Bài 1 : tìm x
a) |x-6|=6-x
b) x - 25 < 15
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1: =>x=8-35=-27
2: =>15-4+x=6
=>x+11=6
hay x=-5
3: =>-30+25-x=-1
=>x+5=1
hay x=-4
4: =>x-(-13)=-8
=>x+13=-8
hay x=-21
5: =>x-29-17+38=-9
=>x-8=-9
hay x=-1
a)|x+9|-7=15
|x+9|=15+7
|x+9|=22
Suy ra x+9=22 hoặc x+9=-22
Suy ra x=13;-31
b)24-(x-6)=-3+25
24-(x-6)=22
x-6=24-22
x-6=2
x=2+6
x=8
15-(4-x)=6
⇒15-4+x=6
⇒11+x=6
⇒x=-5
x-(12-25)=-8
⇒x+13=-8
⇒x=-21
(x-29)-(17-38)=-9
⇒x-29+21=-9
⇒x-8=-9
⇒x=-1
-30+(25-x)=-1
⇒-30+25-x=-1
⇒-5-x=-1
⇒x=-4
a) x-(-25+x)=13-x
⇒ x+25-x=13-x
⇒ 13-x=25
⇒ x= -12
b) 15-(30+x)=x-(27-l-8l)
⇒ 15-30-x=x-(27-8)
⇒ -15-x=x-19
⇒ -x-x=-19+15
⇒ -2x+4=0
⇒ -2x=-4
⇒ x=2
c) (12x-43).83=4.84
⇒ 12x.83-43.83=4.84
⇒ 6144x-32768=16384
⇒ 6144x=49152
⇒ x=8
a: Ta có: \(x-\left(-25+x\right)=13-x\)
\(\Leftrightarrow13-x=25\)
hay x=-12
b: Ta có: \(15-\left(30+x\right)=x-\left(27-\left|-8\right|\right)\)
\(\Leftrightarrow-15-x=x-19\)
\(\Leftrightarrow-2x=-4\)
hay x=2
c: Ta có: \(\left(12x-4^3\right)\cdot8^3=4\cdot8^4\)
\(\Leftrightarrow\left(12x-64\right)\cdot512=16384\)
\(\Leftrightarrow12x-64=32\)
hay x=8
Bài 2:
\(=\dfrac{28}{25}\cdot\dfrac{15}{7}\cdot5=\dfrac{75}{25}\cdot4=12\)
Bài 1:
a: \(x+\dfrac{7}{8}=\dfrac{13}{2}:4=\dfrac{13}{8}\)
nên x=13/8-7/8=6/8=3/4
b: \(x:\dfrac{5}{3}=\dfrac{6}{5}-\dfrac{2}{3}=\dfrac{18-10}{15}=\dfrac{8}{15}\)
nên \(x=\dfrac{8}{15}\cdot\dfrac{5}{3}=\dfrac{8}{9}\)
Bài 2 :
a,\(\frac{x-1}{3}=2-\frac{x}{-2}\)
\(\Leftrightarrow\frac{x-1}{3}=\frac{-4-x}{-2}\Leftrightarrow-2x+2=-12-3x\Leftrightarrow x=-14\)
b, \(\frac{x-1}{x+5}=\frac{6}{7}\Leftrightarrow7x-7=6x+30\Leftrightarrow x=37\)
c, \(\frac{2x-1}{4}=\frac{4}{2x-1}\Leftrightarrow\left(2x-1\right)^2=16\)
\(\Leftrightarrow\left(2x-1\right)^2-4^2=0\Leftrightarrow\left(2x-5\right)\left(2x+3\right)=0\Leftrightarrow x=\frac{5}{2};-\frac{3}{2}\)
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 \(\in\) {\(0;-2;2;-4;4;-6;14;-16\)}
\(\text{a)}\left|x-6\right|=6-x=-x+6=-\left(x-6\right)\Leftrightarrow x-6<0\Leftrightarrow x<6\)
Vậy x<6
\(\text{b)}x-25<15\Leftrightarrow x<15+25=40\)
Vậy x<40