Tìm x
a. ( 27-x ) . 2018= 0
b. 27. x + 23 . x = 100
Trình bảy luôn nha
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\(a,\Leftrightarrow\left(x-2\right)^3-3x\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-2-3x\right)=0\\ \Leftrightarrow\left(x-2\right)\left(-2x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\\ b,\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\\ \Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
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.
$71+(26-3x):5=75$
$(26-3x):5=75-71=4$
$26-3x=4.5=20$
$3x=26-20=6$
$x=6:2=3$
a, <=> (x-5/100) -1 +(x-4/101) -1 +(x-3/102) -1= (x-100/5) -1+(x-101/4) -1 +(x-102/3) -1
<=> (x-105)(1/100 +1/101 +1/102)= (x-105)(1/5+1/4+1/3)
<=> (x-105)(1/100+1/101+1/102-1/5-1/4-1/3)=0
vì 1/100+1/101+1/102-1/5-1/4-1/3 khác 0 <=> x-105=0
<=> x=105
b, 29-x/21 +1+27-x/23 +1+25-x/25 +1+23-x/27 +1+21-x/29 +1=0
<=> 50-x/21 +50-x/23 +50-x/25 +50-x/27 +50-x/29=0
<=> (50-x)(1/21 +1/23 +1/25 +1/27 +1/29)=0
vì 1/21+1/23+1/25+1/27+1/29 lớn hơn 0
nên 50-x=0
<=> x=50
a) \(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
\(\Leftrightarrow\left(\frac{x-5}{100}-1\right)+\left(\frac{x-4}{101}-1\right)+\left(\frac{x-3}{102}-1\right)=\left(\frac{x-100}{5}-1\right)+\left(\frac{x-101}{4}-1\right)+\left(\frac{x-102}{3}-1\right)\)
\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)
\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
\(\Leftrightarrow x=105\)
b) \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)
\(\Leftrightarrow\left(\frac{29-x}{21}+1\right)+\left(\frac{27-x}{23}+1\right)+\left(\frac{25-x}{25}+1\right)+\left(\frac{23-x}{27}+1\right)+\left(\frac{21-x}{29}+1\right)=0\)
\(\Leftrightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
\(\Leftrightarrow x=50\)
a) 2016 - 100.(x+31)=27:23
<=> 2016 - 100.(x+31) = 24 = 16
<=> 100.(x+31) = 2016 - 16 = 2000
<=> x+31 = 2000:100 = 20
<=> x = 20 - 31 = -11
b) (x+8)3 = 125
<=> (x+8)3 = 53
<=> x+8 = 5
<=> x = 5 - 8
<=> x = -3
mk chi chep thua so thu 3 thoi nhe con 2 thua so con lai giu nguyen
42/30.75/23-19/23.210/38
=7/5.75.1/23-1/23.19.210/38
=7.15.1/23-1/23.210/2
=105.1/23-1/23.105=0
gia tri bieu thuc=0
a: =>x+28=0
=>x=-28
b: =>27-x=0 hoặc x+9=0
=>x=27 hoặc x=-9
c: =>x=0 hoặc x-43=0
=>x=0 hoặc x=43
a) (2x-6)3 = (2x-6)2018
=> (2x-6)3 - (2x-6)2018 = 0
(2x-6)3.[1-(2x-6)2015 ] = 0
=> (2x-6)3 = 0 =>...
1 - (2x-6)2015 = 0 => (2x-6)2015 = 1 => ...
b) (2x-1)3 = 27 = 33
=> 2x - 1 = 3
=> ...
c) (x + 1) + (x+2) + (x+3) + ...+ (x+100) = 5750
x.100 + (1+2+3+...+100) = 5750
x.100 + [(1+100).100:2] = 5750
x.100 + 5050 = 5750
x.100 = 700
x = 7
a. \(\left(27-x\right)\cdot2018=0\)
\(\left(27-x\right)=0\text{ : }2018\)
\(\left(27-x\right)=0\)
\(x=0+27\)
\(x=27\)
b. \(27\cdot x+23\cdot x=100\)
\(\left(27+23\right)\cdot x=100\)
\(50\cdot x=100\)
\(x=100\text{ : }50\)
\(x=2\)
a, (27-x).2018=0
27-x=0:2018
27-x=0
x=27-0
x=27
b,27.x+23.x=100
x.(27+23)=100
x.50=100
x=100:50
x=2
tk mk nha