Tìm \(x,y\in N\):
a) 32x+1 . 7y = 9 . 21x
b) \(\dfrac{27^x}{3^{2x-y}}=243\) và \(\dfrac{25^x}{5^{x+y}}=125\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
27 tháng 11 2022
Bài 2:
\(\Leftrightarrow\left\{{}\begin{matrix}3^{3x-2x+y}=3^5\\5^{2x-x-y}=5^3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-y=5\\x-y=3\end{matrix}\right.\)
=>x=1;y=-2
HN
31 tháng 10 2018
a) => 5x.52 + 5x.53=750
=> 5x . (52+53) =750
=> 5x . 150 =750
=> 5x = 750 : 150
=> 5x = 5
=> x =1
Vậy x = 1
b) => 32x+1 . 7y = 32 . (3.7)x
=> 32x+1 . 7y = 3x+2 . 7x
=> \(\dfrac{3^{2x+1}}{3^{x+2}}\) =\(\dfrac{7^x}{7^y}\)
=> 3(2x+1)-(x+2) = 7x-y
=> 3x-1 = 7x-y
=>\(\left\{{}\begin{matrix}x-1=0\\x-y=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\x=y\end{matrix}\right.\)
=>x=y=1
Vậy x=y=1
c)
=>\(\dfrac{3^{3x}}{3^{2x-y}}\) =35 và =>\(\dfrac{5^{2x}}{5^{x+y}}\) =53
=> 3(3x)-(2x-y) =35 =>5(2x)-(x+y) =53
=> 33x-2x+y =35 => 52x-x-y =53
=> 3x+y =35 => 5x-y =53
=> x+y =5 (1) => x-y =3 (2)
Từ (1) và (2) có :
+x = (5+3):2 =4
+y = (5-3):2 =1
Vậy x=4 ; y=1
- Nếu làm đúng cho mình xin cái tick ! Tks
MH
15 tháng 9 2021
\(2x=3y\text{⇒}\dfrac{x}{3}=\dfrac{y}{2}\text{⇒}\dfrac{x}{21}=\dfrac{y}{14}\)
\(5y=7z\text{⇒}\dfrac{y}{7}=\dfrac{z}{5}\text{⇒}\dfrac{y}{14}=\dfrac{z}{10}\)
⇒\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}\)⇒\(\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}=\dfrac{3x-7y+5z}{63-98+50}=\dfrac{30}{15}=2\)
⇒x=42,y=28,z=20
MH
15 tháng 9 2021
\(\dfrac{x}{3}=\dfrac{y}{2}\)⇒\(\dfrac{x}{15}=\dfrac{y}{10}\)
\(\dfrac{x}{5}=\dfrac{z}{7}\text{⇒}\dfrac{x}{15}=\dfrac{z}{21}\)
⇒\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{21}\)⇒\(\dfrac{x}{15}=\dfrac{2y}{20}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{15}=\dfrac{2y}{20}=\dfrac{x+2y}{15+20}=\dfrac{-112}{35}=\dfrac{-16}{5}\)
⇒x=48,y=32,z=336/5
Q
21 tháng 6 2017
1)
a) \(0,25^x\cdot12^x=243\)
\(\Leftrightarrow\left(0,25\cdot12\right)^x=3^5\)
\(\Leftrightarrow3^x=3^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
b) \(38^y:19^y=512\)
\(\Leftrightarrow2y\cdot y=512\)
\(\Leftrightarrow2y^2=512\)
\(\Leftrightarrow y^2=256\)
\(\Leftrightarrow\left[{}\begin{matrix}y=16\\y=-16\end{matrix}\right.\)
Vậy \(y_1=-16;y_2=16\)
2)
a) \(3^x+3^{x+2}=2430\)
\(\Leftrightarrow\left(1+3^2\right)\cdot3^x=2430\)
\(\Leftrightarrow\left(1+9\right)\cdot3^x=2430\)
\(\Leftrightarrow10\cdot3^x=2430\)
\(\Leftrightarrow3^x=243\)
\(\Leftrightarrow3^x=3^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
b) \(2^{x+3}-2^x=224\)
\(\Leftrightarrow\left(2^3-1\right)\cdot2^x=224\)
\(\Leftrightarrow\left(8-1\right)\cdot2^x=224\)
\(\Leftrightarrow7\cdot2^x=224\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
3)
a) \(\left(x-\dfrac{1}{4}\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow x-\dfrac{1}{4}=\pm\dfrac{2}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{4}=\dfrac{2}{3}\\x-\dfrac{1}{4}=-\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}+\dfrac{1}{4}\\x=-\dfrac{2}{3}+\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{12}\\x=-\dfrac{5}{12}\end{matrix}\right.\)
Vậy \(x_1=\dfrac{11}{12};x_2=-\dfrac{5}{12}\)
b) \(\left(x+0,7\right)^3=-27\)
\(\Leftrightarrow\left(x+\dfrac{3}{10}\right)^3=\left(-3\right)^3\)
\(\Leftrightarrow x+\dfrac{3}{10}=-3\)
\(\Leftrightarrow x=-3-\dfrac{3}{10}\)
\(\Leftrightarrow x=-\dfrac{37}{10}\)
Vậy \(x=-\dfrac{37}{10}\)
4)
a) \(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\dfrac{2}{5}-3x=\pm\dfrac{3}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\\dfrac{2}{5}-3x=-\dfrac{3}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{1}{15};x_2=\dfrac{1}{3}\)
b) \(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\dfrac{1}{243}\)
\(\Leftrightarrow\dfrac{2}{3}x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(\Leftrightarrow2x-1=1\)
\(\Leftrightarrow2x=1+1\)
\(\Leftrightarrow2x=2\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
21 tháng 6 2017
1. a) \(0,25^x.12^x=243\)
\(\Rightarrow\left(0,25.12\right)^x=243\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)
b) \(38^y:19^y=512\)
\(\Rightarrow\left(38:19\right)^y=512\)
\(\Rightarrow2^y=2^9\)
\(\Rightarrow y=9\)
Vậy \(y=9.\)
2) a) \(3^x+3^{x+2}=2430\)
\(\Rightarrow3^x\left(1+9\right)=2430\)
\(\Rightarrow3^x=243=3^5\)
\(\Rightarrow x=5\)
Vậy x=5.
b) \(2^{x+3}-2^x=224\)
\(\Rightarrow2^x\left(8-1\right)=224\)
\(\Rightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
Vậy x=5.
Bài 3: dễ tự làm.
Lời giải:
a)
$3^{2x+1}.7^y=9.21^x=3^2.(3.7)^x=3^{2+x}.7^x$
Vì $x,y$ là số tự nhiên nên suy ra $2x+1=2+x$ và $y=x$
$\Rightarrow x=y=1$
b) \(\frac{27^x}{3^{2x-y}}=\frac{3^{3x}}{3^{2x-y}}=3^{x+y}=243=3^5\Rightarrow x+y=5(1)\)
\(\frac{25^x}{5^{x+y}}=\frac{5^{2x}}{5^{x+y}}=5^{x-y}=125=5^3\Rightarrow x-y=3\) $(2)$
Từ $(1);(2)\Rightarrow x=4; y=1$