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a)Vì 35\(⋮\)(2x-1)
=>2x-1 \(\in\)Ư(35)
=>2x-1\(\in\){1;-1;5;-5;7;-7;35;-35}
=>2x\(\in\){2;0;6;-4;8;-6;36;-34}
=>x\(\in\){1;0;3;-2;4;-3;18;-17}
b)18\(⋮\)(x+1)
=>x+1\(\in\)Ư(18)
=>x+1\(\in\){1;-1;2;-2;3;-3;6;-6;9;-9;18;-18}
=>x\(\in\){0;-2;1;-3;2;-4;5;-7;8;-10;17;-19}
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\(\frac{2}{x}=\frac{x}{8}\Rightarrow x^2=16=4^2\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
\(A=\frac{4116-14}{10290-35}=\frac{4102}{10255}=\frac{2}{5}\)
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a) \(\frac{3}{7}x-\frac{1}{35}=\frac{3}{5}\)
\(\frac{3}{7}x=\frac{3}{5}+\frac{1}{35}\)
\(\frac{3}{7}x=\frac{22}{35}\)
\(x=\frac{49}{35}=1,4\)
b) \(1,5-x:\frac{1}{2}=\frac{1}{4}\)
\(x:\frac{1}{2}=1,5-\frac{1}{4}\)
\(x:\frac{1}{2}=\frac{5}{4}\)
\(x=\frac{5}{4}.\frac{1}{2}\)
\(x=\frac{5}{8}\)
Vậy ..
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a/ \(\left(x-35\right)-120=0\)
\(\Leftrightarrow x-35=0+120\)
\(\Leftrightarrow x-35=120\)
\(\Leftrightarrow x=120+35\)
\(\Leftrightarrow x=155\)
Vậy ...
b/ \(124+\left(188-x\right)=217\)
\(\Leftrightarrow188-x=217-124\)
\(\Leftrightarrow188-x=93\)
\(\Leftrightarrow x=95\)
Vậy ...
b/ \(156-\left(x+61\right)=82\)
\(\Leftrightarrow x+61=74\)
\(\Leftrightarrow x=13\)
Vậy ..
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Bài 1:tính
a)65.(-19)+19.(-35)
=65.(-19)+(-19).35
=(-19).(65+35)
=(-19).100
=-1900
b)85.(35-27)-35.(85-27)
=85.35-85.27-35.85+35.27
=(85.35-35.85)+(-85.27+35.27)
=27.(-85+35)
=27.(-50)
=1350
c)47.(45-15)-47.(45+15)
=47.[(45-15)-(45+15)]
=47.[30-60]
=47.(-30)
=-1410
Bài2: Tìm các số nguyên x biết
a)(-2).(x+6)+6.(x-10)=8
-2x-12+6x-60=8
4x-72=8
4x=72+8
4x=50
x=\(\frac{25}{2}\)
b)(-4).(2x+9)-(-8x+3)-(x+13)=0
-6x-36+8x-3-x-13=0
x-41=0
x=41
Bài 1: Tính
a) \(65.\left(-19\right)+19.\left(-35\right)\)
= \(-1235+-665\)
= \(-1900\)
b) \(85.\left(35-27\right)-35.\left(85-27\right)\)
= \(-1350\)
c) \(47.\left(45-15\right)-47.\left(45+15\right)\)
=\(-1410\)
Bài 2: Tìm x:
\(\left(-2\right).\left(x+6\right)+6.\left(x-10\right)=8\)
\(x=20\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\dfrac{-6}{7}+\dfrac{2}{5}=\dfrac{\left(-30\right)+14}{35}=\dfrac{-16}{35}\)
\(\dfrac{-3}{105}+\dfrac{17}{35}=\dfrac{1+17}{35}=\dfrac{18}{35}\)
\(\Rightarrow-16< 7x< 18\)
\(\Rightarrow7x\in\left\{-14;14;7;-7;0\right\}\)
\(\Rightarrow x\in\left\{-2;2;1;-1;0\right\}\)
Vậy \(x\in\left\{-2;2;1;-1;0\right\}\)
\(\dfrac{21}{105}=\dfrac{35-x}{105}\Rightarrow35-x=21\Leftrightarrow x=14\)