(x-5)x30/100=200x/100+5
Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
[x-5] x 30/100= 200x : 100+5
(x-5) x \(\frac{30}{100}\)= 200x : 100 + 5
(x-5) x \(\frac{3}{10}\)= 200x : 100 + 5
\(\frac{x.3-15}{10}\)= 2x + 5
\(\frac{x.3-15}{10}-2x=5\)
\(\frac{x.3-15}{10}-\frac{20x}{10}=5\)
\(\frac{3x-15-20x}{10}=5\)
3x - 20x - 15 = 50
-17x - 15 = 50
-17x = 65
x = 65 : -17
x = \(\frac{-65}{17}\)
( x - 5 ) . \(\dfrac{30}{100}\)= \(\dfrac{200x}{100}\)+5
=>3/10(x-5)=2x+5
=>3/10x-3/2=2x+5
=>-17/10x=5+3/2=6/2+3/2=11/2
=>x=-55/17
TÌM x:
\(\left(x-5\right)\times\frac{30}{100}=\frac{200x}{100}+5\)
tim x biet
a)2^x+2 -2^x=96
b)x-(50x/100+25x/200)=111/4
c) (x-5).30/100=200x/100+5
a) \(2^{x+2}-2^2=96\)
<=> \(2^x.2^2-2^x=96\)
<=> \(2^x\left(4-1\right)=96\)
<=> \(3.2^x=96\)
<=> \(2^x=32\)
<=> \(2^x=2^5\)
<=> x = 5
b, \(x-\left(\frac{50x}{100}+\frac{25x}{200}\right)=11\frac{1}{4}\)
\(\Rightarrow x-\left(\frac{1x}{2}+\frac{1x}{8}\right)=\frac{45}{4}\)
\(\Rightarrow x-\left(\frac{4x}{8}+\frac{1x}{8}\right)=\frac{45}{4}\)
\(\Rightarrow x-\frac{5x}{8}=\frac{45}{4}\)
\(\Rightarrow\frac{8x}{8}-\frac{5x}{8}=\frac{45}{4}\)
\(\Rightarrow\frac{3x}{8}=\frac{45}{4}\Rightarrow x=\frac{45}{4}\div\frac{3}{8}=30\)
Vậy x = 30
tìm x biết:
a,(3x+ 1) x ( x-2000)x (3x +6000) = 0
b, (3x-1)^2 = 169
c,|2x+5|=1/2
d, x-6 / x-7= x+1 / x-3
e, (x-5)x30 / 100=20x/100 + 5
a) pt a <=> 3x+1=0 hoặc x-2000=0 hoặc 3x+6000=0
<=> x=-1/3 hoặc x=2000 hoặc x=-2000
Tìm x:
a) x - (\(\frac{50x}{100}\) + \(\frac{25x}{100}\)) = \(11\frac{1}{4}\)
b) (x-5) . \(\frac{30}{100}\) = \(\frac{200x}{100}\) + 5
- Help me!.
Bài 1: Tìm X
a) (X-5)x30%= \(\dfrac{20xX}{100}\)+5
b) \(\dfrac{X+3}{X-2}\)∈ Z
a, \(\left(x-5\right).30\%=\dfrac{20x}{100}+5\Rightarrow\left(x-5\right).\dfrac{3}{10}=\dfrac{x}{5}+5\)
\(\Rightarrow\dfrac{3}{10}x+\dfrac{3}{2}=\dfrac{x}{5}+5\Rightarrow\dfrac{3}{10}x-\dfrac{x}{5}=5-\dfrac{3}{2}\)
\(\Rightarrow x\left(\dfrac{3}{10}-\dfrac{1}{5}\right)=\dfrac{7}{2}\Rightarrow\dfrac{1}{10}x=\dfrac{7}{2}\Rightarrow x=35\)
b, \(\dfrac{x+3}{x-2}\in Z\)
Ta có: \(\dfrac{x-2+5}{x-2}=\dfrac{x-2}{x-2}+\dfrac{5}{x-2}=1+\dfrac{5}{x-2}\)
Để \(\dfrac{x+3}{x-2}\in Z\Leftrightarrow\dfrac{5}{x-2}\Leftrightarrow5⋮\left(x-2\right)\Rightarrow\left(x-2\right)\in U\left(5\right)\)
\(\Rightarrow\left(x-2\right)\in\left\{1;5\right\}\Rightarrow x\in\left\{-1;3\right\}\)
Kiến thức lâu k học :> sai gì bỏ qua giùm :<<
Tìm x:
a) x - (\(\frac{50x}{100}+\frac{25x}{200}\)) = \(11\frac{1}{4}\)
b) (x - 5) . \(\frac{30}{100}\) = \(\frac{200x}{100}\) + 5
Tìm x;y thuộc N:
a/ \(\frac{4}{x}\)- \(\frac{y}{3}\)= \(\frac{5}{6}\)
b/ ( x-5) . \(\frac{30}{100}\)= \(\frac{200x}{100}\)+ 5