Tìm x biết: (x-7)^x+1 - (x-7)^x+11 =0
Giải giúc mình với...////!
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
c: \(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=-\sqrt{7}\\x=-5\\x=5\end{matrix}\right.\)
\(\frac{1}{x}=\frac{2}{y}\Rightarrow\frac{x}{1}=\frac{y}{2}=\frac{x+y}{1+2}=\frac{4}{3}\)
=> x=4/3.1=4/3
y=4/3.2=8/3
(x + 3) + (x + 7) + (x + 11) + ... + (x + 79) = 860
=> x + 3 + x + 7 + x + 11 + ... + x + 79 = 860
=> (x + x + x + ... + x) + (3 + 7 + 11 + ... + 79) = 860
=> 20x + (79 + 3).20 : 2 = 860
=> 20x + 82.20 : 2 = 860
=> 20x + 82.10 = 860
=> 20x + 820 = 860
=> 20x = 40
=> x = 2
vậy_
#)Giải :
\(\left(x+3\right)+\left(x+7\right)+...+\left(x+79\right)=860\)
\(\left(x+x+...+x\right)+\left(3+7+...+79\right)=860\)(trong mỗi ngoặc có 20 số hạng)
\(x\times20+\frac{\left(79+3\right)\times20}{2}=860\)
\(x\times20+820=860\)
\(x\times20=860-820\)
\(x\times20=40\)
\(x=40\div20\)
\(x=2\)
Bài 7:
a, \(x\) = \(\dfrac{1}{5}\) + \(\dfrac{2}{11}\)
\(x\) = \(\dfrac{11}{55}\) + \(\dfrac{10}{55}\)
\(x=\dfrac{21}{55}\)
b, \(\dfrac{x}{15}\) = \(\dfrac{3}{5}\) - \(\dfrac{2}{3}\)
\(\dfrac{x}{15}\) = \(\dfrac{9}{15}\) - \(\dfrac{10}{15}\)
\(\dfrac{x}{15}\) = \(\dfrac{1}{15}\)
\(x\) = 1
c, \(\dfrac{11}{8}\) + \(\dfrac{13}{6}\)= \(\dfrac{85}{x}\)
\(\dfrac{33}{24}\) + \(\dfrac{52}{24}\) = \(\dfrac{85}{x}\)
\(\dfrac{85}{24}\) = \(\dfrac{85}{x}\)
24 = \(x\)
a) Ta có: \(x-12=\left(-9\right)-15\)
\(\Rightarrow x-12=-24\)
\(\Rightarrow x=-12\)
b) \(2-x=17-\left(-5\right)\)
\(\Rightarrow2-x=22\)
\(\Rightarrow x=2-22\)
\(\Rightarrow x=-20\)
c) \(11-\left(15+11\right)=x-\left(25-9\right)\)
\(\Rightarrow-15=x-16\)
\(\Rightarrow x=1\)
d) \(x-\left(17-x\right)=x-7\)
\(\Rightarrow x-17+x=x-7\)
\(\Rightarrow x+x-x=-7+17\)
\(\Rightarrow x=10\)
e) \(9-25=\left(7-x\right)-\left(25+7\right)\)
\(\Rightarrow9-25=7-x-25-7\)
\(\Rightarrow-16=-25-x\)
\(\Rightarrow x=-9\)
g) \(x-13=21-x\)
\(\Rightarrow x+x=21+13\)
\(\Rightarrow2x=34\)
\(\Rightarrow x=17\)
a, x - 12 = ( -9 ) - 15
x - 12 = -24
x = -24 + 12
x = -12
b, 2 - x = 17 - ( -5 )
2 - x = 22
x = 2 - 22
x = -20
c, 11 - ( 15 + 11 ) = x - ( 25 - 9 )
11 - 26 = x - 16
-15 = x - 16
x = -15 + 16
x = 1
d, x - ( 17 - x ) = x - 7
x - 17 + x = x - 7
x + x - 17 = x - 7
x + x - x = -7 + 17
x = 10
e.9 - 25 = ( 7- x ) - ( 25 + 7 )
-16 = ( 7 - x ) -32
7 - x = -32 + 16
7 - x = 16
x = 7 - 16
x= -9
g, x - 13 = 21 - x
x + x = 21 + 13
2x = 44
x = 44 : 2
x=22
a,\(\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1\)
b, \(\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)
c, \(\left(x^2-4\right)\left(2x-3\right)-\left(x^2-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x-3-x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\Leftrightarrow x=-2;x=2\)
d, \(\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9\)
a.
\(A=B\)
\(\Leftrightarrow\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=\dfrac{-16}{x^2-4}\);ĐK:\(x\ne\pm2\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-16}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\left(x+2\right)^2-\left(x-2\right)^2=-16\)
\(\Leftrightarrow x^2+4x+4-x^2+4x-4+16=0\)
\(\Leftrightarrow8x+16=0\)
\(\Leftrightarrow8\left(x+2\right)=0\)
\(\Leftrightarrow x=-2\left(ktm\right)\)
Vậy không có giá trị x thỏa mãn A=B
b.
\(A:B=\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}:\dfrac{-16}{\left(x-2\right)\left(x+2\right)}< 0\)
\(\Leftrightarrow\dfrac{x^2+4x+4-x^2+4x-4}{-16}< 0\)
\(\Leftrightarrow\dfrac{8x}{-16}< 0\)
\(\Leftrightarrow\dfrac{8x}{16}>0\)
\(\Leftrightarrow\dfrac{x}{2}>0\)
\(\Leftrightarrow x>0\)
\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+1}\cdot\left(x-7\right)^{10}=0\)
\(\left(x-7\right)^{x+1}\cdot\left[1-\left(x-7\right)^{10}\right]=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=7\\\left(x-7\right)^{10}=1\end{cases}}\)
Bạn xét nốt TH \(\left(x-7\right)^{10}=1\)nhé
\(\Rightarrow\hept{\begin{cases}x=7\\\hept{\begin{cases}x-7=1\\x-7=-1\end{cases}}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=7\\\hept{\begin{cases}x=7\\x=6\end{cases}}\end{cases}}\)
( x - 7 )x + 1 - ( x - 7 )x + 11 = 0
=> ( x - 7 )x + 1 .[ 1 - ( x - 7 )10] = 0
=>( x - 7 )x + 1 = 0 hoặc 1 - ( x - 7 )10 = 0
x - 7 = 0 hoặc ( x - 7 )10 = 0
x - 7 = 0 hoặc / x - 7 / = 1
x - 7 = 0 hoặc x - 7 = 1 hoặc x - 7 = -1
=> x = 7 hoặc x = 8 hoặc x = 6
Hok tốt !
\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
=> (x-7)x . (x-7) - (x-7)x . (x-7)11 = 0
=> \(\left(x-7\right)^x.\left[\left(x-7\right)-\left(x-7\right)^{11}\right]=0\)
=> [(x-7) - (x-7)11 ] = 0
=> \(\left\{\left(x-7\right).\left[1-\left(x-7\right)^{10}\right]\right\}=0\)
\(\Rightarrow\orbr{\begin{cases}x-7=0\\1-\left(x-7\right)^{10}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\\left(x-7\right)^{10}=1\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=7\\\left(x-7\right)^{10}=\left(-1\right)^{10}=1^{10}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=7\\x-7=1\\x-7=-1\end{cases}}\Rightarrow\hept{\begin{cases}x=7\\x=8\\x=6\end{cases}}\)
Vậy x thuộc { 6,7,8}
cố đúng k vậy bạn được bn phần trăn ạ