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12 tháng 10 2021

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

28 tháng 11 2021
Lol .ngudoots
24 tháng 9 2021

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

6 tháng 11 2021

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

25 tháng 10 2021

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\\ =\left(x^2+7x+11\right)^2-1-24\\ =\left(x^2+7x+11\right)^2-25\\ =\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\\ =\left(x^2+7x+6\right)\left(x^2+7x+16\right)\\ =\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

25 tháng 10 2021

chữ đẹp thế :>>

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

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

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)

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

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

9 tháng 10 2021

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

Đặt \(x^2+7x+11=y\)

\(\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\\ =\left(y+1\right)\left(y-1\right)-24\\ =y^2-1-24\\ =y^2-25\\ =\left(y-5\right)\left(y+5\right)\\ =\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\\ =\left(x^2+7x+6\right)\left(x^2+7x+16\right)\\ =\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

14 tháng 12 2016

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

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

Đặt \(t=x^2+7x+10\) ta có:

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

\(=t^2-4t+6t-24\)\(=t\left(t-4\right)+6\left(t-4\right)\)

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

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

 

27 tháng 3 2017

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

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

Đặt x^2+7x+10=a

a(a+2)-24

=a^2+2a-24

=(a-4)(a+6)

=(x^2+7x+6)(x^2+7x+16)

=(x+1)(x+6)(x^2+7x+16)

28 tháng 8 2018

A  =   ( x + 2 ) ( x + 3 ) ( x + 4 ) ( x + 5 ) - 48

= ( x2 + 7x + 10 ) ( x2 + 7x + 12 ) - 48

Đặt x2 + 7x + 10 = t 

=>  A = t. ( t + 2 ) - 48

         = t2 + 2t  + 1 - 49

        =  ( t + 1 )2 - 72

        = ( t + 1 - 7 ) ( t + 1 + 7 )

     =   ( t - 6 ) ( t + 8 )

Thay t = x2 + 7x + 10

=> A = ( x2 + 7x + 4 )( x2 + 7x  +  18 ) 

4 tháng 10 2020

Ta có: \(\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)

\(=\left[\left(x-2\right)\left(x-5\right)\right]\cdot\left[\left(x-3\right)\left(x-4\right)\right]+1\)

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

\(=\left[\left(x^2-7x+11\right)-1\right]\cdot\left[\left(x^2-7x+11\right)+1\right]\)

\(=\left(x^2-7x+11\right)^2-1+1\)

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

4 tháng 10 2020

\(\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)   

\(=\left(x-2\right)\left(x-5\right)\left(x-4\right)\left(x-3\right)+1\)   

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

Đặt t = \(x^2-7x\)   

\(t\left(t+2\right)+1\)   

\(=t^2+2t+1\)   

\(=\left(t+1\right)^2\)   

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

30 tháng 8 2015

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

= (x2+7x+10)(x2+7x+12)-24

đặt y=x2+7x+10

ta có biểu thức:

y.(y+2)-24

= y2+2y-24

= y2+6y-4y-24

= y(y+6)-4(y+6)

= (y+6)(y-4)

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

= (x2+7x+16)(x2+7x+6)

30 tháng 8 2015

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

= (x2 + 3x + 2x + 6)(x2 + 5x + 4x + 20) - 24

= (x2 + 5x + 6)(x2 + 9x + 20) - 24

= x4 + 9x3 + 20x2 + 5x3 + 45x2 + 100x + 6x2 + 54x + 120 - 24

= x4 + 14x3 + 71x2 + 100x + 96

18 tháng 10 2015

Đặt A = (x^2+5x+4)(x^2+5x+6)-24 và  x^2+5x+5=a 

Do đó A= (a-1)(a+1)-24

= a^2- 25

= a^2-5^2

=(a-5)(a+5)

= ( x^2+5x+5-5)( x^2+5x+5+5)

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

19 tháng 10 2015

Đặt A = (x^2+5x+4)(x^2+5x+6)-24 và  x^2+5x+5=a 

Do đó A= (a-1)(a+1)-24

= a^2- 25

= a^2-5^2

=(a-5)(a+5)

= ( x^2+5x+5-5)( x^2+5x+5+5)

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