tim x thuoc Z biet \(\frac{x}{4}\)= \(\frac{18}{x+1}\)
giup voi
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\(\frac{x-1}{2}\)= \(\frac{2y-4}{6}\)=\(\frac{3z-9}{12}\)=\(\frac{x-1-2y+4+3z-9}{2-6+12}\)= \(\frac{14-1+4-9}{8}\)= 1
=> x =2+1=3
y= (6+4) : 2=5
z=(12+9) : 3=7
1/18 < x/12 < y/9 < 1/4.
Ta quy dong mau len co mau chung la 36: 2/36 < x.3/36 < y.4/36 < 9/36.
Suy ra: Vi 2<x<y<9 nen phai bang 3;4;5;6;7;8:
x.3 3 4 5 6 7 8
x 1 loai loai 2 loai loai
y.4 3 4 5 6 7 8
y loai 1 loai loai loai 2
Suy ra ta co 3 truong hop:
TH1: x=1;y=1: 2/36 < 1.3/36 < 1.4/36 < 9/36
TH2: x=2;y=2: 2/36 < 2.3/36 < 2.4/36 < 9/36
TH3: x=1;y=1: 2/36 < 1.3/36 < 2.4/36 < 9/36
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=\frac{2009}{2011}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}=\frac{2009}{4022}\)(nhân mỗi vế với 1/2)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2009}{4022}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}=\frac{1}{2011}\)
\(\Rightarrow x+1=2011\Rightarrow x=2010\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2009}{2011}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}\right)=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)\(=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)\(=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2011}\)
\(\Rightarrow x+1=2011\)
\(\Rightarrow x=2010\)
\(\frac{x}{4}=\frac{18}{x+1}\)
\(\Leftrightarrow\)\(x\left(x+1\right)=4.18\)
\(\Leftrightarrow\)\(x\left(x+1\right)=72\)
\(\Leftrightarrow\)\(x^2+x-72=0\)
\(\Leftrightarrow\)\(\left(x-8\right)\left(x+9\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-8=0\\x+9=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=8\\x=-9\end{cases}}\)
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
⇔[
⇔[
Vậy \(x=8\)hoặc \(-9\)