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3 tháng 11 2019

Ta có:

(x + 2)(x + 3)(x + 4)(x + 5) - 24

= [(x + 2)(x + 5)][(x + 3)(x + 4)] - 24

= (x2 + 5x + 2x + 10)(x2 + 4x + 3x + 12) - 24

= (x2 + 7x + 10)(x2 + 7x + 12) - 24

Đặt x2 + 7x + 10 = k 

=> k(k + 2) - 24 = k2 + 2k - 24 = k2 + 6x - 4x - 24 

                            = k(k + 6)  - 4(k  + 6)

                          = (k - 4)(k + 6)

=> (x + 2)(x + 3)(x + 4)(x + 5) - 24

= (x2 + 7x + 10 - 4)(x2 + 7x + 10 + 6)

= (x2 + 7x + 6)(x2 + 7x + 16)

3 tháng 11 2019

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)(1)

Đặt \(x^2+7x+11=t\)thay vào (1) ta được:
\(\left(t-1\right)\left(t+1\right)-24\)

\(=t^2-1-24\)

\(=t^2-25\)

\(=\left(t-5\right)\left(t+5\right)\)Thay \(t=x^2+7x+11\)ta được:

\(\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x^2+x+6x+6\right)\left(x^2+7x+16\right)\)

\(=\left[x\left(x+1\right)+6\left(x+1\right)\right]\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

25 tháng 2 2017

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Let \(t=x^2+7x+10\) we have:

\(=t\left(t+2\right)-24=t^2+2t-24\)

\(=\left(t-4\right)\left(t+6\right)=\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

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?

27 tháng 10 2018

Đặt \(A=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(A=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(x^2+7x+10=y\)

\(\Rightarrow\)\(A=y.\left(y+2\right)-24\)

\(A=y^2+2y+1-25\)

\(A=\left(y+1\right)^2-5^2\)

\(A=\left(y+1-5\right)\left(y+1+5\right)\)

\(A=\left(y-4\right)\left(y+6\right)\)

\(\Rightarrow A=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(A=\left[\left(x^2+x\right)+\left(6x+6\right)\right].\left(x^2+7x+16\right)\)

\(A=\left[x.\left(x+1\right)+6.\left(x+1\right)\right].\left(x^2+7x+16\right)\)

\(A=\left(x+1\right).\left(x+6\right).\left(x^2+7x+16\right)\)

27 tháng 10 2018

Đặt \(B=\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)

\(B=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)

Đặt \(12x^2+11x-1=a\)

\(\Rightarrow B=a.\left(a+3\right)-4\)

\(B=a^2+3a-4\)

\(B=\left(a^2-a\right)+\left(4a-4\right)\)

\(B=a.\left(a-1\right)+4.\left(a-1\right)\)

\(B=\left(a-1\right)\left(a+4\right)\)

\(\Rightarrow B=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)

11 tháng 9 2018

(x + 1)(x + 2)(x + 3)(x + 4) - 24

= x4 + 10x3 + 35x2 + 50x + 24 - 24

= x4 + 10x3 + 35x2 + 50x

11 tháng 9 2018

( x + 1 ). ( x + 2 ) ( x + 3 ) ( x + 4 ) - 24

= ( x2 + 5x + 4 ) .( x2 + 5x + 6 ) - 24

Đặt t = x2 + 5x + 5 

=> ( t - 1 ). ( t + 1 ) - 24

= t2 - 1 - 24 

= t2 - 25

= ( t - 5 ). ( t + 5 )

= ( x2 + 5x + 5 - 5 ) . ( x2 + 5x + 5 + 5 )

= ( x2 + 5x ) . ( x2 + 5x + 10 )

= x. ( x + 5 ) . ( x2 + 5x + 10 )

10 tháng 7 2021

`(x+3)^4+(x+5)^4-2`

`={[(x+3)^2]^2-1^2}+{[(x+5)^2]^2 -1^2}`

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

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

`=(x+2)(x+4)[(x+3)^2+1]+(x+4)(x+6)[(x+5)^2+1]`

`=(x+4){(x+2)[(x+3)^2+1]+(x+6)[(x+5)^2+1]}`

`=(x+4)(2x^3+24x^2+108x+176)`

10 tháng 7 2021

Bạn gì ơi hình như phải ra \(2\left(t+4\right)^2\left(x^2+8x+22\right)\)chứ nhỉ???

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)

 

17 tháng 10 2023

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)-24\)

\(=\left(x^2+7x+11\right)^2-25\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

17 tháng 10 2023

\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)

Đặt \(y=x^2+7x+10\), khi đó biểu thức trở thành:

\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(x^2+7x+10+1-5)(x^2+7x+10+1+5)\\=(x^2+7x+6)(x^2+7x+16)\\=(x^2+x+6x+6)(x^2+7x+16)\\=[x(x+1)+6(x+1)](x^2+7x+16)\\=(x+1)(x+6)(x^2+7x+16)\\Toru\)

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 )