Tìm x biết
4x2-36x+56=0
Bài 4: Tìm x, biết.
a) 4x(x - 7) - 4x2 = 56
b) 12x(3x - 2) - (4 - 6x) = 0
c) 4(x - 5) - (5 - x)2 = 0
d) x(x +1) - x(x - 3) = 0
e) - 6x + 8 = 0 f) 2 + 2x + = 0
c: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
tìm x biết a) ( x + 3 )2 - ( 2x + 1 ).( x+3 ) = 0 ; b) x3 - 12x2 + 36x = 0
\(a,\Leftrightarrow\left(x+3\right)\left(x+3-2x-1\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-12x+36\right)=0\\ \Leftrightarrow x\left(x-6\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
a, (x+3)2 - ( 2x + 1 ).( x+3)=0 b, x3-12x2+36x =0
=> (x+3).(x+3-2x-1) => x(x2-12x+36) = 0
=>(x+3).(-x+2) => x(x-6)2 = 0
=> x+3=0 <=> x=-3 => x=0 <=> x=0
-x+2=0 <=> x=-2 x-6= 0 <=> x=6
Rút gọn phân thức A = 3|x − 2| − 5|x − 6| 4x 2 − 36x + 81 với 2 < x < 6 ta được?
A. A = 4 x − 9
B. A = 4 9 − 2x
C. A = 4 2x − 9
D. A = 8 2x − 9
tìm x biết: 20x - 4x2 = 0
`20x - 4x^2=0`
`<=>4x(5-x)=0`
\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\5-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0:4\\x=5-0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Vậy `x ∈ { 0; 5 }`
20x - 4x2 = 0
4x( 5 - x ) = 0
TH1: 4x = 0
x = 0 : 4
x = 0
TH2: 5 - x = 0
x = 5 - 0
x = 5
Vậy x ∈ { 0; 5 }
tìm x , biết rằng : \(36x-x^2=0\)
36x - x2 = 0
<=> x(36 - x) = 0
<=> x = 0 hoặc 36 - x = 0
<=> x = 0 hoặc x = 36
Vậy x = 0 hoặc x = 36
ung ho minh len 200 nha
Tìm x biết : 4x2-4x-1=0
\(\left(2x-1\right)^2=0\)
\(2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)
\(\Rightarrow\left(4x^2-4x+1\right)-2=0\\ \Rightarrow\left(2x-1\right)^2=2\\ \Rightarrow\left[{}\begin{matrix}2x-1=\sqrt{2}\\2x-1=-\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{2}+1}{2}\\x=\dfrac{1-\sqrt{2}}{2}\end{matrix}\right.\)
Tìm x,biết
36x2 - 49 = 0
36x^2 - 49=0 =>(6x-7)(6x+7)=0
6x-7=0 =>x=7/6
6x+7=0=>x=-7/6
\(\left(6x\right)^2=7^2\)
6x=+-7
\(x=+-\frac{7}{6}\)
Tìm x biết :
(4x2-3)2+8=0
`(4x^2-3)^2+8=0`
`(4x^2-3)^2=-8`
Vì `(4x^2-3)^2 >=0> -8` với mọi `x` nên PT trên vô nghiệm.
tìm x biết : x2(x - 1)2 - 4x2 + 8x - 4 = 0
\(\Leftrightarrow x^2\left(x-1\right)^2-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)^2\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.\)