Tìm x biết
x+5/2005 + x+6/2004 + x+7/2003 = -3
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\(\left(84,6-2\cdot x\right):3,02=5,1\)
\(\Rightarrow84,6-2\cdot x=15,402\)
\(\Rightarrow2\cdot x=69,198\)
\(\Rightarrow x=69,198:2\)
\(\Rightarrow x=34,599\)
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\(\left(15\cdot24-x\right):0,25=100:0,25\)
\(\Rightarrow\left(360-x\right):0,25=400\)
\(\Rightarrow360-x=100\)
\(\Rightarrow x-360-100\)
\(\Rightarrow x=260\)
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\(128\cdot x-12\cdot x-16\cdot x=5200\)
\(\Rightarrow x\cdot\left(128-12-16\right)=5200\)
\(\Rightarrow x\cdot100=5200\)
\(\Rightarrow x=5200:100\)
\(\Rightarrow x=52\)
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\(5\cdot x+3,75\cdot x+1,25\cdot x=20\)
\(\Rightarrow x\cdot\left(5+3,75+1,25\right)=20\)
\(\Rightarrow10\cdot x=20\)
\(\Rightarrow x=20:10\)
\(\Rightarrow x=2\)
\(x\cdot3,7+x\cdot6,3=360:120\)
\(\Rightarrow x\cdot\left(3,7+6,3\right)=3\)
\(\Rightarrow x\cdot10=3\)
\(\Rightarrow x=\dfrac{3}{10}\)
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\(x\cdot23-6\cdot23+x\cdot69=320\)
\(\Rightarrow x\cdot\left(23+69\right)=320+6\cdot23\)
\(\Rightarrow x\cdot92=458\)
\(\Rightarrow x=458:92\)
\(\Rightarrow x=\dfrac{229}{46}\)
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\(\left(x+1\right)\left(x+2\right)=72\)
\(\Rightarrow x^2+2x+x+2=72\)
\(\Rightarrow x^2+3x+2=72\)
\(\Rightarrow x^2+3x+2-72=0\)
\(\Rightarrow x^2+3x-70=0\)
\(\Rightarrow x^2+10x-7x-70=0\)
\(\Rightarrow\left(x-7\right)\left(x+10\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=7\\x=-10\end{matrix}\right.\)
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\(\left(x+2\right)\cdot16\cdot x=160x\)
\(\Rightarrow16x^2+32x=160x\)
\(\Rightarrow16x^2+32x-160x=0\)
\(\Rightarrow16x^2-128x=0\)
\(\Rightarrow16x\left(x-8\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)
\(1.x-\dfrac{2}{3}\times\left(x+9\right)=1\)
\(x-\dfrac{2}{3}\times x-6=1\)
\(x\times\left(1-\dfrac{2}{3}\right)=7\)
\(x\times\dfrac{1}{3}=7\)
\(x=21\)
\(2.x-\dfrac{11}{15}=\dfrac{3+x}{5}\)
\(\dfrac{15x}{15}-\dfrac{11}{15}=\dfrac{9+3x}{15}\)
\(15x-11=9+3x\)
\(12x=20\)
\(x=\dfrac{5}{3}\)
bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 => x-1/3=y-2/4=z-3/5
áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1
do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự
Bài 1:
a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )
a) \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\) (ĐK: \(x\ne\pm3\))
\(A=\left[\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2-1}{\left(x+3\right)\left(x-3\right)}\right]:\left(2+\dfrac{x+5}{x+3}\right)\)
\(A=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x+3\right)\left(x-3\right)}:\dfrac{2\left(x+3\right)-\left(x+5\right)}{x+3}\)
\(A=\dfrac{-5x-5}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}{x+1}\)
\(A=\dfrac{-5\left(x+1\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)\left(x+1\right)}\)
\(A=\dfrac{-5}{x-3}\)
b) Ta có: \(\left|x\right|=1\)
TH1: \(\left|x\right|=-x\) với \(x< 0\)
Pt trở thành:
\(-x=1\) (ĐK: \(x< 0\))
\(\Leftrightarrow x=-1\left(tm\right)\)
Thay \(x=-1\) vào A ta có:
\(A=\dfrac{-5}{x-3}=\dfrac{-5}{-1-3}=\dfrac{5}{4}\)
TH2: \(\left|x\right|=x\) với \(x\ge0\)
Pt trở thành:
\(x=1\left(tm\right)\) (ĐK: \(x\ge0\))
Thay \(x=1\) vào A ta có:
\(A=\dfrac{-5}{x-3}=\dfrac{-5}{1-2}=\dfrac{5}{2}\)
c) \(A=\dfrac{1}{2}\) khi:
\(\dfrac{-5}{x-3}=\dfrac{1}{2}\)
\(\Leftrightarrow-10=x-3\)
\(\Leftrightarrow x=-10+3\)
\(\Leftrightarrow x=-7\left(tm\right)\)
d) \(A\) nguyên khi:
\(\dfrac{-5}{x-3}\) nguyên
\(\Rightarrow x-3\inƯ\left(-5\right)\)
\(\Rightarrow x\in\left\{8;-2;2;4\right\}\)
a: \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\)
\(=\dfrac{x\left(x-3\right)-2\left(x+3\right)-x^2+1}{\left(x-3\right)\left(x+3\right)}:\dfrac{2x+6-x-5}{x+3}\)
\(=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+1}\)
\(=\dfrac{-5x-5}{\left(x-3\right)}\cdot\dfrac{1}{x+1}=\dfrac{-5}{x-3}\)
b: |x|=1
=>x=-1(loại) hoặc x=1(nhận)
Khi x=1 thì \(A=\dfrac{-5}{1-3}=-\dfrac{5}{-2}=\dfrac{5}{2}\)
c: A=1/2
=>x-3=-10
=>x=-7
d: A nguyên
=>-5 chia hết cho x-3
=>x-3 thuộc {1;-1;5;-5}
=>x thuộc {4;2;8;-2}
\(\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}=-3\)
=>\(\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+6}{2004}+1\right)+\left(\dfrac{x+7}{2003}+1\right)=0\)
=>\(\left(x+2010\right)\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\right)=0\)
=>\(x+2010=0\)(do\(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\)khác 0)
=>x=-2010
Vậy...
Chị Nhung ơi , có phải chị chơi Bang Bang k ???