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a) \(\left(-4\right)\left(-3\right)\left(-125\right).25.\left(-8\right)=\left[25\left(-4\right)\right]\left[\left(-125\right)\left(-8\right)\right]\left(-3\right)\)
\(=\left(-100\right).1000.\left(-3\right)=300.1000=300000\)
b) \(\left(-4\right).9\left(-125\right).25.\left(-8\right)=\left[25\left(-4\right)\right]\left[\left(-125\right)\left(-8\right)\right].9\)
\(=\left(-100\right).1000.9=\left(-100\right).9000=-900000\)
c) \(7.\left(-25\right)\left(-3\right).2.\left(-4\right)=\left[\left(-25\right)\left(-4\right)\left(-3\right)\right].7.2=-4200\)
d) \(\left[93-\left(20-7\right)\right]\div16=\left(93-13\right)\div16=80\div16=5\)
e) \(53-\left(-51\right)+\left(-53\right)+49=\left(53-53\right)+\left(51+49\right)=100\)
f) \(168-49+\left(-68\right)+4=168-49-68+4=55\)
Bài 2:
a: =>x=0 hoặc x+3=0
=>x=0 hoặc x=-3
b: =>x-2=0 hoặc 5-x=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
Lời giải:
a.
$(x-15).27=0$
$x-15=0:27=0$
$x=15+0=15$
b.
$23(42-x)=0$
$42-x=0$
$x=42$
c.
$(9x+2).3=60$
$9x+2=60:3=20$
$9x=18$
$x=2$
d.
$71+(26-3x):5=75$
$(26-3x):5=75-71=4$
$26-3x=4.5=20$
$3x=26-20=6$
$x=6:2=3$
1, \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\) ( Trừ mỗi vế cho 2 ta được phương trình như này nhé ! )
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Do \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên \(x-2010=0\Leftrightarrow x=2010\)
2, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)
\(\Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\) \(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\) Do \(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\ne0\) nên \(100-x=0\Leftrightarrow x=100\)
2:
a: \(126⋮x;144⋮x\)
=>x thuộc ƯC(126;144)
mà x lớn nhất
nên x=UCLN(126;144)=18
b: 121 chia x dư 1
=>121-1 chia hết cho x
=>120 chia hết cho x(1)
183 chia x dư 3
=>183-3 chia hết cho 3
=>180 chia hết cho x(2)
Từ (1), (2) suy ra \(x\inƯC\left(120;180\right)\)
mà x lớn nhất
nên x=ƯCLN(120;180)=60
c: 240 và 384 đều chia hết cho x
=>\(x\inƯC\left(240;384\right)\)
=>\(x\inƯ\left(48\right)\)
mà x>6
nên \(x\in\left\{8;12;16;24;48\right\}\)
Bài 2:
a: =>x-1=1 hoặc x-1=-1
=>x=2 hoặc x=0
b: =>x+1=-1
hay x=-2
c: =>(135-7x):9=8
=>135-7x=72
=>7x=63
hay x=9
d: =>(x+7)(x-3)<0
=>-7<x<3
e: \(\Leftrightarrow3^{x-3}=18+9=27\)
=>x-3=3
hay x=6
f: =>4-2x=0
hay x=2
a) (x + 49) - 115 = 0
x + 49 = 0 + 115
x + 49 = 115
x = 115 - 49
x = 66
Vậy x = 66
b) 125 . (2x - 3) = 55 : 53
125 . (2x - 3) = 52 = 25
2x - 3 = 25 : 125
2x - 3 = 1/5
2x = 1/5 + 3
2x = 1/5 + 15/5
2x = 16/5
x = 16/5 : 2
x = 16/5 . 1/2
x = 8/5
Vậy x = 8/5
a) (x+49)-115=0
=>x+49=0+115=115
=>x=115-49=66
b)125.(2x-3)=55:53
=>125.(2x-3)=52
=>125(2x-3)=25
=>2x-3=25:125
=>2x-3=\(\frac{1}{3}\)
=>2x=\(\frac{1}{3}\)+3
=>2x=\(\frac{10}{3}\)
=>x=\(\frac{10}{3}:2\)
=>x=\(\frac{5}{3}\)