Tìm x, biết:
\(3^{x+1}\) \(+\) \(3^{x-1}\) \(=90\)
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sao nhìn nó lạ lắm ko giống x đâu bn nên bn ghi lại đi để mik nhìn rõ hơn nha :))
\(x\in\left\lbrace45,46\right\rbrace\) nhé
a) \(\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
=> \(\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x+\frac{1}{3}=0\)
=> \(\left(\frac{2}{3}x-\frac{1}{2}x\right)+\left(-\frac{2}{5}+\frac{1}{3}\right)=0\)
=> \(\frac{1}{6}x-\frac{1}{15}=0\Rightarrow\frac{1}{6}x=\frac{1}{15}\Rightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{2}{5}\)
Vậy x = 2/5
b) \(\frac{1}{3}x+\frac{2}{5}\left(x+1\right)=0\)
=> \(\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
=> \(\frac{11}{15}x+\frac{2}{5}=0\Rightarrow\frac{11}{15}x=-\frac{2}{5}\)
=> \(x=\left(-\frac{2}{5}\right):\frac{11}{15}=\left(-\frac{2}{5}\right)\cdot\frac{15}{11}=-\frac{6}{11}\)
Vậy x = -6/11
c) \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
=> \(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
=> \(\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}\right)+\left(-\frac{1}{3}x-x\right)=5\)
=> \(\frac{2}{3}-\frac{4}{3}x=5\)
=> \(\frac{4}{3}x=-\frac{13}{3}\Rightarrow x=\left(-\frac{13}{3}\right):\frac{4}{3}=\left(-\frac{13}{3}\right)\cdot\frac{3}{4}=-\frac{13}{4}\)
Vậy x = -13/4
d) \(\frac{11}{5}-\left(\frac{7}{9}-x\right)\cdot\frac{3}{8}=\frac{61}{90}+\frac{x}{3}\)
=> \(\frac{11}{5}-\frac{3}{8}\left(\frac{7}{9}-x\right)=\frac{61}{90}+\frac{30x}{90}\)
=> \(\frac{11}{5}-\frac{7}{24}+\frac{3}{8}x=\frac{61+30x}{90}\)
=> \(\frac{229}{120}+\frac{3}{8}x=\frac{61+30x}{90}\)
=> \(\frac{229}{120}+\frac{3x}{8}=\frac{61+30x}{90}\)
=> \(\frac{229}{120}+\frac{45x}{120}=\frac{61+30x}{90}\)
=> \(\frac{229+45x}{120}=\frac{61+30x}{90}\)
=> \(\frac{3\left(229+45x\right)}{360}=\frac{4\left(61+30x\right)}{360}\)
=> \(3\left(229+45x\right)=4\left(61+30x\right)\)
=> \(687+135x=244+120x\)
=> \(687+135x-244-120x=0\)
=> \(\left(687-244\right)+\left(135x-120x\right)=0\)
=> \(443+15x=0\)
=> \(15x=-443\Rightarrow x=-\frac{443}{15}\)
Vậy x = -443/15
Ta có:
1/30+1/42+1/56+1/72+1/90+1/110+1/132 -x = 2/3
=>1/(5.6) + 1/(6.7) + 1/(7.8) + (1/8.9) + 1/(9.10) + 1/(10.11) + 1/(11.12) - x = 2/3
=>1/5-1/6+1/6-1/7+...+1/11-1/12 - x = 2/3
=>1/5-1/12 - x = 2/3
=>7/60 - x = 2/3
=> x = 7/60 - 2/3
=> x = -11/20
Ta có:
1/30+1/42+1/56+1/72+1/90+1/110+1/132 -x = 2/3
=>1/(5.6) + 1/(6.7) + 1/(7.8) + (1/8.9) + 1/(9.10) + 1/(10.11) + 1/(11.12) - x = 2/3
=>1/5-1/6+1/6-1/7+...+1/11-1/12 - x = 2/3
=>1/5-1/12 - x = 2/3
=>7/60 - x = 2/3
=> x = 7/60 - 2/3
=> x = -11/20
sai đề
(x+1 ) + (x+3)+..........+(x+99)=0
=> x + 1 + x + 3 + x + 5 + ........... + x + 99 = 0
Từ 1-->99 có số số hạng là:
(99-1):2+1=50(số)
Tổng từ 1 -->99 là:
(99+1) x 50:2=2500
=> 50x + 2500=0
=>50x=-2500
=>x=-50
Câu 2 mình k hiểu đề
a: \(\left(1-\frac12\right)\times\left(1-\frac13\right)\times\ldots\times\left(1-\frac{1}{99}\right)\)
\(=\frac12\times\frac23\times\ldots\times\frac{98}{99}\)
\(=\frac{1}{99}\)
b: \(\left(x+\frac12\right)+\left(x+\frac16\right)+\left(x+\frac{1}{12}\right)+\left(x+\frac{1}{20}\right)+\cdots+\left(x+\frac{1}{90}\right)=\frac{99}{10}\)
=>\(\left(x+\frac{1}{1\times2}\right)+\left(x+\frac{1}{2\times3}\right)+\cdots+\left(x+\frac{1}{9\times10}\right)=\frac{99}{10}\)
=>\(9\times x+1-\frac12+\frac12-\frac13+\cdots+\frac19-\frac{1}{10}=\frac{99}{10}\)
=>\(9\times x+1-\frac{1}{10}=\frac{99}{10}\)
=>\(9\times x=\frac{99}{10}+\frac{1}{10}-1=\frac{100}{10}-1=10-1=9\)
=>x=1
3\(^{x+1}\) + 3\(^{x-1}\) = 90
3\(^{x-1}\).(3\(^2\) + 1) = 90
3\(^{x-1}\).(9 + 1) = 90
3\(^{x-1}\).10 = 90
3\(^{x-1}\) = 90 : 10
3\(^{x-1}\) = 9
3\(^{x-1}\) = 3\(^2\)
\(x-1=2\)
\(x=2+1\)
\(x=3\)
Vậy \(x=3\)
3^x . 3^1 + 3^x: 3^1 = 90
3^x . 3 + 3^x. 1/3 =90
3^x . (3+1/3) = 90
3^x . 10/3 = 90
3^x= 90:10/3
3^x= 27
3^x= 3^3
⇒x=3
Vậy x=3