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HQ
Hà Quang Minh
Giáo viên
10 tháng 8 2023
\(\left(5-x\right)\left(x-2\right)+\left(x-7\right)\left(x+7\right)=\left(3x-1\right)^2-\left(3x-2\right)\left(3x+2\right)\\ \Leftrightarrow-x^2+7x-10+x^2-49=9x^2-6x+1-9x^2+4\\\Leftrightarrow7x-59=-6x+5\\ \Leftrightarrow13x=44\\ \Leftrightarrow x=\dfrac{64}{13} \)
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)
8 tháng 9 2023
2:
a: =>x^2+3x-4x-12-(x^2-5x+x-5)=8
=>x^2-x-12-x^2+4x+5=8
=>3x-7=8
=>3x=15
=>x=5
b: =>3x^2+3x-2x-2-3x^2-21x=13
=>-20x=15
=>x=-3/4
c: =>x^2-25-x^2-2x=9
=>-2x=25+9=34
=>x=-17
d: =>x^3-1-x^3+3x=1
=>3x-1=1
=>3x=2
=>x=2/3
H9
HT.Phong (9A5)
CTVHS
6 tháng 10 2023
Bài 5.5:
\(\left(2x-3\right)\left(x+1\right)+\left(4x^3-6x^2-6x\right):\left(-2x\right)=18\)
\(\Leftrightarrow\left(2x^2+2x-3x-3\right)+2x\cdot\left(2x^2-3x-3\right):\left(-2x\right)=18\)
\(\Leftrightarrow2x^2-x-3-2x^2+3x+3=18\)
\(\Leftrightarrow2x=18\)
\(\Leftrightarrow x=\dfrac{18}{2}\)
\(\Leftrightarrow x=9\)
TV
3
5 tháng 10 2021
\(\left(x+7\right)\left(3x-15\right)=0\\ \Rightarrow3\left(x-5\right)\left(x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\\ 4x\left(x+7\right)=2\left(x+7\right)\\ \Rightarrow4x\left(x+7\right)-2\left(x+7\right)=0\\ \Rightarrow2\left(2x-1\right)\left(x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-7\end{matrix}\right.\\ \left(x-3\right)^2-x\left(x-4\right)=5\\ \Rightarrow x^2-6x+9-x^2+4x-5=0\\ \Rightarrow-2x+4=0\\ \Rightarrow-2x=-4\Rightarrow x=2\)
hưng phúc đầy đủ chưa bạn nhỉ?
MH
5 tháng 10 2021
1) \(\left(x+7\right)\left(3x-15\right)=0\)
⇔\(\left[{}\begin{matrix}x+7=0\\3x-15=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)
2) \(4x\left(x+7\right)=2\left(x+7\right)\)
\(2\left(2x+1\right)\left(x+7\right)=0\)
⇔\(\left[{}\begin{matrix}2x+1=0\\x+7=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-7\end{matrix}\right.\)
1 tháng 6 2016
Ta có: (x - 3)(3x + 7) - 3x2 = 0
=> 3x2 - 2x - 21 - 3x2 = 0
=> -2x - 21 = 0
=> -2x = 21
=> x = -10,5
TV
1
21 tháng 3 2021
Ta có: \(M=\dfrac{x^5+3x^3-x^2+3x-7}{x^2+2}\)
\(=\dfrac{x^5+2x^3+x^3+2x-x^2-2+x-5}{x^2+2}\)
\(=\dfrac{x^3\left(x^2+2\right)+x\left(x^2+2\right)-\left(x^2+2\right)+\left(x-5\right)}{x^2+2}\)
\(=\dfrac{\left(x^2+2\right)\left(x^3+x-1\right)+\left(x-5\right)}{\left(x^2+2\right)}\)
\(=x^3+x-1+\dfrac{x-5}{x^2+2}\)
Để M nguyên thì \(x-5⋮x^2+2\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)⋮x^2+2\)
\(\Leftrightarrow x^2-25⋮x^2+2\)
\(\Leftrightarrow x^2+2-27⋮x^2+2\)
mà \(x^2+2⋮x^2+2\)
nên \(-27⋮x^2+2\)
\(\Leftrightarrow x^2+2\inƯ\left(-27\right)\)
\(\Leftrightarrow x^2+2\in\left\{1;-1;3;-3;9;-9;27;-27\right\}\)
\(\Leftrightarrow x^2+2\in\left\{3;9;27\right\}\)(Vì \(x^2+2\ge2\forall x\))
\(\Leftrightarrow x^2\in\left\{1;7;25\right\}\)
hay \(x\in\left\{1;-1;\sqrt{7};-\sqrt{7};5;-5\right\}\)
Vậy: Để M nguyên thì \(x\in\left\{1;-1;\sqrt{7};-\sqrt{7};5;-5\right\}\)
Bai lam
\(5x+7=3x+2\Leftrightarrow2x=-5\Leftrightarrow x=-\frac{5}{2}\)