Tìm x, y biết: \(8.2^{3x}.7^y=56^{2x}.5^{x-1}\)
Giúp mik nha!!!!!!
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
24 tháng 10 2017
8.2^3x.7^y=56^2x.5^x-1
=>8.8^x.7^y=(8.7)^2x.5^x-1
=>8^1+x.7^y=8^2x.7^2x.5^x-1
=>8^1+x.7^y / 8^2x.7^2x=5^x-1
=>8^x+1-2x . 7^y-2x = 5^x-1
=>8^1-x.7^y-2x=5^-(1-x)
=>8^1-x.7^y-2x=1/5^1-x
=>8^1-x.7^y-2x.5^1-x=1
=>(8.5)^x-1.7^y-2x=1
=>40^x-1.7^y-2x=1
=>x-1=0 và y-2x=0
=>x=1 và y=2
27 tháng 7 2018
\(8.2^{3x}.7^y=56^{2x}.5^{x-1}\)
\(2^3.2^{3x}.7^y=7^{2x}.8^{2x}.5^{x-1}\)
\(2^{3+3x}.7^y=7^{2x}.2^{6x}.5^{x-1}\)
\(7^{2x}:7^y=2^{6x}:2^{3+3x}.5^{x-1}\)
\(7^{2x-y}=2^{6x-3-3x}.5^{x-1}\)
\(7^{2x-y}=2^{3x-3}.5^{x-1}\)
\(7^{2x-y}=2^{3x}:8.5^x:5\)
\(7^{2x-y}=8^x.5^x:40\)
\(7^{2x-y}=40^x:40\)
\(7^{2x-y}=40^{x-1}\)
\(\Rightarrow x=y=1\)
31 tháng 7 2022
\(\Leftrightarrow2^{3x+3}\cdot7^y=2^{6x}\cdot7^{2x}\cdot5^{x-1}\)
=>3x+3=6x; y=2x; x-1=0
=>\(\left(x,y\right)\in\varnothing\)
TN
0
ND
Nguyễn Đức Trí
VIP
19 tháng 7 2023
a) \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7};x+y+z=56\)
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{2+5+7}=\dfrac{56}{14}=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.2=8\\y=4.5=20\\z=4.7=28\end{matrix}\right.\)
b) \(\dfrac{x}{1,1}=\dfrac{y}{1,3}=\dfrac{z}{1,4}\left(1\right);2x-y=5,5\)
\(\left(1\right)\Rightarrow\dfrac{2x-y}{1,1.2-1,3}=\dfrac{5,5}{0,9}\)
\(\Rightarrow\left\{{}\begin{matrix}x=1,1.\dfrac{5,5}{0,9}=\dfrac{6,05}{0,9}\\y=1,3.\dfrac{5,5}{0,9}=\dfrac{7,15}{0,9}\\z=\dfrac{1,4}{1,1}.x=\dfrac{1,4}{1,1}.\dfrac{6,05}{0,9}=\dfrac{8,47}{0,99}\end{matrix}\right.\)
d) \(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5};xyz=-30\)
\(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5}=\dfrac{xyz}{2.3.5}=\dfrac{-30}{30}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(-1\right)=-2\\y=3.\left(-1\right)=-3\\z=5.\left(-1\right)=-5\end{matrix}\right.\)
3 tháng 7 2018
a. Ta có: \(x^2-10x+26+y^2+2y=0\Leftrightarrow\left(x^2-10x+25\right)+\left(y^2+2y+1\right)=0\\ \)
\(\Leftrightarrow\left(x+5\right)^2+\left(y+1\right)^2=0\Rightarrow\hept{\begin{cases}x+5=0\\y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-1\end{cases}}}\)
b. \(\left(2x+5\right)^2-\left(x-7\right)^2=0\Leftrightarrow\left(2x+5+x-7\right).\left(2x+5-x+7\right)=0\)
\(\Leftrightarrow\left(3x-2\right).\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}3x-2=0\\x+12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-12\end{cases}}}\)
c. \(25.\left(x-3\right)^2=49.\left(1-2x\right)^2\Leftrightarrow\left(5x-15\right)^2=\left(7-14x\right)^2\Leftrightarrow\left(5x-15\right)^2-\left(7-14x\right)^2=0\)
\(\Leftrightarrow\left(5x-15-7+14x\right).\left(5x-15+7-14x\right)=0\Leftrightarrow\left(19x-22\right).\left(-9x-8\right)=0\)
\(\Leftrightarrow\left(19x-22\right).\left(9x+8\right)=0\Leftrightarrow\orbr{\begin{cases}19x-22=0\\9x+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{22}{19}\\x=-\frac{8}{9}\end{cases}}}\)
d. \(\left(x+2\right)^2=\left(3x-5\right)^2\Leftrightarrow\left(x+2\right)^2-\left(3x-5\right)^2=0\Leftrightarrow\left(x+2+3x-5\right).\left(x+3-3x+5\right)=0\)
\(\Leftrightarrow\left(4x-3\right).\left(8-2x\right)=0\Leftrightarrow\orbr{\begin{cases}4x-3=0\\8-2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=4\end{cases}}}\)
e. \(x^2-2x+1=16\Leftrightarrow\left(x-1\right)^2-16=0\Leftrightarrow\left(x-1-4\right).\left(x-1+4\right)=0\)
\(\Leftrightarrow\left(x-5\right).\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x-5=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)