tìm x
x2+5x-2250=0
tìm x biết:
a)x2 + 3x = 0 b) x3 – 4x = 0
c) 5x(x-1) = x-1 d) 2(x+5) - x2-5x = 0
e) 2x(x-5)-x(3+2x)=26 f) 5x.(x – 2012) – x + 2012 = 0
a) \(\Rightarrow x\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
b) \(\Rightarrow x\left(x^2-4\right)=0\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)
Tìm x biết:
a/ 5x( x- 3) = x – 3 b/ x3 - x = 0 c/ x2 – 7x + 6 = 0
d/ x2 – 4 + ( x – 2)2 = 0 e/ x2 – 16 –( x +4) = 0 f/ x2 + x – 2 = 0
a: \(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
b: \(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Câu 2.(1,5 điểm) Tìm x, biết:
a) 5x(x2 – 9) = 0. b) 3(x+3) - x2 - 3x =0. c) x2 – 9x – 10 = 0
\(a,5x\left(x^2-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,3\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow3\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\ c,x^2-9x-10=0\\ \Leftrightarrow x^2+x-10x-10=0\\ \Leftrightarrow x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)
a, 5\(x\)(\(x^2\) - 9) = 0
\(\left[{}\begin{matrix}x=0\\x^2-9=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy \(x\) \(\in\) { -3; 0; 3}
b, 3.(\(x+3\)) - \(x^2\) - 3\(x\) = 0
3.(\(x+3\)) - \(x\).( \(x\) + 3) = 0
(\(x+3\))( 3 - \(x\)) = 0
\(\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
Vậy \(x\) \(\in\){ -3; 3}
c, \(x^2\) - 9\(x\) - 10 = 0
\(x^2\) + \(x\) - 10\(x\) - 10 = 0
\(x.\left(x+1\right)\) - 10.( \(x-1\)) = 0
(\(x+1\))(\(x-10\)) = 0
\(\left[{}\begin{matrix}x+1=0\\x-10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)
Vậy \(x\) \(\in\){ -1; 10}
a) 5x(x2-9)=0
=> TH1 5x=0
<=> x= 0
TH2: 2x-9=0
<=> 2x=9
<=> x = \(\dfrac{9}{2}\)
b, 3(x+3) - x2- 3x = 0
<=> 3x + 9 - x2 -3x = 0
<=> - x2 +9 = 0
<=> - x2 = -9
<=> x = 3
c, x2 -9x -10 = 0
<=> x2 -x + 10x -10 = 0
<=> x(x-1)+10(x-1)=0
<=> (x-1)(x+10)=0
=> TH1: x-1=0
<=> x=1
TH2: x +10=0
<=> x=-10
Bài 2: Tìm x , biết
a) 6x3 + x2 : 2x - 3x + 2 = 0 b) 5x4 - 3x2 : x2 - x(5x + 6) = 0
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Bài 2: Tìm x , biết
a) 6x3 + x2 : 2x - 3x + 2 = 0 b) 5x4 - 3x2 : x2 - x(5x + 6) = 0
Tìm x biết:
a) 5x.(x-1)-(x+2).(5x-7)=6
b) (x+2)2-(x2-4)=0
`a)5x(x-1)-(x+2)(5x-7)=6`
`<=>5x^2-5x-(5x^2-7x+10x-14)=6`
`<=>5x^2-5x-(5x^2+3x-14)=6`
`<=>-8x+14=6`
`<=>8x=8<=>x=1`
Vậy `x=1`
`b)(x+2)^2-(x^2-4)=0`
`<=>x^2+4x+4-x^2+4=0`
`<=>4x+8=0`
`<=>4x=-8`
`<=>x=-2`
Vậy `x=-2`
a)5x.(x-1)-(x+2).(5x-7)=6
<=> 5x2-5x-(5x2-7x+10x-14)=6
<=> 5x2-5x-5x2+7x-10x+14=6
<=> -8x+14=6
<=> -8x=-8 => x=1
Vậy x=1
b) (x+2)2-(x2-4)=0
<=> (x+2)2-(x2-22)=0 <=> (x+2)2-(x-2)(x+2)=0
<=> (x+2)[(x+2)-(x-2)]=0
<=> (x+2)(x+2-x+2)=0
<=> (x+2).4=0
=> x+2=0
=> x=-2
Vậy x=-2
Tìm x:
a)x2-4x=3.(x-4)
b)x2-5x-24=0
a: \(x^2-4x=3\left(x-4\right)\)
\(\Leftrightarrow\left(x-4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)
b: \(x^2-5x-24=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)
a) pt <=> (x - 4)(x - 3) = 0
<=> x = 4 hoặc x = 3
b) pt <=> (x - 8)(x + 3) = 0
<=> x = 8 hoặc x = -3
2(x + 5) – x2 – 5x = 0 (Tìm x)
\(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
tìm x, biết:
2(x+5)+x2+5x=0
=>(x+5)(x+2)=0
=>x=-5 hoặc x=-2
Đề sai nên mình sửa lại giữa x2 với 5x thành dấu nhân nhé bạn :))
\(2\left(x+5\right)+x\left(x+5\right)=0\)
\(\left(2+x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2+x=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-5\end{matrix}\right.\)
\(2\left(x+5\right)+x\left(x+5\right)=0\\ \left\{{}\begin{matrix}x+5=0\\2+x=0\end{matrix}\right.\left\{{}\begin{matrix}x=0-5=-5\\x=0-2=-2\end{matrix}\right.\)