2∧x+2∧x+1+2∧x+2=56
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\(2^x+2^{x+1}+2^{x+2}=56\\ 2^x\left(1+2^1+2^2\right)=56\\ 2^x.7=56\\ 2^x=56:7\\ 2^x=8=2^3\\ Vậy:x=3\)
`#3107.101107`
a)
\(27< 3^x< 243\\ \Rightarrow3^3< 3^x< 3^5\\ \Rightarrow3< x< 5\\ \Rightarrow x=4\)
Vậy, `x = 4`
b)
\(2^x+2^{x+1}+2^{x+2}=56?\\ \Rightarrow2^x+2^x\cdot2+2^x\cdot4=56\\ \Rightarrow2^x\cdot\left(1+2+4\right)=56\\ \Rightarrow2^x\cdot7=56\\ \Rightarrow2^x=8\\ \Rightarrow2^x=2^3\\ \Rightarrow x=3\)
Vậy, `x = 3`
c)
\(3^x+3^{x+2}=810\\ \Rightarrow3^x+3^x\cdot9=810\\ \Rightarrow3^x\cdot\left(1+9\right)=810\\ \Rightarrow3^x\cdot10=810\\ \Rightarrow3^x=81\\ \Rightarrow3^x=3^4\\ \Rightarrow x=4\)
Vậy, `x = 4.`
a) \(27< 3^x< 243\)
\(\Rightarrow3^3< 3^x< 3^5\)
\(\Rightarrow3< x< 5\)
c) \(3^x+3^{x+2}=810\)
\(\Rightarrow3^x\left(1+3^2\right)=810\)
\(\Rightarrow3^x.10=810\)
\(\Rightarrow3^x=810:10\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
Ta có \(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2}{9}\)
\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}\div2\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\left(1\right)\)
Có \(\left(1\right)\Leftrightarrow\left(x+1\right).1=1.18\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=18-1\)
\(\Rightarrow x=17\)
\(\frac{1}{6x7}+\frac{1}{7x8}+\frac{1}{8x9}+...+\frac{1}{Xx\left(X+1\right)}=\frac{1}{9}\)
\(\frac{7-6}{6x7}+\frac{8-7}{7x8}+\frac{9-8}{8x9}+...+\frac{\left(X+1\right)-X}{Xx\left(X+1\right)}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{X}-\frac{1}{X+1}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{X+1}=\frac{1}{9}\)
\(\frac{1}{X+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)
X+1=18
X=17
(x+1)+(x+2)+(x+3)+...+(x+9)+(x+10)=56
⇒ x.10+(1+2+3+...+9+10)=56
⇒ x.10+[(10+1).10:2] = 56
⇒ x.10+55=56
⇒ x.10 = 56-55 = 1
⇒ x = 1:10=0,1
(\(x+1\)) + (\(x+2\)) + (\(x\) + 3)+...+ (\(x\) + 10) = 56
xét dãy số \(x+1;x+2;x+3;...;x+10\)
Dãy số trên có khoảng cách là: \(x+2-x-1\) = 1
Dãy số trên có số số hạng là: (\(x+10-x-1\)) : 1 + 1 = 10
Vậy:
(\(x+1\))+(\(x+2\))+(\(x+3\))+...+(\(x+10\)) = (\(x+10\) + \(x+1\)).10 : 2
⇒(2\(x\) + 11).5 = 56
2\(x\) + 11 = 56 : 5
2\(x\) + 11 = 11,2
2\(x\) = 11,2 - 11
2\(x\) = 0,2
\(x\) = 0,2 : 2
\(x\) = 0,1
3/5 X + 2X + 40 = 53
13/5 X = 53 - 40
13/5 X = 13
X = 13 : 13/5
X = 5
2\(^x\) + 2\(^{x+1}\) + 2\(^{x+2}\) = 56
2\(^x\).(1 + 2 + 22) = 56
2\(^x\).(1 + 2 + 4) = 56
2\(^x\).7 = 56
2\(^x\) = 56 : 7
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)
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