\left\{0;25;50;75\right\}\)
d) Ta có: \(x+16⋮x+1\)
nên \(15⋮x+1\)
\(\Leftrightarrow x+1\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)
hay \(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)
a) 35 \(⋮\) x \(\Rightarrow\) x \(\in\) Ư(35) = {\(\pm1;\pm5;\pm7;\pm35\)}
b) 15 \(⋮\) x \(\Rightarrow\) x \(\in\) Ư(15) = {\(\pm1;\pm3;\pm5;\pm15\)}
c) x \(⋮\) 25 \(\Rightarrow\) x \(\in\) B(25) = {\(0;25;50;75;100;...\)}
Mà x < 100 \(\Rightarrow\) x \(\in\){\(0;25;50;75\)}
d) x + 16 \(⋮\) x + 1
\(\Rightarrow\) x + 16 - (x + 1) \(⋮\) x + 1
\(\Rightarrow\) 15 \(⋮\) x + 1
\(\Rightarrow\) x + 1 \(\in\) Ư(15) = {\(\pm1;\pm3;\pm5;\pm15\)}
\(\Rightarrow\) x \(\in\) {\(0;-2;2;-4;4;-6;14;-16\)}
a, bổ sung đề
\(\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)
\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{50-x}{27}+\dfrac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\ne0\right)=0\Leftrightarrow x=50\)
18-25=2(7-x)-(25+24)
-7=2(7-x)-49
2(7-x)=(-7)+49
2(7-x)=42
7-x=42:2
7-x=21
x=21+7
x=28
Vậy x=28
\(a,\dfrac{6}{5}=\dfrac{18}{x}\\ \Rightarrow x=18:\dfrac{6}{5}\\ \Rightarrow x=15\\ b,\dfrac{3}{4}=\dfrac{-21}{x}\\ \Rightarrow x=-21:\dfrac{3}{4}\\ \Rightarrow x=-28\\ c,\dfrac{2}{-7}=\dfrac{18}{x}\\ \Rightarrow x=18:\dfrac{2}{-7}\\ \Rightarrow x=-63\\ d,\dfrac{-5}{2}=\dfrac{10}{-x}\\ \Rightarrow x=-10:\dfrac{-5}{2}\\ \Rightarrow x=4\)
\(a,\dfrac{6}{5}=\dfrac{18}{x}\Rightarrow6.x=5.18=90\\ \Rightarrow6.x=90\\ \Rightarrow x=15\\ b,\dfrac{3}{4}=\dfrac{-21}{x}\Rightarrow3.x=4.21=84\\ \Rightarrow x=28\)
\(a,\left(x+3\right)\left(5-x\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x+3=0\\5-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
\(c,x+17⋮x+3\\ x+3+14⋮x+3\\ 14⋮x+3\\ x+3\inƯ\left(14\right)=\left\{\pm14;\pm7\pm2;\pm1\right\}\)
Từ đó bạn tìm những giá trị của x nha!