timx,y thuoc z biet:\(\frac{1}{18}<\frac{x}{12}<\frac{y}{9}<\frac{1}{4}\)
[x mu2 -8].[x mu2 -12]<0 Timx biet x thuoc z
timx,y thuoc z thoa x^2+xy+y^2=x^2y^2
Tim x,y thuoc Z biet
x/4=18/x+1
x/4=18/x+1
=>x(x+1)=4.18=72=8.9=8.(8+1)
hoặc x(x+1)=72=(-9).(-8)=(-9).[(-9)+1]
=>x=8 hoặc x=-9
\(\frac{x}{4}=\frac{18}{x+1}\)
=> \(x\left(x+1\right)=4.18\)
=> x (x + 1) = 72
=> x (x + 1) = 8 . 9
=> x = 8
\(\frac{x}{4}=\frac{18}{x+1}\)
=> \(18.4=x\left(x+1\right)=72\)
Mà: 8 . 9 = 72 => x = 8
k có ý ;)
tim x thuoc Z biet \(\frac{x}{4}\)= \(\frac{18}{x+1}\)
giup voi
\(\frac{x}{4}=\frac{18}{x+1}\)
\(\Leftrightarrow\)\(x\left(x+1\right)=4.18\)
\(\Leftrightarrow\)\(x\left(x+1\right)=72\)
\(\Leftrightarrow\)\(x^2+x-72=0\)
\(\Leftrightarrow\)\(\left(x-8\right)\left(x+9\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-8=0\\x+9=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=8\\x=-9\end{cases}}\)
Vậy....
tim x thuoc Z biet x4 = 18x+1
giup voi
Toán lớp 6
Đường Quỳnh Giang 14 giây trước (18:57)
Thống kê hỏi đáp
Báo cáo sai phạm
x4 =18x+1
⇔x(x+1)=4.18
⇔x(x+1)=72
⇔x2+x−72=0
⇔(x−8)(x+9)=0
⇔[
x−8=0 |
x+9=0 |
⇔[
x=8 |
x=−9 |
Vậy \(x=8\)hoặc \(-9\)
Ta có:
x/4=18/x+1
<=>x.(x+1)=18.4
<=>x.(x+1)=72
<=>x.(x+1)=8.(8+1)
<=>x=8
tim x thuoc Z biet : 2x+\(\frac{1}{7}=\frac{1}{y}\)
cho p=(x+3)/(x^2+9). timx thuoc z de p thuoc z
tim x,y thuoc Z biet -3/6=x/-2=-18/y=3/24
ta có : \(\dfrac{-3}{6}=\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{3}{24}\)
\(\Rightarrow\dfrac{-3}{6}=\dfrac{3}{24}\) (vô lí)
\(\Rightarrow\) đề sai
Tim x,y thuoc Z biet: \(2x+\frac{1}{7}=\frac{1}{y}\)
Timx,y biết:
a,\(|x+2|=|2019x+2020|\)
b,\(\frac{x}{18}=\frac{y}{24}\)và -2x+3y=-72
Tìm x,y thuộc Z:
\(\frac{x}{6}-\frac{1}{y}=\frac{1}{2}\)
a)
TH1: x+2 =2019x+2020
x-2019x=2020-2
x(1-2019)=2018
x. (-2018)=2018
x=2018:(-2018)
x=-1
TH2: x+2 = -(2019x+2020)
x+2 =-2019x -2020
x+2019x = -2020-2
2020x=-2022
x=-2022:2020= - 1011/1010