\(\frac{\left(1+3y\right)}{12x}=\frac{\left(1+5y\right)}{5x}=\frac{\left(1+7y\right)}{4x}\)
tìm x và y
Tìm x biết:
a) \(\left(x-3\right)^{x+5}-\left(x-3\right)^{x+15}=0\)
b)\(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}\)
\(a)\) \(\left(x-3\right)^{x+5}-\left(x-3\right)^{x+15}=0\)
\(\Leftrightarrow\)\(\left(x-3\right)^{x+5}-\left(x-3\right)^{x+5}.\left(x-3\right)^{10}=0\)
\(\Leftrightarrow\)\(\left(x-3\right)^{x+5}.\left[1-\left(x-3\right)^{10}\right]=0\)
Trường hợp 1 :
\(\left(x-3\right)^{x+5}=0\)
\(\Leftrightarrow\)\(\left(x-3\right)^{x+5}=0^{x+5}\)
\(\Leftrightarrow\)\(x-3=0\)
\(\Leftrightarrow\)\(x=3\)
Trường hợp 2 :
\(1-\left(x-3\right)^{10}=0\)
\(\Leftrightarrow\)\(\left(x-3\right)^{10}=1\)
\(\Leftrightarrow\)\(\left(x-3\right)^{10}=1^{10}\)
\(\Leftrightarrow\)\(x-3=1\)
\(\Leftrightarrow\)\(x=4\)
Vậy \(x=3\) hoặc \(x=4\)
Chúc bạn học tốt ~
Tìm x,y biết
a, \(\left(x-3\right)^{x+5}-\left(x-3\right)^{x+15}=0\)
b, \(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}\)
c, \(\frac{1}{2018}:2017x=-\frac{1+7y}{4x}\)
Giúp mình nhé, thứ 4 nộp r
a) \(\left(x-3\right)^{x+5}-\left(x-3\right)^{x+15}=0\)
\(\left(x-3\right)^{x+5}-\left(x-3\right)^{x+5}\cdot\left(x-3\right)^{10}=0\)
\(\left(x-3\right)^{x+5}\cdot\left[1-\left(x-3\right)^{10}\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-3\right)^{x+5}=0\\1-\left(x-3\right)^{10}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\\left(x-3\right)^{10}=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\\left(x-3\right)^{10}=\left(\pm1\right)^{10}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=\left\{4;2\right\}\end{cases}}\)
Vậy........
Câu 1:Cho a,b,c đôi một khác nhau và thõa mãn \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}\)
Tính giá trị của biểu thức:\(P=\left(1+\frac{a}{b}\right).\left(1+\frac{b}{c}\right).\left(1+\frac{c}{a}\right)\)
Câu 2 Tìm x,y\(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}\)
tìm x,y biết:
a \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
b \(\frac{5x-1}{3}=\frac{7y-6}{5}=\frac{5x+7y-7}{4x}\)
c \(\left|x+5\right|+\left(3y-4\right)^{2010}=0\)
Tìm x,y,z biết:
a, \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+1}=0\)
b, \(\frac{5x-1}{3}=\frac{7y-6}{5}=\frac{5x+7y-7}{4x}\)
c,\(\left|x+5\right|+\left(3y-4\right)^{2010}-0\)
\(.a.\)
\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+1}=0\)
\(\Leftrightarrow\left(x-7\right)^{x+1}.\left[1-\left(x-7\right)^{10}\right]=0\)
\(\Leftrightarrow\left[\begin{matrix}\left(x-7\right)^{x+1}=0\\\left[1-\left(x-7\right)^{10}\right]=0\end{matrix}\right.\)
+ Nếu \(\left(x-7\right)^{x+1}=0\)
\(\Rightarrow x-7=0\)
\(\Rightarrow x=0+7\)
\(\Rightarrow x=7\)
+ Nếu \(1-\left(x-7\right)^{10}=0\)
\(\Rightarrow\left(x-7\right)^{10}=1\)
\(\Rightarrow\left(x-7\right)^{10}=\left(\pm1\right)^{10}\)
\(\Rightarrow\left[\begin{matrix}x-7=1\\x-7=-1\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=1+7\\x=-1+7\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=8\\x=6\end{matrix}\right.\)
Vậy : \(x\in\left\{6;7;8\right\}\)
Tìm x; y biết:
a,\(\left|x\right|+3\sqrt{x^2+9}+x^{2010}=9\)9
b, \(\frac{1+3y}{12}=\frac{1+5y}{5x}+\frac{1+7y}{4x}\)
tìm x;y\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+1}\)=0
\(\frac{5x-1}{3}=\frac{7y-6}{5}=\frac{6x+7y-7}{4x}\)
(x - 7)x+1 - (x - 7)x+1 = 0
<=> 0 = 0
Vậy phương trình có nghiệm với mọi x thuộc R
b/ Chi cần áp dụng tính chất dãy tỷ số bằng nhau thì ra thôi
Tính
a, \(\frac{1}{\left(y-1\right)\left(y-2\right)}+\frac{2}{\left(2-y\right)\left(3y-y\right)}+\frac{3}{\left(1-y\right)\left(y-3\right)}\)
b, \(\frac{x^2}{x+1}+\frac{2x}{x^2-1}-\frac{1}{1-x}+1\)
c, \(\frac{1}{x^2+3x+2}-\frac{2x}{x^2+4x^2+4x}+\frac{1}{x^2+5x+6}\)
a, A= \(\left(15x+2y\right)-\left[\left(2x+3\right)-\left(5x+y\right)\right]\) tại x=1 ; y= -1
b, B= -\(\left(12x+3y\right)+\left(5x-2y\right)-\left[13x+\left(2y+5\right)\right]\) tại x= \(\frac{-1}{2}\); y= \(\frac{1}{7}\)
a,thay x=1,y=-1
=>A=(15.1+2.-1)-[(2.1+3)-(5.1+-1)]=13-[5-4]=12
b,thay=-1/2,y=1/7
=>B=4