Tìm \(x\) biết \(7^{x+2}+2.7^{x-1}=345\)
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\(7^{x+2}+2.7^{x+1}=345\)
\(7^{x-1}.7^3+2.7^{x-1}=345\)
\(7^{x-1}.\left(7^3+2\right)=345\)
\(7^{x-1}.345=345\)
\(7^{x-1}=1\)
\(7^{x-1}=7^0\)
\(\Rightarrow x-1=0\)
\(x=1\)
\(1,\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{21}{7}=3\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=15\end{matrix}\right.\\ 2,7x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{16}{-4}=-4\\ \Rightarrow\left\{{}\begin{matrix}x=-12\\y=-28\end{matrix}\right.\\ 3,\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-6-7}=\dfrac{36}{-8}=-\dfrac{9}{2}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{45}{2}\\y=-27\\z=-\dfrac{63}{2}\end{matrix}\right.\\ 4,x:y:z=3:5:7\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{6+15-7}=\dfrac{-14}{14}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-7\end{matrix}\right.\)
3. Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-6-7}=\dfrac{36}{-8}=\dfrac{-9}{2}\)
\(x=\dfrac{-45}{2}\)
\(y=-27\)
\(z=\dfrac{-63}{2}\)
7x+2+2.7x-1=345
=>7x-1(73+2)=345
=>7x-1=345:345=1
=>x-1=0
=>x=1
7^x-1 . 7^x+3 + 2. 7^x-1 = 345
7^x-1 .(7^x+3 + 2) =345
7^x-1. (7^x+3 + 2) = 7^x+3 -2
7^x-1 =(7^x+3 - 2) : (7^x+3 +2)
7^x-1 = 0
=> x-1 = 1
chắc chắn luôn
a) 72.62 - 72.16 + 72.8
= 72.36 - 72.16 + 72.8
= 72 . ( 36 - 16 + 8 )
= 72 . 28 = 2016
b) 8.2x-1 = 256
16x-1 = 162
x - 1 = 2
x = 2 + 1
x = 3
c) 3.5x - 35 = 5.23
3.5x - 35 = 40
3.5x = 40 + 35
3.5x = 75
5x = 75 : 3 = 25
5x = 52
=> x = 2
d) 345 - ( 7x + 23 ) = 1722 : 6 = 287
7x + 23 = 345 - 287
7x + 23 = 58
7x = 58 - 23 = 35
7x = 35
x = 35 : 7
x = 5
b) 2^x-1 = 256 : 8
2^x-1 = 32
ta thấy : 32 = 2^5
cơ số = cơ số
=> x - 1 = 5
x = 5 = 1
x = 6
a:1/5 + a:0,25 + a = 2,7x 0,1
=>a:(1/5 + 0.25 + 1) = 0.27
=>a:1.45 = 0.27
=> a = 0.27 x 1.45
a.
\(7^{x+2}+2\times7^{x-1}=345\)
\(7^x\times7^2+2\times7^x\div7=345\)
\(7^x\times\left(7^2+\frac{2}{7}\right)=345\)
\(7^x\times\frac{345}{7}=345\)
\(7^x=345\div\frac{345}{7}\)
\(7^x=345\times\frac{7}{345}\)
\(7^x=7\)
\(x=1\)
b.
\(2^{x+2}-2^x=96\)
\(2^x\times\left(2^2-1\right)=96\)
\(2^x\times3=96\)
\(2^x=\frac{96}{3}\)
\(2^x=32\)
\(2^x=2^5\)
\(x=5\)
\(a,7^{x+2}+2.7^{x-1}=345\Rightarrow7^x.49+\frac{2}{7}.7^x=345\Rightarrow7^x\left(49+\frac{2}{7}\right)=345\Rightarrow7^x.\frac{345}{7}=345\Rightarrow7^x=345:\frac{345}{7}=7^1\Rightarrow x=1\)
\(b,2^{x+2}-2^x=96\Rightarrow2^x.4-2^x=96\Rightarrow2^x\left(4-1\right)=96\Rightarrow2^x.3=96\Rightarrow2^x=96:3=32\Rightarrow2^x=2^5\Rightarrow x=5\)
\(a,7^{x+2}+2.7^{x-1}=345=>7^{x-1+3}+2.7^{x-1}=345=>7^{x-1}.7^3+2.7^{x-1}=345\)
\(=>\left(7^3+2\right).7^{x-1}=345=>345.7^{x-1}=345=>7^{x-1}=1=7^0=>x-1=0=>x=1\)
\(b,2^{x+2}-2^x=96=>2^x.2^2-2^x=96=>2^x.\left(4-1\right)=96=>2^x.3=96=>2^x=32=2^5=>x=5\)
\(7^{x-1+3}+2.7^{x-1}=345\Leftrightarrow7^{x-1}\left(7^3+2.1\right)=345\)
\(\Leftrightarrow7^{x-1}.345=345\Leftrightarrow7^{x-1}=1\Leftrightarrow7^{x-1}=7^0\Leftrightarrow x-1=0\Rightarrow x=1\)
Vậy x= 1
=> \(7^{x-1}.7^3+2.7^{x-1}=345\)
=> \(7^{x-1}.\left(7^3+2\right)=345\)
=> \(7^{x-1}.345=345\Rightarrow7^{x-1}=1\Rightarrow x-1=0\) => x = 1