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6 tháng 2 2016

mik chỉ biết (x+2)(x+3)(x+4)(x+5)-24=(x+6)(x+1)(x2+7x+16 bằng cách đặt ẩn phụ thui còn lại ko biết sorry nha

 

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 \(12x^2+11x-1=a\)

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

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

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

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

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

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

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:

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

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

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

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

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

AH
Akai Haruma
Giáo viên
18 tháng 12 2023

Bài 1:

a. $2x^3+3x^2-2x=2x(x^2+3x-2)=2x[(x^2-2x)+(x-2)]$

$=2x[x(x-2)+(x-2)]=2x(x-2)(x+1)$

b.

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

$=[(x+1)(x+4)][(x+2)(x+3)]-24$

$=(x^2+5x+4)(x^2+5x+6)-24$

$=a(a+2)-24$ (đặt $x^2+5x+4=a$)

$=a^2+2a-24=(a^2-4a)+(6a-24)$

$=a(a-4)+6(a-4)=(a-4)(a+6)=(x^2+5x)(x^2+5x+10)$

$=x(x+5)(x^2+5x+10)$

AH
Akai Haruma
Giáo viên
18 tháng 12 2023

Bài 2:

a. ĐKXĐ: $x\neq 3; 4$

\(A=\frac{2x+1-(x+3)(x-3)+(2x-1)(x-4)}{(x-3)(x-4)}\\ =\frac{2x+1-(x^2-9)+(2x^2-9x+4)}{(x-3)(x-4)}\\ =\frac{x^2-7x+14}{(x-3)(x-4)}\)

b. $x^2+20=9x$

$\Leftrightarrow x^2-9x+20=0$

$\Leftrightarrow (x-4)(x-5)=0$

$\Rightarrow x=5$ (do $x\neq 4$)

Khi đó: $A=\frac{5^2-7.5+14}{(5-4)(5-3)}=2$

2 tháng 11 2016

a)x4+2x3+5x2+4x-12

=(x4+2x3+x2)+(4x2+4x)-12

=(x2+x)2+4(x2+x)-12

Đặt t=x2+x

=t2+4t-12=(t-2)(t+6)

=(x2+x-2)(x2+x+6)

=(x-1)(x+2)(x2+x+6)

b)(x+1)(x+2)(x+3)(x+4)+1

=(x2+5x+4)(x2+5x+6)+1

Đặt x2+5x+4=t

t(t+2)+1=t2+2t+1

=(t+1)2=(x2+5x+4+1)2

=(x2+5x+5)2

c)(x+1)(x+3)(x+5)(x+7)+15

=(x2+8x+7)(x2+8x+15)+15

Đặt t=x2+8x+7

t(t+8)+15=(t+3)(t+5)

=(x2+8x+7+3)(x2+8x+7+5)

=(x2+8x+10)(x+2)(x+6)

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

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

Đặt t=x2+5x+4 

t(t+2)-24=(t-4)(t+6)

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

=x(x+5)(x2+5x+10)

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?

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)

 

(x+2).(x+3).(x+4).(x+5)−24

=(x2+7x+10).(x2+7x+12)−24

=(x2+7x+10).(x2+7x+10+2)−24

Đặt x2+7x+10=t, ta có

t.(t+2)−24

=t2+2t−24

=t2+2t+1−25

=(t−1)2−25

=(t−1−5)(t−1+5)

=(t−6)(t+4)

=(x2+7x+10−6)(x2+7x+10+4)

(x2+7x+4)(x2+7x+14)

P/s tham khảo nha

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

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

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

Đặt \(x^2+7x+10=t\), ta có

\(t.\left(t+2\right)-24\)

\(\Leftrightarrow t^2+2t-24\)

\(\Leftrightarrow t^2+2t+1-25\)

\(\Leftrightarrow\left(t-1\right)^2-25\)

\(\Leftrightarrow\left(t-1-5\right)\left(t-1+5\right)\)

\(\Leftrightarrow\left(t-6\right)\left(t+4\right)\)

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

\(\Leftrightarrow\left(x^2+7x+4\right)\left(x^2+7x+14\right)\)

P/s tham khảo nha

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 )