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1 tháng 8 2019

#)Giải :

\(4x^4+4x^3+5x^2+2x+1\)

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

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

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

1 tháng 8 2019

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

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

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

~ Rất vui vì giúp đc bn ~

20 tháng 10 2021

giúp tui 

 

20 tháng 10 2021

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

a: =4(x-2)(x+1)+4(x-2)^2+(x+1)^2

=(2x-4)^2+2*(2x-4)(x+1)+(x+1)^2

=(2x-4+x+1)^2=(3x-3)^2=9(x-1)^2

b: =x^7(x^2-1)-x^5(x+1)+x^3(x+1)+(x^2-1)

=(x+1)[x^7(x-1)-x^5+x^3+x-1]

=(x+1)[x^7(x-1)-x^3(x-1)(x+1)+(x-1)]

=(x+1)(x-1)(x^7-x^4-x^3+1)

=(x+1)(x-1)(x^3-1)(x^4-1)

=(x+1)(x-1)^2*(x^2+x+1)(x^2+1)(x-1)(x+1)

=(x+1)^2*(x-1)^3*(x^2+1)(x^2+x+1)

 

16 tháng 9 2023

Mình bổ sung nhé:

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

\(=\left(x+1\right)\left[x^2\left(x^2+x+1\right)-\left(x^3-1\right)\right]\)

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

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

=x^3(x^2+x+1)+(x^2+x+1)

=(x^2+x+1)(x^3+1)

=(x^2+x+1)(x+1)(x^2-x+1)

15 tháng 9 2015

  (x+2).(x+3)(x+4)(x+5)-24

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

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

=(x2+7x+11-1)(x2+7x+11+1)-24

Đặt x2+7x+11=a thì

=(a-1)(a+1)-24

=a2-1-24=a2-25=a2-52

=(a+5)(a-5)

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

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 :>>

22 tháng 7 2015

 

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

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

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

Đặt t=x2+7x+10 ta được:

(t+2)t-24

=t2+2t-24

=t2+4t-6t-24

=t.(t+4)-6.(t+4)

=(t+4)(t-6)

thay t=x2+7t+10 ta được:

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

Vậy  ( x+3) (x+2) (x+4) (x+5) -24=(x2+7x+14)(x2+7x+4)

 

22 tháng 7 2015

song la phai nhuong nhin

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

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

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

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

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

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 \(x^2+7x+11=y\)

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

24 tháng 9 2021

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