Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Tìm \(n\in N\), biết:
\(3.5^{2n+1}-3.25^n=300\)
b) Tìm x để:
\(f\left(x\right)=6x^{^{ }4}-2x^3+5=5\)
a)\(3\cdot5^{2n+1}-3\cdot25^n=300\)
\(3\cdot5^{2n}\cdot5-3\cdot25^n=300\)
\(15\cdot25^n-3\cdot25^n=300\)
\(25^n\cdot12=300\)
\(25^n=25\)
\(\Rightarrow n=1\)
b)\(f\left(x\right)=6x^4-2x^3+5=5\)
\(6x^4-2x^3=0\)
\(6x^4=2x^3\)
\(3x^4=x^3\)
\(3x^4-x^3=0\)
\(x^3\left(3x-1\right)=0\)
\(\Rightarrow x^3=0\) hoặc 3x-1=0
\(\Rightarrow x=0,3x=1\)
\(\Rightarrow x=0,x=\frac{1}{3}\)(loại vì \(x\in N\))
Vậy x=0
Đáp án:
x=1
Cách làm:tách 5^2x+1 ra rồi nhóm 25^x ra
xl k giải chi tiết đc :((
#Châu's ngốc
\(3-5^{2x+1}-3.25^x=300\)
\(\Rightarrow3-5^{2x}.5-3.25^x=300\)
\(\Rightarrow3-\left(5^2\right)^x.5-3.25^x=300\)
\(\Rightarrow3-25^x.5-3.25^x=300\)
\(\Rightarrow25^x\left(5-3\right)=3-300\)
\(\Rightarrow25^x.2=-297\)
\(\Rightarrow25^x=\frac{-297}{2}\)
.......................
a/ \(P=\dfrac{x-\dfrac{1}{3}}{3.25-x}>0\)
=> \(\left[{}\begin{matrix}x-\dfrac{1}{3}>0;75-x>0\\x-\dfrac{1}{3}< 0;75-x< 0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x>\dfrac{1}{3};x< 75\\x< \dfrac{1}{3};x>75\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\dfrac{1}{3}< x< 75\\75< x< \dfrac{1}{3}\left(vôlý\right)\end{matrix}\right.\)
Vậy \(\dfrac{1}{3}< x< 75\) để P > 0
b/ \(Q=\dfrac{x+3}{2x-5}< 0\)
TH1: \(\left\{{}\begin{matrix}x+3>0\\2x-5< 0\Rightarrow2x< 5\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x>-3\\x< \dfrac{5}{2}\end{matrix}\right.\) => \(-3< x< \dfrac{5}{2}\)
TH2: \(\left\{{}\begin{matrix}x+3< 0\\2x-5>0\Rightarrow2x>5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x< -3\\x>\dfrac{5}{2}\end{matrix}\right.\) (vô lý)
Vậy \(-3< x< \dfrac{5}{2}\) để Q < 0
a) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.........+\frac{2}{x\left(x+1\right)}=\frac{1998}{2000}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.......+\frac{2}{x\left(x+1\right)}=\frac{1998}{2000}\)
\(\Leftrightarrow2.\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+......+\frac{1}{x\left(x+1\right)}\right]=\frac{1998}{2000}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{x\left(x+1\right)}=\frac{999}{2000}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{x}-\frac{1}{x+1}=\frac{999}{2000}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{999}{2000}\)\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2000}\)
\(\Leftrightarrow x+1=2000\)\(\Leftrightarrow x=1999\)
Vậy \(x=1999\)
b) \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+......+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
\(\Leftrightarrow\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{15.2}{93}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+......+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Leftrightarrow\frac{1}{2x+3}=\frac{1}{93}\)\(\Leftrightarrow2x+3=93\)
\(\Leftrightarrow2x=90\)\(\Leftrightarrow x=45\)
Vậy \(x=45\)
32.1+1-3.251
=375-75
=300
=>x=1