(3x-10):10=50
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cac ban oi truoc cho 3x co dau ngoac va sau so 10 cung co dau ngoac
3x - 10: 10 = 50
=>. 3x -1 = 50
=> 3x = 50 +1
=> 3x = 51
=> x = 51: 3
=> x = 17
\(\dfrac{3x-40}{50}+\dfrac{3x-10+2x+60+4x-360}{40}=0\)
=> \(\dfrac{3x-40}{50}+\dfrac{9x-310}{40}=0\)
=> \(\dfrac{3x-40}{50}=\dfrac{-9x+310}{40}\)
=> \(40\left(3x-40\right)=50\left(-9x+310\right)\)
=> \(120x-1600=-450x+15500\)
=> \(120x+450x=15500+1600\)
Hay \(570x=17100\)
=>x = 30
Hơi dài nhé bạn
\(\dfrac{3x-40}{50}\)+\(\dfrac{3x-10+2x+60+4x-360}{40}\)=0
⇒\(\dfrac{3x-40}{50}\)+\(\dfrac{9x-310}{40}\)=0
⇒\(\dfrac{3x-40}{50}\)=\(\dfrac{9x-310}{40}\)
⇒40(3x -40) = 50(-9x+310)
⇒120x - 1600 = -450x + 15500
⇒120x + 450x = 15500 + 1600
Mặt khác: 570x = 17100
⇒x = 30
LÀM LUN NHA
2/3x-1/2+x=1/10
5/3x=1/10+1/2
5/3x=3/5
X=9/25
**** CHO MK NHA
a) Ta có: \(\left|x\right|-\left(-9\right)^3=-100\)
\(\Leftrightarrow\left|x\right|-\left(-729\right)=-100\)
\(\Leftrightarrow\left|x\right|+729=-100\)
\(\Leftrightarrow\left|x\right|=-100-729=-829\)(vô lý)
Vậy: \(x\in\varnothing\)
b) Ta có: \(3x-\left(-10\right)=2x+50\)
\(\Leftrightarrow3x+10-2x-50=0\)
\(\Leftrightarrow x-40=0\)
\(\Leftrightarrow x=40\)
Vậy: x=40
\(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+10=-50\)
\(\Leftrightarrow-62x+12=-50\)
\(\Leftrightarrow-62x=-62\)
\(\Leftrightarrow x=1\)
a)
\(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)
\(\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}\)
\(\frac{1}{3}:x=\frac{-1}{15}\)
\(x=\frac{1}{3}:\frac{-1}{15}\)
\(x=-5\)
b)
\(\frac{10}{3x}+\frac{67}{4}=-13.25\)
\(\frac{10}{3x}+\frac{67}{4}=\frac{-53}{4}\)
\(\frac{10}{3x}=\frac{-53}{4}+\frac{67}{4}\)
\(\frac{10}{3x}=\frac{7}{2}\)
\(\Rightarrow10.2=7.3x\)
\(20=21x\)
\(x=\frac{20}{21}\)
c)
x+30%x=-1.3
x+0.3x=-1.3
x (1+0.3) = -1.3
x . 1.3 = -1.3
x = -1.3 : 1.3
x = 1
d)
\(\left(\frac{14}{5x}-50\right):\frac{2}{3}=51\)
\(\frac{14}{5x}-50=51.\frac{2}{3}\)
\(\frac{14}{5x}-50=34\)
\(\frac{14}{5x}=34+50\)
\(\frac{14}{5x}=84\)
\(84.5x=14\)
420x = 14
\(x=\frac{1}{30}\)
a, \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)
=> \(\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}\)
=>\(\frac{1}{3}:x=\frac{-1}{15}\)
=> \(x=\frac{1}{3}:\frac{-1}{15}\)
=> \(x=\frac{-1}{5}\)
e) Ta có: \(E=\left(3x+2\right)\left(3x-5\right)\left(x-1\right)\left(9x+10\right)+24x^2\)
\(=\left(9x^2-15x+6x-10\right)\left(9x^2+10x-9x-10\right)+24x^2\)
\(=\left(9x^2-10-9x\right)\left(9x^2-10+x\right)+24x^2\)
\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)-9x^2+24x^2\)
\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)+15x^2\)
\(=\left(9x^2-10\right)^2-3x\left(9x^2-10\right)-5x\left(9x^2-10\right)+15x^2\)
\(=\left(9x^2-10\right)\left(9x^2-3x-10\right)-5x\left(9x^2-10-3x\right)\)
\(=\left(9x^2-3x-10\right)\left(9x^2-5x-10\right)\)
c)
\(4\left(3x-4\right)-2=18\)
<=> \(12x-16-2=18\)
<=> \(12x=36\)
<=> \(x=3\)
Vậy x=3
d)
\(\left(3x-10\right):10=50\)
<=> \(3x-10=500\)
<=> \(3x=510\)
<=> x= \(170\)
Vậy x= 170
f)
\(x-\left[42+\left(-25\right)\right]=-8\)
<=> \(x-17=-8\)
<=> x= \(9\)
Vậy x=9
h)
\(x+5=20-\left(12-7\right)\)
<=> \(x+5=15\)
<=> \(x=10\)
Vậy x= 10
k)
\(\left|x-5\right|=7-\left(-3\right)\)
<=> \(\left|x-5\right|=10\)
* Với \(x>=5\) ; ta được:
\(x-5=10\)
<=> x= 15 (thoả mãn điều kiện )
*Với \(x< 5\) ; ta được:
\(-\left(x-5\right)=10\)
<=> \(-x+5=10\)
<=> \(-x=5\)
<=> \(x=-5\) (thoả mãn điều kiện)
Vậy x=15 ; x= -5
i)
\(\left|x-5\right|=\left|7\right|\)
<=> \(\left|x-5\right|=7\)
*Với \(x>=5\) ; ta được:
\(x-5=7\)
<=> \(x=12\) (thoả mãn)
*Với \(x< 5\) ; ta được:
\(-\left(x-5\right)=7\)
<=> \(-x=2\)
<=> \(x=-2\) (thoả mãn)
Vậy x= 12; x= -2
m)
\(2^{x+1}.2^{2009}=2^{2010}\)
<=> \(2^{x+1+2009}=2^{2010}\)
<=> \(2^{x+2010}=2^{2010}\)
=> \(x+2010=2010\)
=> \(x=0\)
Vậy x=0
n)
\(10-2x=25-3x\)
<=>\(x=15\)
Vậy x=15
(3x-10):10=50
3x-10=500
3x=510
x=170
3x-10=500
3x=510
x=170