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\(\left(x+2015\right)^7=\left(x+2015\right)^5\)
\(\left(x+2015\right)^7-\left(x+2015\right)^5=0\)
\(\left(x+2015\right)^5.\left(x+2015\right)^2-\left(x+2015\right)^5=0\)
\(\left(x+2015\right)^5\left[\left(x+2015\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-2015\right)^5=0\\\left(x+2015\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2015\\x=-2016;-2014\end{cases}}}\)
Vậy x = { - 2016; - 2014; 2015 }
1.
\(\left(\frac{3}{1\times3}+\frac{3}{3\times5}+\frac{3}{5\times7}+...+\frac{3}{97\times99}\right)-x:\frac{3}{2}=\frac{7}{3}\\
\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{97\times99}\right):\frac{3}{2}-x:\frac{3}{2}=\frac{7}{3}\\\left[\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-x\right]:\frac{3}{2}=\frac{7}{3}\\
\left(1-\frac{1}{99}\right)-x=\frac{7}{3}\times\frac{3}{2}\\
\frac{98}{99}-x=\frac{7}{2}\\
x=\frac{98}{99}-\frac{7}{2}=\frac{-497}{198}\)
2.\(\frac{x}{y}=\frac{4}{3}\Rightarrow\hept{\begin{cases}x=4a\\y=3a\\x-y=4a-3a=a\end{cases}}\\ \left(x-y\right)^{2015}=5^{2015}\Rightarrow x-y=5\\ \Rightarrow a=5\Rightarrow\hept{\begin{cases}x=4\times5=20\\y=3\times5=15\end{cases}}\)
a) \(x+5=2015-\left(12-7\right)\)
\(\Leftrightarrow x+5=2015-5\)\(\Leftrightarrow x+5=2010\)
\(\Leftrightarrow x=2005\)
Vậy \(x=2005\)
b) \(-7\left|x+3\right|=-49\)\(\Leftrightarrow\left|x+3\right|=7\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=-7\\x+3=7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-10\\x=4\end{cases}}\)
Vậy \(x=-10\)hoặc \(x=4\)
c) \(\left(105-x\right):2^5=2015^0+1\)
\(\Leftrightarrow\left(105-x\right):32=1+1\)\(\Leftrightarrow\left(105-x\right):32=2\)
\(\Leftrightarrow105-x=64\)\(\Leftrightarrow x=41\)
Vậy \(x=41\)
a. x+5 = 2015-(12-7)
x+5= 2015-5
x+5 = 2010
x=2005
b, -7/ x+3/ =-49
/x+3/ = 7
x+3 = 7 suy ra x=4
hoặc x+3= -7 suy ra x=-10
c. (105-x) : 2^5= 2015^0 + 1
(105-x) : 32= 1+1
(105-x) : 32= 2
105-x= 64
x= 41
Cái này kiến thức cơ bản thôi ạ
Do |x+2015| lớn hoặc = 0 với mọi x nên A bé hơn hoặc bằng -2016
Dấu "=" xảy ra khi và chỉ khi x+2015=0
=> x=-2015
0.x + 2015 = 2015
0x = 2015 - 2015
0x = 0 (luôn đúng với mọi x)
⇒ B = ℕ
Số dấu ngoặc = (2015 - 1):2 +1 = 1008
=> 1008x +(2015+ 1).1008:2 = 1023120
=> 1008x + 1008.1008 = 1008.1015
=> x +1008 = 1015
x = 7
\(\left(x+2015\right)^7=\left(x+2015\right)^5\)
\(\Rightarrow\left(x+2015\right)\in\left\{0;1;-1\right\}\)
\(\Rightarrow x\in\left\{-2015;-2014;-2016\right\}\)
Vậy \(x\in\left\{-2015;-2014;-2016\right\}\)
\(\left(x+2015\right)^7=\left(x+2015\right)^5\)
\(\Leftrightarrow\left(x+2015\right)^7-\left(x+2015\right)^5=0\)
\(\Leftrightarrow\left(x+2015\right)^5\left[\left(x+2015\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x+2015\right)^5\left(x+2014\right)\left(x+2016\right)=0\)
\(\Leftrightarrow x=-2016;x=-2015;x=-2014\)