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a: =>\(4\cdot3^x\cdot\dfrac{1}{3}+2\cdot3^x\cdot9=4\cdot3^6+2\cdot3^9\)
=>3^x(4*1/3+2*9)=3^6(4+2*3^3)
=>3^x*58/3=3^6*58
=>3^x/3^6=3
=>x-6=1
=>x=7
b: =>\(2^x\cdot\left(\dfrac{1}{5}+\dfrac{1}{3}\cdot2\right)=2^7\left(\dfrac{1}{5}+\dfrac{1}{3}\cdot2\right)\)
=>2^x=2^7
=>x=7
\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)
`4^(x+2) .3^x = 16.12^5`
`4^(x+2) .3^x = 4^2 . (3.4)^5`
`4^(x+2) .3^x = 4^2 . 3^5. 4^5`
`4^(x+2) .3^x = 4^7 . 3^5`
`4^(x+2) .3^x = 4^(5+2) . 3^5`
`x=5`
Ta có:4x+2.3x=16.125
⇔ 4x+2.3x=417.315
⇔ 4x+2:417.3x:315=1
⇔ 4x-15.3x-15=1
⇔ 12x-15=120
⇔ x-15=0
⇔ x=15
+ 2 mặt đáy : ABC, MNP
+ 3 mặt bên : ACPM, BAMN, BCPN
+ Cạnh đáy : NM, MP, NP, AB, BC, CA
+ Cạnh bên : AM, BN, CP
\(\left(3x+1\right)^2=25\)
\(\Rightarrow\left(3x+1\right)^2=5^2=\left(-5\right)^2\)
\(\Rightarrow\orbr{\begin{cases}3x+1=5\\3x+1=-5\end{cases}\Rightarrow\orbr{\begin{cases}3x=5-1=4\\3x=-5-1=-6\end{cases}}}\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=-2\end{cases}}\)
\(\left[x-\frac{1}{2}\right]+\frac{1}{2}=\frac{5}{8}\)
\(\Rightarrow x-0=\frac{5}{8}\)
\(x=\frac{5}{8}\)
\(\left[x+\frac{3}{4}\right]-\frac{1}{3}=0\)
\(x+\frac{3}{4}=0+\frac{1}{3}=\frac{1}{3}\)
\(x=\frac{1}{3}-\frac{3}{4}\)
\(x=\frac{-5}{12}\)
a)Viết dưới dạng phân số rồi sử dụng tích chéo ý
b)\(\frac{-1}{7}.2^3-2x:1\frac{4}{3}=-2^{x-1}\)
\(\Rightarrow\frac{-8}{7}-2x:\frac{7}{3}=-2^{x-1}\)
\(\Rightarrow\frac{-8}{7}-\frac{6x}{7}=-2^{x-1}\)
\(\Rightarrow\frac{-8-6x}{7}=\frac{2^{x-1}}{-1}\)
\(\Rightarrow-1\left(-8-6x\right)=7.2^{x-1}\)
\(\Rightarrow6x+8=7.2^{x-1}\)
.........
\(2.3^{x+2}+4.3^{x+1}=10.3\)
\(\Leftrightarrow2.3^{x+1}\left(3+2\right)=10.3\)
\(\Leftrightarrow3^{x+1}=3\)
\(\Leftrightarrow x+1=1\)
\(\Leftrightarrow x=0\)