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a) Áp dụng tính chất dãy tỉ số bằng nhau ta dc:
\(\frac{ab+1}{9}=\frac{ac+2}{15}=\frac{bc+3}{27}=\frac{ab+ac+bc+6}{51}=\frac{17}{51}=\frac{1}{3}\)
=> \(\frac{ab+1}{9}=\frac{1}{3}\)=> ab = 2 (1)
Tương tự nha vậy ta dc: ac = 3 (2) và bc = 6 (3)
Khi đó: (abc)2 = 36 => \(\orbr{\begin{cases}abc=6\\abc=-6\end{cases}}\)
* Với abc = 6
Từ (1), (2), (3) ta có: \(\hept{\begin{cases}c=3\\b=2\\a=1\end{cases}}\)
* Với abc = - 6
Từ (1), (2), (3) ta có: \(\hept{\begin{cases}c=-3\\b=-2\\a=-1\end{cases}}\)
Vậy ...
b) x + 2xy + y = 0
<=> 2x + 4xy + 2y = 0
<=> 2x(1 + 2y) + (1 + 2y) = 1
<=> (2x + 1)(2y + 1) = 1
Tới đây bạn giải theo pt ước số nha
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a)\(\left(\frac{-1}{3}\right)^3\cdot x=\frac{1}{81}\) \(< =>\frac{-1}{27}x=\frac{1}{81}\)\(< =>x=\frac{-1}{3}\)
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a) 3x - / 2x + 1/=2
Ta co: /2x+1/ lon hon hoac bang 0
ma 3x- / 2x+1/ = 2
=> 3x la so tu nhien
=>3x-/2x+1/ = 3x - 2x+1 = 2
=>3x - 2x = 1
=>x(3-2) = 1
=>x . 1 = 1
=> x=1
KL........\
Tich cho minh nhe ! Cau b dang suy nghi .
a) Ta co: /2x+1/ lon hon hoac bang 0
ma 3x - /2x+1/ = 2
=> 3x la so tu nhien
=> 3x - /2x+1/ = 3x -2x +1 = 2\
=> 3x -2x =1
=>x=1
tick cho minh nha!!!!! Thank you nhieuuuuuuuuu !!!!
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a) \(\left(2x-3\right)^2=25\)
\(\Leftrightarrow\left(2x-3\right)^2=\left(\pm5\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=5\\2x-3=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-1\end{cases}}\)
Vậy: x = 4 hoặc x = -1
b) \(\frac{27}{3x}=3\Leftrightarrow\frac{9}{x}=3\Leftrightarrow9=3x\Leftrightarrow3x=9\Leftrightarrow x=\frac{9}{3}=3\)
=> x = 3
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a, \(2^{x+1}.3^y=12^x\Rightarrow2^{x+1}.3^y=3^x.4^x\Rightarrow2^{x+1}.3^y=2^{2x}.3^x\)
=> x + 1 = 2x ; y = x
=> x = 1 ; y = x = 1
b, \(10^x:5^y=20^y\Rightarrow2^x.5^x:5^y=4^y.5^y\Rightarrow2^x.5^{x-y}=2^{2y}.5^y\)
=> x = 2y ; x- y = y => x = 2y
VẬy mọi số tự nhiên x,y đều thỏa mãn miễn x = 2y ( thử xem)
c, \(2^x=4^{y-1}\Rightarrow2^x=2^{2\left(y-1\right)}\Rightarrow x=2\left(y-1\right)\Rightarrow x=2y-2\)
\(27^y=3^{x+8}\Rightarrow3^{3y}=3^{x+8}\Rightarrow3y=x+8\Rightarrow3y=2y-2+6\)
=> 2y + 4 = 3y => y = 4 ;
x = 2.4 - 2 = 6
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=> \(\left(\frac{x+4}{2011}+1\right)+\left(\frac{x+3}{2012}+1\right)=\left(\frac{x+2}{2013}+1\right)+\left(\frac{x+1}{2014}+1\right)\)
=> \(\frac{x+5}{2011}+\frac{x+2015}{2012}=\frac{x+2015}{2013}+\frac{x+2015}{2014}\)
=> \(\left(x+2015\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
=> x = -2015 Vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\)
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\(\left(2x-3\right)^2=25\)
\(\Rightarrow\left(2x-3\right)^2=5^2\)
\(\Rightarrow2x-3=5\)
\(\Rightarrow2x=5+3\)
\(\Rightarrow2x=8\)
\(\Rightarrow x=4\)
Bài 1
A = \(\frac{3}{7}.\left(\frac{3}{7}\right)^{19}\)= \(\left(\frac{3}{7}\right)^{20}\)
B = \(\left[\left(-\frac{3}{7}\right)^5\right]^4\)= \(\left(-\frac{3}{7}\right)^{20}\)
Bài 2
a. (2x - 3)2 = 25
<=> \(\orbr{\begin{cases}2x-3=5\\2x-3=-5\end{cases}}\)
<=> \(\orbr{\begin{cases}x=4\\x=-1\end{cases}}\)
Vậy ...
b. \(\frac{27}{3^x}\)= 3
<=> 27 = 31+x
<=> 33 = 31+x
<=> 3 = 1 + x
<=> x = 2
\(\left(1-2x\right)3=27\)
\(3-6x=27\)
\(6x=3-27\)
\(6x=-24\)
\(x=-24:6\)
\(x=-4\)
( 1 - 2x ) x 3 = 27
( 1 - 2x ) = 27 : 3
( 1 - 2x ) = 9
<=> 1 - 2x = 9
<=> 2x = ( 9 + 1 )
<=> 2x = 10
<=> x = 10 : 2 = 5
=> x = 5