tìm x:
a) ( x + 5) ( x-1) - ( x-5) ( x-2) =0
b) ( x-3) (x+7) - ( x-2) ( x+4) = 0
c) ( x- 6) (x-7) - x( x+4) -5(x-1) = 0
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\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)
\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)
bài 1:
a) (x+1)^2-(x-1)^2-3(x+1)(x-1)
=(x+1+x-1)(x+1-x+1)-3x^2-3
=2x^2-3x^2-3
=-x^2-3
xem lại câu b
a)(x+4)^2-(x+1)(x-1)=0
<=>x2+8x+16-(x2-1)=0
<=>x2+8x+16-x2+1=0
<=>8x+17=0
<=>8x=-17
<=>x=-17/8
c)3(x-7)+(5-x)=2^3(x-1)
<=>3x-21+5+x=8x-8
<=>3x+x-8x=-8-5+21
<=>-4x=8
<=>x=-2
1/(2.x-5)+17=6
=> 2x - 5 = -11
=> 2x = -6
=> x = 3
vậy_
2/10-2.(4-3x)=-4
=> 2(4 - 3x) = 14
=> 4 - 3x = 7
=> 3x = -3
=> x = -1
3/-12+3.(-x+7)=-18
=> 3(-x+7) = -6
=> -x+7 = -2
=> -x = -9
=> x = 9
4/24:(3.x-2)=-3
=> 3x - 2 = -8
=> 3x = -6
=> x = -2
5/-45:5.(-3-2.x)=3
=> 5(-3 - 2x) = -15
=> -3 - 2x = -3
=> - 2x = 0
=> x = 0
6/x.(x+7)=0
=> x = 0 hoặc x + 7 = 0
=> x = 0 hoặc x = -7
7/(x+12).(x-3)=0
=> x + 12 = 0 hoặc x - 3 = 0
=> x = -12 hoặc x = 3
8/(-x+5).(3-x)=0
=> -x + 5 = 0 hoặc 3 - x = 0
=> x = 5 hoặc x = 3
9/x.(2+x).(7-x)=0
=> x = 0 hoặc 2 + x = 0 hoặc 7 - x = 0
=> x = 0 hoặc x = -2 hoặc x = 7
10/(x-1).(x+2).(-x-3)=0
=> x - 1 = 0 hoặc x + 2 = 0 hoặc -x-3 = 0
=> x = 1 hoặc x = -2 hoặc x = -3
Bài 1 :
\(\frac{x-1}{x-5}=\frac{6}{7}\Leftrightarrow7x-7=6x-30\)
\(\Leftrightarrow x=-23\)
\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)ĐK : \(x\ne1;-7\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-1\right)\)
\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)
\(\Leftrightarrow2x-10=0\Leftrightarrow x=5\)
Lời giải:
a)
PT $\Leftrightarrow (x^2+4x-5)-(x^2-7x+10)=0$
$\Leftrightarrow 11x-15=0$
$\Leftrightarrow x=\frac{15}{11}$
b)
PT $\Leftrightarrow (x^2+4x-21)-(x^2+2x-8)=0$
$\Leftrightarrow 2x-13=0$
$x=\frac{13}{2}$
c)
PT $\Leftrightarrow (x^2-13x+42)-(x^2+4x)-(5x-5)=0$
$\Leftrightarrow -22x+47=0\Rightarrow x=\frac{47}{22}$