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\(1\)) \(5-\left(10-x\right)=7\)
\(10-x=5-7\)
\(10-x=-2\)
\(x=10-\left(-2\right)\)
\(x=12\)
\(2\)) \(-32-\left(x-5\right)=0\)
\(x-5=-32-0\)
\(x-5=-32\)
\(x=-32+5\)
\(x=-27\)
e: =>-40+3+33+40-x=47
=>36-x=47
=>x=-11
f: =>x(x-3)(11-x)(11+x)=0
hay \(x\in\left\{0;3;11;-11\right\}\)
g: =>-62-38-x+2x=-100
=>x-100=-100
hay x=0
i: =>x-12-2x-31=6
=>-x-43=6
=>x+43=-6
hay x=-49
h: =>(x+1)=0
=>x=-1
f: =>x(x-3)(x+11)(x-11)=0
hay \(x\in\left\{0;3;-11;11\right\}\)
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 \(\dfrac{124}{x+16}\in Z\)
\(\Rightarrow124\) ⋮ \(x+16\)
\(\Rightarrow x+16\inƯ\left(124\right)=\left\{1;-1;2;-2;4;-4;31;-31;62;-62;124;-124\right\}\)
\(\Rightarrow x\in\left\{-15;-17;-14;-18;-12;-20;15;-47;46;-78;108;-140\right\}\)
a: \(x+7⋮x+2\)
=>\(x+2+5⋮x+2\)
=>\(5⋮x+2\)
=>\(x+2\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{-1;-3;3;-7\right\}\)
b: \(2x+5⋮x+1\)
=>\(2x+2+3⋮x+1\)
=>\(3⋮x+1\)
=>\(x+1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{0;-2;2;-4\right\}\)
c: \(3x-2⋮x+3\)
=>\(3x+9-11⋮x+3\)
=>\(-11⋮x+3\)
=>\(x+3\in\left\{1;-1;11;-11\right\}\)
=>\(x\in\left\{-2;-4;8;-14\right\}\)
d: \(12x+1⋮3x+2\)
=>\(12x+8-7⋮3x+2\)
=>\(-7⋮3x+2\)
=>\(3x+2\in\left\{1;-1;7;-7\right\}\)
=>\(3x\in\left\{-1;-3;5;-9\right\}\)
=>\(x\in\left\{-\dfrac{1}{3};-1;\dfrac{5}{3};-3\right\}\)
e: \(x^2+3x+5⋮x+3\)
=>\(x\left(x+3\right)+5⋮x+3\)
=>\(5⋮x+3\)
=>\(x+3\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{-2;-4;2;-8\right\}\)
f: \(x^2-2x+3⋮x+2\)
=>\(x^2+2x-4x-8+11⋮x+2\)
=>\(11⋮x+2\)
=>\(x+2\in\left\{1;-1;11;-11\right\}\)
=>\(x\in\left\{-1;-3;9;-13\right\}\)
a)
<=> 3x - 3 + x - 2 = 2x - 2 - x + 1
<=> 3x + x - 2x + x = -2 + 1 + 3 + 2
<=> 3x = 4
<=> x = 4/3
Các câu sau làm tương tự
\(\left(3x-3\right)+\left(x-2\right)=\left(2x-2\right)-\left(x-1\right)\)
<=> \(3x-3+x-2=2x-2-x+1\)
<=> \(4x-5=x-1\)
<=> \(3x=4\)
<=> \(x=\frac{4}{3}\)
Vậy....
x*(2/3-1/2=5/12
x*1/6=5/12
x=5/12:1/6
x=5/2
Vậy x=5/2