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\(\dfrac{4^3.5^4}{10^3.2^3}=\dfrac{2^6.5^4}{5^3.2^3.2^3}=\dfrac{2^6.5^4}{5^3.2^6}=5\)
a) Ta có: \(3\cdot5^x=3\cdot5^4\)
\(\Leftrightarrow5^x=5^4\)
\(\Leftrightarrow x=4\)
Vậy: x=4
b) Ta có: \(9\cdot7^x=6\cdot7^5+3\cdot7^5\)
\(\Leftrightarrow9\cdot7^x=\left(6+3\right)\cdot7^5\)
\(\Leftrightarrow9\cdot7^x=9\cdot7^5\)
\(\Leftrightarrow7^x=7^5\)
\(\Leftrightarrow x=5\)
Vậy: x=5
c) Ta có: \(2\cdot3^{x+2}+4\cdot3^{x+1}=10\cdot3^7\)
\(\Leftrightarrow2\cdot3^{x+2}+2\cdot6^{x+1}=10\cdot3^7\)
\(\Leftrightarrow2\cdot\left(3^{x+2}+6^{x+1}\right)=10\cdot3^7\)
\(\Leftrightarrow3^{x+2}+6^{x+1}=\frac{10\cdot3^7}{2}\)
\(\Leftrightarrow3^{x+1}\cdot3+3^{x+1}\cdot2=5\cdot3^7\)
\(\Leftrightarrow3^{x+1}\cdot\left(3+2\right)=5\cdot3^7\)
\(\Leftrightarrow5\cdot3^{x+1}=5\cdot3^7\)
\(\Leftrightarrow3^{x+1}=3^7\)
\(\Leftrightarrow x+1=7\)
\(\Leftrightarrow x=6\)
Vậy: x=6
a) \(3.5^x=3.5^4\)
\(\Rightarrow3.5^x-3.5^4=0\)
\(\Rightarrow3.\left(5^x-5^4\right)=0\)
Vì \(3\ne0.\)
\(\Rightarrow5^x-5^4=0\)
\(\Rightarrow5^x=0+5^4\)
\(\Rightarrow5^x=5^4\)
\(\Rightarrow x=4\left(TM\right).\)
Vậy \(x=4.\)
Chúc bạn học tốt!
a,?????
b, Với mọi giá trị của x;y ta có:
\(\left|x-\dfrac{1}{2}\right|+\left|x+y\right|\ge0\)
Để \(\left|x-\dfrac{1}{2}\right|+\left|x+y\right|=0\) thì:
\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|=0\\\left|x+y\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\\dfrac{1}{2}+y=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy..........
c, \(\left|2x\right|-\left|3,5\right|=\left|-6,5\right|\)
\(\Rightarrow\left|2x\right|=6,5+3,5=10\)
\(\Rightarrow\left\{{}\begin{matrix}2x=10\\2x=-10\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
Vậy..........
d, \(\left|x-1,7\right|=2,3\)
\(\Rightarrow\left\{{}\begin{matrix}x-1,7=2,3\\x-1,7=-2,3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\x=-0,6\end{matrix}\right.\)
Vậy.........
Chúc bạn học tốt!!!
#)Giải :
a)\(2^6.5^6=\left(2.5\right)^6=10^6\)
b)\(8^2.5^2=\left(8.5\right)^2=40^2\)
c)\(4^3.5^3=\left(4.5\right)^3=20^3\)
d)\(5^2.6^2.3^2=\left(5.6.2\right)^2=60^2\)
e)\(\frac{625^5}{25^8}=\frac{\left(25^2\right)^5}{25^8}=\frac{25^{10}}{25^8}=25^2\)
g)\(\frac{3^9}{7}.\frac{7^9}{3}=\frac{\left(3.7\right)^9}{7.3}=\frac{21^9}{21}=21^8\)
a) 3,5(15) = 3,5 + 0,0(15) = 3,5 + 1,5. 0,(01) = 3,5 + 1,5.1/99 = 3,5 + 1/66 = 116/33
b) Ta có: \(\frac{2x-y}{x+y}=\frac{2}{3}\)
=> (2x - y).3 = 2(x + y)
=> 6x - 3y = 2x + 2y
=> 6x - 2x = 2y + 3y
=> 4x = 5y
=> \(\frac{x}{y}=\frac{5}{4}\)
c) Đặt : \(\frac{a}{b}=\frac{c}{d}=k\) => \(\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó, ta có:
\(\frac{\left(bk\right)^2+bk.dk}{\left(dk\right)^2+dk.bk}=\frac{b^2k^2+bdk^2}{d^2k^2+bdk^2}=\frac{k^2\left(b^2+bd\right)}{k^2\left(d^2+bd\right)}=\frac{b^2+bd}{d^2+bd}\)
=> Đpcm
a,\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(-\dfrac{3}{5}+\dfrac{3}{5}\right)+.....+\left(-\dfrac{11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}\)
\(=0+0+...0+0+\dfrac{13}{15}=\dfrac{13}{15}\)
câu b và c xem lại đề nha
Chúc bạn học tốt!!!
\(\dfrac{2^2\cdot3^3\cdot5}{3\cdot2^3\cdot5^3}=\dfrac{1}{2}\cdot3^2\cdot\dfrac{1}{5^2}=\dfrac{1}{2}\cdot\dfrac{9}{25}=\dfrac{9}{50}\)