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

cac ban giup minh nha

đề câu a khó hiểu thế

b/ 4x²-25=0

 <=>(2x)²-5²=0

<=>(2x-5)(2x+5)=0

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x+5=0\\2x-5=0\end{array}\right.\)

\(\Leftrightarrow x=\pm\frac{5}{2}\)

c/ x³-4x²+4x=0

<=>x(x²-4x+4)=0

<=>x(x-2)²=0

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-2=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=2\end{array}\right.\)

\(x.\left(x-2009\right)-2010x+2009.2010=0\)

\(x.\left(x-2009\right)-2010\left(x-2009\right)=0\)

\(\left(x-2009\right)\left(x-2010\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2009=0\\x-2010=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2009\\x=2010\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=2009\\x=2010\end{cases}}\)

x+(x+1)+...+(x+2009) dãy số trên có:x+2009-x+1=2010 số

=>x+(x+1)+...+(x+2009)=(x+x+2009).2010:2=(2x+2009).1005

=>(2x+2009).1005=2009.2010

<=>2x+2009=2009.2

<=>2x=2009.2-2009

<=>2x=2009

<=>x=2009/2

Ta có tất cả 2009 số hạng x

Từ 1 đến 2009 có tất cả 2009 số hạng

x. 1 + ( x+x+x+...+x+x ) + ( 1+2+3+..+2009) = 4038090

x. 1+ x . 2009 + (( 1+ 2009 ) x 2009 : 2) = 4038090

x . 1 + x . 2009 + 2019045 = 4038090

x . ( 1+ 2009 ) + 2019045 = 4038090

x . 2010 + 2019045 = 4038090

x . 2010 = 4038090 - 2019045

x . 2010 = 2019045

x = 2019045 : 2010

x = 1004,5

=> 2010.x + 1+2+3+...+2009 = 2009.2010

=>2010.x  + 2009.2010 : 2 = 2009.2010

=> 2010 .x = 2009.2010:2

= 2019045

=> x = 1004,5

hihi mik cũng làm zậy

??????????????????...???????????????????>.<

Thay 2010 = x + 1 vào P ( x ),ta có :

\(^{x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-\left(x+1\right)x^7+...+\left(x+1\right)x^2-\left(x+1\right)x-1}\)

= x¹⁰ - x¹⁰ - x⁹ + x⁹ + x⁸ - x⁸ - x⁷ + ... + x³ + x² - x² + x - 1

= x + 1

= 2009 + 1

= 2010
 

Thay 2010 = x+ 1 vào P( x) ,có :

\(x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-\left(x+1\right)x^7+...+\left(x+1\right)x^2-\left(x+1\right)x-1\)

= \(x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...+x^3+x^2-x^2+x-1\) 

= x+1 

= 2009 + 1

= 2010

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

X+(X+1)+(x+2)+++++++(x+2009)=2009.2010

=>2010.X+(1+2+3+...........+2009)=2009.2010

1+2+3+...........+2009 có 2009 số hạng

=>2010.X+[(2009+1).2009:2]=2009.2010

2010.X+(2010.2009:2)=2009.2010

2010.X=2009.2010-(2009.2010:2)=2010.2009:2

=>X=2009:2=1004,5

Vậy X=1004,5

x+(x+1)+(x+2)..........(x+2009)=4038090

=>x.(2009+1)=4038090.

x.2010=4038090.

x=4038090:10.

x=2009.

Nhớ-k-cho-mình-nhé!Chúc-bạn-học-tốt.