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a) \(10⋮\left(x-1\right)\) (đkxđ \(x\ne1\))
\(\Rightarrow x-1\in\left\{-1;1;-2;2;-5;5;-10;10\right\}\)
\(\Rightarrow x\in\left\{0;2;-1;3;-4;6;-9;11\right\}\)
b) \(\left(x+5\right)⋮\left(x-2\right)\left(đkxđ,x\ne2\right)\)
\(\Rightarrow\left(x+5\right)-\left(x-2\right)⋮\left(x-2\right)\)
\(\Rightarrow x+5-x+2⋮\left(x-2\right)\)
\(\Rightarrow7⋮\left(x-2\right)\)
\(\Rightarrow x-2\in\left\{-1;1;-7;7\right\}\)
\(\Rightarrow x\in\left\{1;3;-5;9\right\}\)
c) \(\left(3x+8\right)⋮\left(x-1\right)\left(đkxd,x\ne1\right)\)
\(\Rightarrow\left(3x+8\right)-3\left(x-1\right)⋮\left(x-1\right)\)
\(\Rightarrow3x+8-3x+3⋮\left(x-1\right)\)
\(\Rightarrow11⋮\left(x-1\right)\)
\(\Rightarrow x-1\in\left\{-1;1;-11;11\right\}\)
\(\Rightarrow x\in\left\{0;2;-10;12\right\}\)
a) x∈{0;2;−1;3;−4;6;−9;11}
b) x∈{1;3;−5;9}
c) x ∈ {0;2;−10;12}
a) <=> 10 - 2x + 5 = 1 - 3x
<=> -2x + 3x = 1 - 10 - 5
<=> x = -14
b) \(\Leftrightarrow\frac{x-1}{x+1}-\frac{8}{9}=0\)
\(\Leftrightarrow\frac{9\left(x-1\right)-8\left(x+1\right)}{9\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{9x-9-8x-8}{9x+9}=0\)
\(\Leftrightarrow\frac{x-17}{9x+9}=0\)
ĐK: 9x + 9 \(\ne\)0 => x \(\ne\)-1
\(\Leftrightarrow x-17=0\)
\(\Leftrightarrow x=17\)
Ps: Câu b không chắc lắm.
phan b Vu Nhu Mai sai roi phai the nay moi dung
x-1/x+1=8/9
(x-1).9=(x+1).8
9x-9=8x+8
9x-8x=8+9
1x=17
x=17;1
x=17