\(x^4-y^4\)

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

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

\(\left(a-b\right)^3-\left(a-b\right)^3\)

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

\(\left(a^2+2ab+b^2\right)+\left(a+b\right)^3\)

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

\(=\left(a+b\right)^2\left(a+b+1\right)\)

                                                          ......giải ....

  a. \(\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)                                            

 b ...ko cần làm .. =0

c.. =(a+b)^2 +(a+b)^3=(a+b)[ (a+b)+ (a+b)^2  ]

 ... check mk đó ..  The end•••

Bài 3:

a: =>6x(x^2-4)=0

=>x(x-2)(x+2)=0

hay \(x\in\left\{0;2;-2\right\}\)

b: \(\Leftrightarrow9\left(x^2-1\right)-9x^2+6x-1=2\)

=>9x^2-9-9x^2+6x-1=2

=>6x-10=2

=>6x=12

=>x=2

tao chịu ko hiểu mới học lớp 6 nhé very sorrrrrrrrrrrrrryyyyyyyyyyyyyyyyyyyyyy

k nha

ai km ình k lai có 21 nick đó

a: \(=5xy\left(x^2-2xy+y^2\right)=5xy\left(x-y\right)^2\)

b: \(=2x^2+10x-3x-15\)

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

thanks

/ (4x−2)(10x+4)(5x+7)(2x+1)+17=0(4x−2)(10x+4)(5x+7)(2x+1)+17=0

⇔(4x−2)(5x+7)(10x+4)(2x+1)+17=0⇔(4x−2)(5x+7)(10x+4)(2x+1)+17=0

⇔(20x2+18x−14)(20x2+18x+4)+17=0⇔(20x2+18x−14)(20x2+18x+4)+17=0

Đặt t= 20x2+18x+4(t≥0)20x2+18x+4(t≥0) ta có:

(t-18).t +17=0

⇔t2−18t+17=0⇔t2−18t+17=0

⇔(t−17)(t−1)=0⇔(t−17)(t−1)=0

⇔[t=17(tm)t=1(tm)⇔[t=17(tm)t=1(tm) ⇔[20x2+18x+4=1720x2+18x+4=1⇔[20x2+18x−13=020x2+18+3=0⇔[20x2+18x+4=1720x2+18x+4=1⇔[20x2+18x−13=020x2+18+3=0

⇔[(20x+9−341−−−√)(20x+9+341−−−√)=0(20x+9−21−−√)(20x+9+21−−√)=0⇔[(20x+9−341)(20x+9+341)=0(20x+9−21)(20x+9+21)=0

⇔⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢x=−9+341−−−√20x=−9−341−−−√20x=−9+21−−√20x=−9−21−−√20

\(a,\)\(\left(4x-2\right)\left(10x+4\right)\left(5x+7\right)\left(2x+1\right)+17\)

\(=\left(4x-2\right)\left(5x+7\right)\left(10x+4\right)\left(2x+1\right)+17\)

\(=\left(20x^2+18x-5\right)\left(20x^2+18x+4\right)+17\)

Đặt ....

\(-3xy^2+x^2y^2-5x^2y\)

\(=-xy\left(3y+xy-5x\right)\)

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

\(=3y^3+6y+3+xy-x\)

Xem lại nhé ko phân tích được

\(12xy^2-12xy+3x\)

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

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

\(10x^2\left(x+y\right)-5\left(2x+2y\right)y^2\)

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

\(=10\left(x+y\right)\left(x-y\right)\left(x+y\right)\)

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