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17 tháng 10 2018

\(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4.\)

\(=\left(x+a\right)\left(x+4a\right)\left(x+2a\right)\left(x+3a\right)+a^4.\)

\(=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4.\)

\(=\left(x+5ax+4a^2+a^2\right)^2.\)

\(=\left(x+5ax+5a^2\right)^2.\)

18 tháng 10 2018

\(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4\)

\(=\)\(\left(x+a\right)\left(x+4a\right)\left(x+2a\right)\left(x+3a\right)+a^4\)

\(=\)\(\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4\)

\(=\)\(\left[\left(x^2+5ax+5a^2\right)-a^2\right].\left[\left(x^2+5ax+5a^2\right)-a^2\right]+a^4\)

\(=\)\(\left(x^2+5ax+5a^2\right)^2-a^4+a^4\)

\(=\)\(\left(x^2+5ax+5a^2\right)^2\)

Chúc bạn học tốt ~ 

12 tháng 9 2017

(x + a)(x + 2a)(x + 3a)(x + 4a) + a4

= (x + a)(x + 4a)(x + 2a)(x + 3a) + a4

= (x2 + 4ax + ax + 4a2)(x2 + 3ax + 2ax + 6a2) + a4

= (x2 + 5ax + 4a2)(x2 + 5ax + 6a2) + a4

Đặt x2 + 5ax + 4a2 = t

= t(t + 2a2) + a4

= (t + a2)2

= (x2 + 5ax + 4a2 + a2)2

= (x2 + 5ax + 5a2)2

8 tháng 7 2018

đơn giản

13 tháng 9 2017

Ta có:

(x+a)(x+2a)(x+3a)(x+4a) + a4

=(x+a)(x+4a)(x+3a)(x+2a) +a4

=(x2+5ax+4a2)(x2+5ax+6a2) + a4

Đặt x2+5ax+5a2=y

=>(x2+5ax+4a2)(x2+5ax+6a2) + a4=(y-a2)(y+a2)+a4

=y2-a4+a4

=y2

=(x2+5ax+5a2)2

k mik nha

8 tháng 7 2018

=\(\left(x+a-3\right)\left(x^2-2ax-2x+4a-12\right)\)

1. \(4x^2-17xy+13y^2=4x^2-4xy-13xy+13y^2=4x\left(x-y\right)-13y\left(x-y\right)=\left(x-y\right)\left(4x-13y\right)\)

2. \(2x\left(x-5\right)-x\left(3+2x\right)=26\Leftrightarrow2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)

3. \(A=\left(2a-3b\right)^2+2\left(2a-3b\right)\left(3a-2b\right)+\left(2b-3a\right)^2\)

\(\Leftrightarrow\left(2a-3b\right)^2-2\left(2a-3b\right)\left(2b-3a\right)+\left(2b-3a\right)^2=\left(2a-3b-2b+3a\right)^2=\left(5a-5b\right)^2\)

\(=25\left(a-b\right)^2=25\cdot100=2500\)

13 tháng 12 2023

\((x+5)^2+4(x+5)(x-5)+4(x^2-10x+25)=0\\\Rightarrow(x+5)^2+4(x+5)(x-5)+4(x^2-2\cdot x\cdot5+5^2)=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+4(x-5)^2=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+[2(x-5)]^2=0\\\Rightarrow[(x+5)+2(x-5)]^2=0\\\Rightarrow(x+5+2x-10)^2=0\\\Rightarrow(3x-5)^2=0\\\Rightarrow3x-5=0\\\Rightarrow3x=5\\\Rightarrow x=\frac53\\\text{#}Toru\)

13 tháng 12 2023

Sao đề là phân tích mà lại "= 0" vậy bạn?

6 tháng 12 2023

\(5x(2x+3)+6x+9\\=5x(2x+3)+3(2x+3)\\=(2x+3)(5x+3)\)

a: \(5x\left(2x+3\right)+6x+9\)

\(=5x\left(2x+3\right)+\left(6x+9\right)\)

\(=5x\left(2x+3\right)+3\left(2x+3\right)\)

\(=\left(2x+3\right)\left(5x+3\right)\)

b: \(3x\left(x+4\right)+48\left(x+4\right)+5\left(x+4\right)\)

\(=\left(x+4\right)\left(3x+48+5\right)\)

=(x+4)(3x+53)

 

29 tháng 11 2023

a: \(a\left(x-y\right)-b\left(y-x\right)+c\left(x-y\right)\)

\(=a\left(x-y\right)+b\left(x-y\right)+c\left(x-y\right)\)

\(=\left(x-y\right)\left(a+b+c\right)\)

b: \(a^m-a^{m+2}\)

\(=a^m-a^m\cdot a^2\)

\(=a^m\left(1-a^2\right)\)

\(=a^m\left(1-a\right)\left(1+a\right)\)

16 tháng 10 2020

(x - 5)2 - 4(x - 3)2 + 2(2x - 1)(x - 5) + (2x - 1)2

= [(x - 5)2 + 2(2x - 1)(x - 5) + (2x - 1)2) - [2(x - 3)]2

= (x - 5 + 2x - 1)2 - (2x - 6)2

= (3x - 6)2 - (2x - 6)2

= (3x - 6 - 2x + 6)(3x - 6 + 2x - 6) = x(5x - 12)

16 tháng 10 2020

( x - 5 )2 - 4( x - 3 )2 + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )2

= [ ( x - 5 )2 + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )2 ] - 22( x - 3 )2

= ( x - 5 + 2x - 1 )2 - ( 2x - 6 )2

= ( 3x - 6 )2 - ( 2x - 6 )2

= ( 3x - 6 - 2x + 6 )( 3x - 6 + 2x - 6 )

= x( 5x - 12 )

2 tháng 8 2016

\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)

\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10+16+8\right)+16\)

\(=\left(x^2+10x+16\right)^2+2.\left(x^2+10x+16\right).4+4^2\)

\(=\left(x^2+10x+16+4\right)^2\)

\(=\left(x^2+10+20\right)^2\)

 

2 tháng 8 2016

\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+16\)
\(=\left(x^2+8x+2x+16\right) \left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\left(1\right)\)
\(\text{Đặt }x^2+10x+\frac{16+24}{2}=t\)
\(\text{hay }x^2+10x+20=t\)
\(\left(1\right)\Rightarrow\left(t-4\right)\left(t+4\right)+16\)
\(=t^2-4^2+16\)
\(=t^2-16+16\)
\(=t^2\)
\(=\left(x^2+10x+20\right)^2\)