a) 2x–15=-47
b) (-5)2–(5x–3)=43
c) 25–(x–5)=-415–(15–415)
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a) 3 - (17 - x) = 289 - (36 + 289)
3 - (17 - x) = 289 - 325
3 - (17 - x) = -36
17 - x = -36 - 3
17 - x = -39
x = -39 + 17
x = -22
b) 25 - (x + 5) = -415 - (15 - 415)
25 - (x + 5) = -415 - 400
25 - (x + 5) = -15
- (x + 5) = -15 - 25
- (x + 5) = -40
x + 5 = 40
x = 40 - 5
x = 35
c) 34 + (21 - x) = (3747 - 30) - 3746
34 + 21 - x = (3747 - 30) - 3746
55 - x = -29
x = 55 - (-29)
x = 84
1/
a)28+2x=35-(-13)
28+2x=48
2x=48-28
x=20:2
x=10
b)2(x+1)2=-7+15
2(x+1)2=8
(x+1)2=4
x+12=22
x+1=2
x=1
c)(-25)-(x+5)=415+5(x-83)
-25-5-x=415+5x-415
-30-x=5x
-30=x+5x=6x
x=-30:6
x=-5
d)|x-2|+(-7)=-3
|x-2|=-3+7
|x-2|=4
=>x-2=4 hoặc x-2=-4
=>x=6 hoặc x=-2
Vậy x=6 hoặc x=-2
a: =2(3h15'+4h45')=2x8h=16h
b: =(24'30s+25'30s):5
=50':5=10'
a) x - 15 = 7 + (-2)
x-15 = 5
x=20
b) | x - 5 | = 15 + (-8)
I x - 5 I = 7
1) x-5 = 7 => x = 12
2) x-5 = -7 => x= -2
Vậy x=12 hoặc x=-2
c) I x | - 25 = 45 + (-15)
I x | - 25 = 30
I xI = 55
Vậy x= 55 hoặc x=-55
a) x - 15 = 7 + ( - 2 )
x - 15 = 5
x = 5 + 15
x = 20
b) | x - 5 | = 15 + ( - 8 )
| x - 5 | = 7
=> x = 7 + 5
x = 12
c) | x | - 25 = 45 + ( - 15 )
| x | - 25 = 30
=> | x | = 30 + 25
| x | = 55
=> x = 55
-45 x 65 - 45 x 30 - 45 x 3
= -45 x (65 + 30 + 3)
= -45 x 98
= -230
a) -5x (-9) = -11\(\Rightarrow-5\chi=\frac{11}{9}\Rightarrow\chi=\frac{11}{9}:\left(-5\right)\Rightarrow\chi=\frac{11}{9}.\frac{-1}{5}\Rightarrow\chi=\frac{-11}{45}\)b) 25 - (x + 5 ) = -415 - ( 15-415)\(\Rightarrow25-\left(\chi+5\right)=-415-15+415\)
\(\Rightarrow25-\left(\chi+5\right)=\left(-415+415\right)-15\)
\(\Rightarrow25-\left(\chi+5\right)=-15\Rightarrow\chi+5=25+15\)
\(\Rightarrow\chi+5=40\Rightarrow\chi=35\)
HTDT
\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x=-\frac{13}{4}\)
\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)
\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)
\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)
\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)
\(\Leftrightarrow x=-\frac{6}{11}\)
d,e,f Tương tự
Đề \(\Leftrightarrow5x^2-15x+5+x-5x^2=x-2\)\(\Leftrightarrow\left(5x^2-5x^2\right)+\left(x-15x-x\right)+\left(5+2\right)=0\)
\(\Leftrightarrow-15x+7=0\)\(\Leftrightarrow15x-7=0\)\(\Leftrightarrow15x=7\)\(\Leftrightarrow x=\frac{7}{15}\)
Vậy \(S=\frac{7}{15}\)
a, \(\Leftrightarrow\)\(2x=-32\)
\(\Leftrightarrow\)\(x=-16\)
b,\(\Leftrightarrow\)\(25-\left(5x-3\right)=64\)
\(\Leftrightarrow\)\(5x-3=-39\)
\(\Leftrightarrow\)\(5x=-36\)
\(\Leftrightarrow\)\(x=-7,2\)
c,\(\Leftrightarrow\)\(25-\left(x-5\right)=15\)
\(\Leftrightarrow\)\(x-5=10\)
\(\Leftrightarrow\)\(x=15\)