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a, \(\frac{x-3}{y-2}=\frac{3}{2}\)và \(x-y=4\)
Theo bài ra ta có :
\(\frac{x-3}{y-2}=\frac{3}{2}\Leftrightarrow2x-6=3y-6\Leftrightarrow2x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{2}\)
Áps dụng tính chất dãy tỉ số bằng nhau ta đc :
\(\frac{x}{3}=\frac{y}{2}=\frac{x-y}{3-2}=\frac{4}{1}=4\)
\(\frac{x}{3}=4\Leftrightarrow x=12\)
\(\frac{y}{2}=4\Leftrightarrow y=8\)
Tương tự với b thôi bn.
\(\frac{x}{4}=\frac{18}{x+1}\)
\(\Leftrightarrow\)\(x\left(x+1\right)=4.18\)
\(\Leftrightarrow\)\(x\left(x+1\right)=72\)
\(\Leftrightarrow\)\(x^2+x-72=0\)
\(\Leftrightarrow\)\(\left(x-8\right)\left(x+9\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-8=0\\x+9=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=8\\x=-9\end{cases}}\)
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 \(-9\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=\frac{2009}{2011}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}=\frac{2009}{4022}\)(nhân mỗi vế với 1/2)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2009}{4022}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}=\frac{1}{2011}\)
\(\Rightarrow x+1=2011\Rightarrow x=2010\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2009}{2011}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}\right)=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)\(=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)\(=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2011}\)
\(\Rightarrow x+1=2011\)
\(\Rightarrow x=2010\)
a) Ta có: \(-x+\frac{4}{7}=\frac{1}{3}\)
\(\Leftrightarrow-x=-\frac{5}{21}\)
\(\Rightarrow x=\frac{5}{21}\)
b) Ta có: \(x\div\left(-\frac{1}{3}\right)^2=-\frac{1}{3}\)
\(\Rightarrow x=\left(-\frac{1}{3}\right)^3=-\frac{1}{27}\)
c) \(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{5}\right)^7\)
\(\Rightarrow x=\left(\frac{3}{5}\right)^2=\frac{9}{25}\)
\(a.-x+\frac{4}{7}=\frac{1}{3}\)
\(-x=\frac{1}{3}-\frac{4}{7} \)
\(-x=\frac{7}{21}-\frac{12}{21}\)
\(-x=\frac{-5}{21}\)
\(x=\frac{5}{21}\)
\(b.x:\left(\frac{-1}{3}\right)^2=\frac{-1}{3}\)
\(x=\frac{-1}{3}.\left(\frac{-1}{3}\right)^2\)
\(x=\frac{-1}{3}.\frac{-1}{3}.\frac{-1}{3}\)
\(x=\frac{-1}{27}\)
\(c.\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{5}\right)^7\)
\(x=\left(\frac{3}{5}\right)^7:\left(\frac{3}{5}\right)^5\)
\(x=\left(\frac{3}{5}\right)^2\)
\(x=\frac{3}{5}.\frac{3}{5}\)
\(x=\frac{9}{25}\)