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b, \(4x^2-25=0\)
\(\Leftrightarrow4x^2=25\)
\(\Leftrightarrow x^2=\frac{25}{4}=\left(\pm\frac{5}{2}\right)^2\)
\(\Leftrightarrow x=\pm\frac{5}{2}\)
Vậy \(x\in\left\{\frac{5}{2};-\frac{5}{2}\right\}\)
c) x3 - 4x2 + 4x = 0
=> x3 - 2x2 - 2x2 + 4x = 0
=> x2.(x - 2) - 2x.(x - 2) = 0
=> (x - 2).(x2 - 2x) = 0
=> (x - 2).x.(x - 2) = 0
=> (x - 2)2.x = 0
=> (x - 2)2 = 0 hoặc x = 0
=> x - 2 = 0 hoặc x = 0
=> x = 2 hoặc x = 0
\(x.\left(x-2009\right)-2010x+2009.2010=0\)
\(x.\left(x-2009\right)-2010\left(x-2009\right)=0\)
\(\left(x-2009\right)\left(x-2010\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2009=0\\x-2010=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2009\\x=2010\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=2009\\x=2010\end{cases}}\)
Bài 1 :
a) Ta có : x2 - 9x + 8 = x2 - x - 8x + 8 = x(x - 1) - 8(x - 1) = (x - 8)(x - 1)
b) Ta có : x2 + 6x + 8 = x2 + 6x + 9 - 1 = (x + 3)2 - 1 = (x + 3 - 1)(x + 3 + 1) = (x + 2)(x + 4)
Bài 2 :
b) 4x2 - 25 = 0
=> 4x2 = 25
=> (2x)2 = 52
=> 2x = -5;5
=> x = -5/2 ; 5/2
b) = x^2 + 2.x.3 + 3^2 - 1
=(x + 3)^2 - 1
=(x + 3 + 1)(x + 3 - 1)
=(x + 4)(x + 2)
Phần a mk nghĩ bn nên tự lm.
Ta có: x = 2011 \(\Rightarrow\) 2010 = x - 1
\(A=x^{2011}-2010x^{2010}-2010x^{2009}-...-2010x+1\)
\(=x^{2011}-\left(x-1\right)x^{2010}-\left(x-1\right)x^{2009}-...-\left(x-1\right)x+1\)
\(=x^{2011}-\left(x-1\right)x^{2010}-\left(x-1\right)x^{2009}-...-\left(x-1\right)x+1\)
\(=x^{2011}-x^{2011}+x^{2010}-x^{2010}+x^{2009}-...-x^2+x+1\)
\(=x+1\)
\(=2011+1\)
\(=2012.\)
x=2011
=> 2010= x-1
A = x^2011- (x-1) x^2010- (x-1).x^2009-.....- (x-1).x+1
= x^2011-x^2011+x^2010- x^2010+x^2009..x^2.-x^2+x+1
= x+1
=(x-1)+2= 2010+2=2012
a ) 2x ( x - 5 ) - x ( 3 + 2x ) = 26
2x2 - 10x - 3x - 2x2 = 26
- 13x = 26
x = 26 : ( -13 )
x = -2
b) 49x2 - 81 = 0
( 7x - 9 )( 7x + 9 ) = 0
Th1 :
7x - 9 = 0
7x = 9
x = \(\frac{9}{7}\)
Th2
7x + 9 = 0
7x = -9
x = \(-\frac{9}{7}\)
Vay x = \(\frac{9}{7}\) hoac x = \(-\frac{9}{7}\)
a/ 3x(12x-5)-6x(6x-5)=0
<=>36x2-15x-36x2+30x=0
<=>15x=0
<=>x=0
b/ x2-8x+6=0
Nghiệm lẻ xem lại câu b
a) \(3x\left(12x-5\right)-6x\left(6x-5\right)=0\)
\(\Rightarrow36x^2-15x-36x^2+30x=0\)
\(\Rightarrow15x=0\)
\(\Rightarrow x=0\)
b) \(x^2-8x+6=0\)
\(\Rightarrow x^2-8x+16-10=0\)
\(\Rightarrow x^2-8x+4^2=10\)
\(\Rightarrow\left(x-4\right)^2=10\)
\(\Rightarrow x-4=\sqrt{10}\)
\(\Rightarrow x=4+\sqrt{10}\)
\(x\left(x-2009\right)-2010x+2009\times2010=0\)
\(x^2-2009x-2010x+2009\times2010=0\)
\(x\left(x-2010\right)-2009\left(x-2010\right)=0\)
\(\left(x-2009\right)\left(x-2010\right)=0\)
nên x - 2009 = 0
x = 2009
x-2010=0
x=2010