\(2x=8^{\left(y+1\right)}\)
\(9^y=3^{x-9}\)
\(x,y\in N\)
Tính x + y
K
Khách
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NH
0
XK
2
25 tháng 6 2017
Sửa đề bài: \(2^x=8^{y+1}\)và \(9^y=3^{x-9}\)
Có: \(2^x=8^{y+1}\)
\(\Leftrightarrow2^x=\left(2^3\right)^{y+1}\)
\(\Leftrightarrow2^x=2^{3y+3}\)
\(\Leftrightarrow x=3y+3\) (1)
Lại có: \(9^y=3^{x-9}\)
\(\Leftrightarrow\left(3^2\right)^y=3^{x-9}\)
\(\Leftrightarrow3^{2y}=3^{x-9}\)
\(\Leftrightarrow2y=x-9\) (2)
Thay (1) vào (2), ta có:
=> 2y = 3y + 3 - 9
=> 2y = 3y - 6
=> 2y - 3y = -6
=> -1y = -6
=> y = 6 \(\left(y\in N\right)\)
Từ x = 3y + 3 (theo điều 1)
=> x = 3.6 + 3 = 21 \(\left(x\in N\right)\)
Vậy x + y = 21 + 6 = 27
12 tháng 11 2023
a:
ĐKXĐ: \(x\notin\left\{\dfrac{3}{2};1\right\}\)
\(y=\dfrac{\left(x-2\right)^2}{\left(2x-3\right)\left(x-1\right)}=\dfrac{x^2-4x+4}{2x^2-2x-3x+3}\)
=>\(y=\dfrac{x^2-4x+4}{2x^2-5x+3}\)
=>\(y'=\dfrac{\left(x^2-4x+4\right)'\left(2x^2-5x+3\right)-\left(x^2-4x+4\right)\left(2x^2-5x+3\right)'}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{\left(2x-4\right)\left(2x^2-5x+3\right)-\left(2x-5\right)\left(x^2-4x+4\right)}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{4x^3-10x^2+6x-8x^2+20x-12-2x^3+8x^2-8x+5x^2-20x+20}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{2x^3-5x^2-2x+8}{\left(2x^2-5x+3\right)^2}\)
b:
ĐKXĐ: x<>-3
\(y=\left(x+3\right)+\dfrac{4}{x+3}\)
=>\(y'=\left(x+3+\dfrac{4}{x+3}\right)'=1+\left(\dfrac{4}{x+3}\right)'\)
\(=1+\dfrac{4'\left(x+3\right)-4\left(x+3\right)'}{\left(x+3\right)^2}\)
=>\(y'=1+\dfrac{-4}{\left(x+3\right)^2}=\dfrac{\left(x+3\right)^2-4}{\left(x+3\right)^2}\)
y'=0
=>\(\left(x+3\right)^2-4=0\)
=>\(\left(x+3+2\right)\left(x+3-2\right)=0\)
=>(x+5)(x+1)=0
=>x=-5 hoặc x=-1
c:
ĐKXĐ: x<>-2
\(y=\dfrac{\left(5x-1\right)\left(x+1\right)}{x+2}\)
=>\(y=\dfrac{5x^2+5x-x-1}{x+2}=\dfrac{5x^2+4x-1}{x+2}\)
=>\(y'=\dfrac{\left(5x^2+4x-1\right)'\left(x+2\right)-\left(5x^2+4x-1\right)\left(x+2\right)'}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{\left(5x+4\right)\left(x+2\right)-\left(5x^2+4x-1\right)}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{5x^2+10x+4x+8-5x^2-4x+1}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{10x+9}{\left(x+2\right)^2}\)
\(y'\left(-1\right)=\dfrac{10\cdot\left(-1\right)+9}{\left(-1+2\right)^2}=\dfrac{-1}{1}=-1\)
d:
ĐKXĐ: x<>2
\(y=x-2+\dfrac{9}{x-2}\)
=>\(y'=\left(x-2+\dfrac{9}{x-2}\right)'=1+\left(\dfrac{9}{x-2}\right)'\)
\(=1+\dfrac{9'\left(x-2\right)-9\left(x-2\right)'}{\left(x-2\right)^2}\)
=>\(y'=1+\dfrac{-9}{\left(x-2\right)^2}=\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}\)
y'=0
=>\(\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}=0\)
=>\(\left(x-2\right)^2-9=0\)
=>(x-2-3)(x-2+3)=0
=>(x-5)(x+1)=0
=>x=5 hoặc x=-1
29 tháng 12 2021
Bài 1:
\(a,=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+2y^2}{2\left(x-y\right)\left(x+y\right)}=\dfrac{2y\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{y}{x-y}\\ b,Sửa:\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\\ =\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}=\dfrac{x^2+3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{-3x\left(x+3\right)}{x^2-3x+9}\\ =\dfrac{-3}{x-3}\)
Bài 2:
\(a,\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\\ b,\Leftrightarrow x^3+x^2+x+a=\left(x+1\right)\cdot a\left(x\right)\\ \text{Thay }x=-1\Leftrightarrow-1+1-1+a=0\Leftrightarrow a=1\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 1
ĐKXĐ:
a.
\(2x-4>0\Rightarrow x>2\Rightarrow D=\left(2;+\infty\right)\)
b.
\(2x+8>0\Rightarrow x>-4\Rightarrow D=\left(-4;+\infty\right)\)
c.
\(4-x>0\Rightarrow x< 4\Rightarrow D=\left(-\infty;4\right)\)
d.
\(\dfrac{1}{x+4}>0\Rightarrow x>-4\Rightarrow D=\left(-4;+\infty\right)\)
e.
\(\left(x-3\right)\left(x+9\right)>0\Rightarrow\left[{}\begin{matrix}x>3\\x< -9\end{matrix}\right.\) \(\Rightarrow D=\left(-\infty;-9\right)\cup\left(3;+\infty\right)\)
12 tháng 1
a: ĐKXĐ: 2x-4>0
=>2x>4
=>x>2
b: ĐKXĐ: 2x+8>0
=>2x>-8
=>x>-4
c: ĐKXĐ: 4-x>0
=>-x>-4
=>x<4
d: ĐKXĐ: \(\dfrac{1}{x+4}>0\)
=>x+4>0
=>x>-4
e: ĐKXĐ: \(\left(x-3\right)\left(x+9\right)>0\)
=>\(\left[{}\begin{matrix}x-3>0\\x+9< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -9\end{matrix}\right.\)
18 tháng 5 2017
ta có:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{1}{x+y+z}=0\)
\(\Leftrightarrow\dfrac{x+y}{xy}+\dfrac{x+y+z-z}{z\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x+y\right)\left(\dfrac{1}{xy}+\dfrac{1}{z\left(x+y+z\right)}\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(\dfrac{xz+yz+z^2+xy}{xyz\left(x+y+z\right)}\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(\dfrac{\left(y+z\right)\left(x+z\right)}{xyz\left(x+y+z\right)}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\\dfrac{\left(y+z\right)\left(x+z\right)}{xyz\left(x+y+z\right)}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\x+z=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^8=\left(-y\right)^8\\y^9=\left(-z\right)^9\\z^{10}=\left(-x\right)^{10}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x^8-y^8=0\\y^9+z^9=0\\x^{10}-z^{10}=0\end{matrix}\right.\)\(\Rightarrow\left(x^8-y^8\right)\left(y^9+z^9\right)\left(z^{10}-x^{10}\right)=0\)
\(\Rightarrow M=\dfrac{3}{4}\)
AH
Akai Haruma
Giáo viên
4 tháng 11 2023
Lời giải:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}$
$\Rightarrow (\frac{1}{x}+\frac{1}{y})+(\frac{1}{z}-\frac{1}{x+y+z})=0$
$\Leftrightarrow \frac{x+y}{xy}+\frac{x+y}{z(x+y+z)}=0$
$\Leftrightarrow (x+y)(\frac{1}{xy}+\frac{1}{z(x+y+z)})=0$
$\Leftrightarrow (x+y).\frac{z(x+y+z)+xy}{xyz(x+y+z)}=0$
$\Leftrightarrow (x+y).\frac{(z+x)(z+y)}{xyz(x+y+z)}=0$
$\Leftrightarrow (x+y)(y+z)(x+z)=0$
$\Leftrightarrow x=-y$ hoặc $y=-z$ hoặc $z=-x$
Nếu $x=-y$ thì:
$P=\frac{3}{4}+[(-y)^8-y^8](y^9+z^9)(z^{10}-x^{10})=\frac{3}{4}+0.(y^9+z^9)(z^{10}-x^{10})=\frac{3}{4}$
Nếu $y=-z$ thì:
$P=\frac{3}{4}+(x^8-y^8)[(-z)^9+z^9](z^{10}-x^{10})=\frac{3}{4}+(x^8-y^8).0.(z^{10}-x^{10})=\frac{3}{4}$
Nếu $z=-x$ thì:
$P=\frac{3}{4}+(x^8-y^8)(y^9+z^9)[(-x)^{10}-x^{10}]=\frac{3}{4}+(x^8-y^8)(y^9+z^9).0=\frac{3}{4}$
NT
5 tháng 6 2018
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\\ \Leftrightarrow\dfrac{x+y}{xy}+\left(\dfrac{1}{z}-\dfrac{1}{x+y+z}\right)=0\\ \Leftrightarrow\dfrac{x+y}{xy}+\dfrac{x+y}{z\left(x+y+z\right)}=0\\ \Leftrightarrow\left(x+y\right)\left(\dfrac{1}{xy}+\dfrac{1}{xz+yz+z^2}\right)=0\\ \)
Nếu x+y=0 => x=-y
Nếu
\(\dfrac{1}{xy}+\dfrac{1}{xz+yz+z^2}=0\\ \Rightarrow xz+yz+z^2+xy=0\\ \Rightarrow\left(x+z\right)\left(y+z\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-z\\y=-z\end{matrix}\right.\)
Tự thế vào :v
\(2x=8^{\left(y+1\right)}=2^{3\left(y+1\right)}\Rightarrow x=3y+3\) ( 1 )
\(9y=3^{2y}=3^{x-9}\Rightarrow2y=x-9\) ( 2 )
\(x+2y=3y+3+x-9\)
\(y=6\)
\(x=3.6+3=21\)
\(\Rightarrow x+y=27\)
Nếu có j thì nói nha ( giúp thì nói t giải cho )
Rồng Đỏ Bảo Lửa cảm ơn a nhìu nha ( hôm sau có j e nhờ a giải )