Tìm A=\(\frac{2x-2}{3+2x}\)
Giải rõ giùm mình nha!
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{2x+2}{5x-3}=\frac{2x+12}{5x+18}\)
=> ( 2x + 2 ) ( 5x + 18 ) = ( 2x + 12 ) ( 5x - 3 )
=> 2x ( 5x + 18 ) + 2 ( 5x + 18 ) = 2x ( 5x - 3 ) + 12 ( 5x - 3 )
=> 10 x 2 + 36x + 10x + 36 = 10 x 2 - 6x + 60 x - 36
=> 36x + 10x + 6x - 60x = - 36 - 36
=> - 8 x = - 72
=> x = 9
\(\left(x+2\right)^2=\left(2x-1\right)^2\\ \Leftrightarrow\left(x+2\right)^2-\left(2x-1\right)^2=0\\\Leftrightarrow\left[x+2-\left(2x-1\right)\right]\left[x+2+2x-1\right]=0\\ \Leftrightarrow\left(x+2-2x+1\right)\left(x+2+2x-1\right)=0\\ \Leftrightarrow\left(-x+3\right)\left(3x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-x+3=0\\3x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-x=-3\\3x=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{3}\end{matrix}\right.\)
\(\left(x+2\right)^2=\left(2x-1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=2x-1\\x+2=-\left(2x-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2x=-1-2\\x+2=-2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=-3\\x+2x=1-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\3x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{3}\end{matrix}\right.\)
\(2x-3=x-\left(\frac{-1}{2}\right)\)
\(\Rightarrow2x-3=x+\frac{1}{2}\)
\(\Rightarrow2x=x+\frac{1}{2}+3\)
\(\Rightarrow2x-x=\frac{1}{2}+3\)
\(\Rightarrow x=\frac{7}{2}\)
2x-x+\(\frac{1}{2}\)=3
2x-x=3-\(\frac{1}{2}\)
2x-x=\(\frac{5}{2}\)
x-x=\(\frac{5}{2}\): 2
x-x=\(\frac{5}{4}\)
\(x+2x+3x+...+100x=2200\)
=>\(x\left(1+2+3+...+100\right)=2200\)
=>\(x.\frac{100.101}{2}=2200\)
=>\(x.5050=2200\)
=>x=2200:5050
=>x=\(\frac{44}{101}\)
\(x+2x+3x+...+100x=220\)
\(\Rightarrow x\left(1+2+3+....+100\right)=2200\)
\(\Rightarrow5050x=2200\)
\(\Rightarrow x=\frac{44}{101}\)
\(a,\Rightarrow2x^2-18x-2x^2=0\\ \Rightarrow-18x=0\Rightarrow x=0\\ b,\Rightarrow2x^2-5x-12+x^2-7x+10=3x^2-17x+20\\ \Rightarrow5x=22\Rightarrow x=\dfrac{22}{5}\)
a, <=> -4x-10-31+4x=-2016+x
<=> -10-31=-2016+x
<=> -41=-2016+x
<=> x=1975
a)\(2x^2\)+\(3\left(x^2-1\right)\)=\(5x\left(x+1\right)\)
\(2x^2\)+\(3x^2\)\(-3\)=\(5x^2+5x\)
\(5x^2-5x^2-5x=3\)
\(-5x=3\)
\(x=\frac{-3}{5}\)
tự ghi dấu suy ra ở đằng trước nhé
b) Vì \(2x\left(5-3x\right)=2x\left(3x-5\right)-3\left(x-7\right)=3\)
nên chỉ cần giải: \(6x^2-10x-3x+21=3\)
\(\Leftrightarrow6x^2-13x+21=3\)
\(\Leftrightarrow6x^2-13x+18=0\)
\(\Rightarrow\)pt vô nghiệm
\(A=\frac{2x-2}{3+2x}\)hay \(A=\frac{2x-2}{2x+3}\)
\(A=\frac{2x-2}{2x+3}=\frac{2x+3-5}{2x+3}=1-\frac{5}{2x+3}\)
A là số nguyên thì 2x + 3 là ước nguyên của 5
\(2x+3=1\Rightarrow x=-1\)
\(2x+3=-1\Rightarrow x=-2\)
\(2x+3=5\Rightarrow x=1\)
\(2x+3=-5\Rightarrow x=-4\)
Vậy \(x\in\){ -1 ; -2 ; 1; ;-4}
Ai thấy đúng thì ủng hộ, tháy sai thì góp ý nha !!!