2 . X + 25 + 3 . X = 50
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a: =25(15+45*3)
=25*150
=3750
b: \(=-10\left(25+75-50\right)=-10\cdot50=-500\)
c: =>3^x-2=27
=>x-2=3
=>x=5
d: =>2x-5=-4
=>2x=1
=>x=1/2
e: =>2(x-1)^2=32
=>(x-1)^2=16
=>x-1=4 hoặc x-1=-4
=>x=-3 hoặc x=5
f: =>25(x+3)=75
=>x+3=3
=>x=0
4 × 125 × 25 × 8
= (4 × 25) × (125 × 8)
= 100 × 1000
= 100 000
2 × 8 × 50 × 25 × 125
= (2 × 50) × (8 × 125) × 25
= 100 × 1000 × 125
= 12 500 000
2 × 3 × 4 × 5 × 50 × 25
= (2 × 50) × (4 × 25) × (3 × 5)
= 100 × 100 × 15
= 150 000
25 × 20 × 125 × 8 - 8 × 20 × 5 × 125
= 8 × 125 × 20 × (25 - 5)
= 1000 × 20 × 20
= 400 000
- (25 x 4) x (125 x8)=100 x 1000 =100000
- (50 x 2) x (8 x 125) x 25=100 x 1000 x 25 =2500000
- (2 x 50 ) x (4 x 25) x (3 x5)=100 x 100 x 15=150000
- 25 x 20 x125 x 8 - 5
= (20 x 5) x (125 x 8) -25=100 x 1000 -5 =100000 - 5 =99995
a) 450 – ( 25 – 10) = 450 – 15
= 435
450 – 25 – 10 = 425 – 10
= 415
b) 180 : 6 : 2 = 30 : 6
= 15
180 : ( 6 : 2 ) = 180 : 3
= 60
c) 410 – (50 +30) = 410 -80
= 330
410 - 50 + 30 = 360 + 30
= 390
d) 16 x (6 : 3) = 16 x 2
= 32
16 x 6 : 3 = 96 : 3
= 32
1/2×(2×x-3)+105/2=-137/2
1/2×(2×x-3)=-137/2-105/2
1/2×(2×x-3)=-242/2
2×x-3=-242/2:1/2
2×x-3=-242
2.x=(-242)+3
2.x=239
x=239:2
x=239/2
A,-12×25-25×75-25×13
=[(-12)-75-13].25
=(-100).25
-2500
B,-50/7×49/10-35/2×-10/7+ -25/3× -9/5
=(-35)-(-25)+(-13)
=-23
C,-354+265-156+125
=[(-354)-156]+(265+125)
=(-510)+277
=-233
D,-74+40-50+16-35
=(-74)+40+(-50)+16+(-35)
=[(-74)+(-35)]+[40+(-50)+16]
=(-109)+26
=-83
E,-2/3×4/5+-4/5×4/3-0,125
=-2/3.4/5+4/5.(-4/3)-0,125
=4/5.[-2/3+(-4/3)]-0,125
=4/5.(-2)-0,125
=-8/5-0,125
=(-1,6)+(-0,125)
=-1,725
Ta có : \(\frac{x}{50}+\frac{x-1}{49}+\frac{x-2}{48}+\frac{x-3}{47}+\frac{x-150}{25}=0\)
=> \(\frac{x}{50}-1+\frac{x-1}{49}-1+\frac{x-2}{48}-1+\frac{x-3}{47}-1+\frac{x-150}{25}+4=0\)
=> \(\frac{x-50}{50}+\frac{x-50}{49}+\frac{x-50}{48}+\frac{x-50}{47}+\frac{x-50}{25}=0\)
=> \(\left(x-50\right)\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{25}\right)=0\)
=> \(x-50=0\)
=> \(x=50\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{50\right\}\)
`@` `\text {Ans}`
`\downarrow`
`2.x + 25 + 3.x = 50`
`\Rightarrow x.(2 + 3) = 50 - 25`
`\Rightarrow 5x = 25`
`\Rightarrow x = 25 \div 5`
`\Rightarrow x = 5`
Vậy, `x = 5.`