Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1) \(3\left(x+4\right)-x^2-4x=3\left(x+4\right)-x\left(x+4\right)=\left(x+4\right)\left(3-x\right)\)
2) \(5x^2-5y^2-10x+10y=5\left(x^2-y^2\right)-10\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-10\left(x-y\right)=\left(x-y\right)\left(5x+5y-10\right)\)
3) \(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)
4) \(ax-bx-a^2+2ab-b^2=x\left(a-b\right)-\left(a^2-2ab+b^2\right)\)
\(=x\left(a-b\right)-\left(a-b\right)^2=\left(a-b\right)\left(x-a+b\right)\)
5) \(x^3-x^2-x+1=x^2\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)\left(x-1\right)\left(x+1\right)=\left(x-1\right)^2\left(x+1\right)\)
6) \(x^2+4x-y^2+4=x^2+4x+4-y^2=\left(x+2\right)^2-y^2\)
\(=\left(x+2-y\right)\left(x+2+y\right)\)
\(a,\)\(x^3-13x-12\)
\(=x^3-x-12x-12\)
\(=x\left(x^2-1\right)-12\left(x+1\right)\)
\(=x\left(x-1\right)\left(x+1\right)-12\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-12\right)\)
\(=\left(x+1\right)\left(x^2-4x+3x-12\right)\)
\(=\left(x+1\right)\left[x\left(x-4\right)+3\left(x+4\right)\right]\)
\(=\left(x+1\right)\left(x-4\right)\left(x+3\right)\)
a) \(x^3-13x-12\)
\(=x^3+x^2-x^2-x-12x-12\)
\(=x^2\left(x+1\right)-x\left(x+1\right)-12\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-12\right)\)
\(=\left(x+1\right)\left(x^2-4x+3x-12\right)\)
\(=\left(x+1\right)\left[x\left(x-4\right)+3\left(x-4\right)\right]\)
\(=\left(x+1\right)\left(x-4\right)\left(x+3\right)\)
b) \(2x^4+3x^3-9x^2-3x+2\)câu này hình như sai đề rồi, bạn xem lại nhen
c) \(x^4-3x^3-6x^2+3x+1\)câu này cx thế, bạn xem lại nha
a, = [(x-2).(x+1)]^2+(x-2)^2
= (x-2)^2.(x+1)^2+(x-2)^2
= (x-2)^2.[(x+1)^2+1]
= (x-2)^2.(x^2+2x+2)
Tk mk nha
b) \(6x^5+15x^4+20x^3+15x^2+6x+1\)
\(=6x^5+3x^4+12x^4+6x^3+14x^3+7x^2+8x^2+4x+2x+1\)
\(=\left(2x+1\right)\left(3x^4+6x^3+7x^2+4x+1\right)\)
\(=\left(2x+1\right)\left(3x^4+3x^3+3x^2+3x^3+3x^2+3x+x^2+x+1\right)\)
\(=\left(2x+1\right)\left(x^2+x+1\right)\left(3x^2+3x+1\right)\)
f)\(\left(x-y\right)^2-4=\left(x-y-4\right)\left(x-y+4\right)\)
h) \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)
i)\(10x-x^2-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)
k)\(4x^2-12xy+9y^2=\left(2x\right)^2-2.2x.3y+\left(3y\right)^2=\left(2x-3y\right)^2\)
mấy bài này cơ bản mà, mở sgk toán 8 ra có các dạng đấy, đăng cũng đăng ít chứ, đăng nhiều quá
a)\(6x^3-9y^2=3\left(2x^3-3y^2\right)\)
b)\(4x^2y-8xy^2+18x^2y^2=2xy\left(2x-4y+9xy\right)\)
c)\(18x^2y-12x^3=6x^2\left(3y-2x\right)\)
d) \(5x\left(x-1\right)-3y\left(x-1\right)=\left(x-1\right)\left(5x-3y\right)\)
e)\(\left(x+1\right)^2-3\left(x+1\right)=\left(x+1\right)\left(x+1-3\right)=\left(x+1\right)\left(x-2\right)\)
g)\(\left(4x^2-4x+4\right)-\left(x+1\right)^2=\left(4x^2-4x+4\right)-\left(x^2+2x+1\right)\)
\(=4x^2-4x+4-x^2-2x-1\)\(=3x^2-6x+3\)\(=3\left(x^2-2x+1\right)\)
\(=3\left(x-1\right)^2\)
Bài 1:
a, \(x^2-x-12\)
\(=x^2-4x+3x-12=\left(x^2-4x\right)+\left(3x-12\right)\)
\(=x.\left(x-4\right)+3.\left(x-4\right)=\left(x-4\right).\left(x+3\right)\)
b, \(x^2+8x+15\)
\(=x^2+3x+5x+15=\left(x^2+3x\right)+\left(5x+15\right)\)
\(=x.\left(x+3\right)+5.\left(x+3\right)=\left(x+3\right).\left(x+5\right)\)
c, \(x^{16}+x^8-2\)
\(=x^{16}-x^8+2x^8-2=\left(x^{16}-x^8\right)+\left(2x^8-2\right)\)
\(=x^8.\left(x^8-1\right)+2.\left(x^8-1\right)=\left(x^8-1\right)\left(x^8+2\right)\)
d, \(x^2+7x+12\)
\(=x^2+3x+4x+12=\left(x^2+3x\right)+\left(4x+12\right)\)
\(=x.\left(x+3\right)+4.\left(x+3\right)=\left(x+3\right).\left(x+4\right)\)
Chúc bạn học tốt!!!
\(x^5+y^5-\left(x+y\right)^5\)
\(=x^5+y^5-\left(x^5+5x^4y+10x^3y^2+10x^2y^3+8xy^4+y^5\right)\)
\(=-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)
\(=-5xy\left[\left(x+y\right)\left(x^2-xy+y^2\right)+2xy\left(x+y\right)\right]\)
\(=-5xy\left(x+y\right)\left(x^2+xy+y^2\right)\)
a. \(=x^3+2^3+1^3-x^3\)
\(=\left(x^3-x^3\right)+8+1\)
\(=0+8+1\)
\(=9\)
Bài 1 :
a) ( x + 2 )( x2 - 2x + 4 ) + (1 - x)(1+x+ + x2 )
= ( x3 - 8 ) + ( 1 - x3 )
= x3 - 8 + 1 - x3
= 7
b) 7x( 4x - 2) - ( x - 3)( x+1 ) + 16x
= 28x2 - 14x - x2 - x + 3x + 3 + 16x
= 27x2 + 3
a, Ta có : \(\frac{x+1}{2}+\frac{x-2}{4}=1-\frac{2\left(x-1\right)}{3}\)
=> \(\frac{6\left(x+1\right)}{12}+\frac{3\left(x-2\right)}{12}=\frac{12}{12}-\frac{8\left(x-1\right)}{12}\)
=> \(6\left(x+1\right)+3\left(x-2\right)=12-8\left(x-1\right)\)
=> \(6x+6+3x-6=12-8x+8\)
=> \(17x=20\)
=> \(x=\frac{20}{17}\)
b, Ta có : \(\frac{5x-1}{6}+x=\frac{6-x}{4}\)
=> \(\frac{5x-1+6x}{6}=\frac{6-x}{4}\)
=> \(4\left(11x-1\right)=6\left(6-x\right)\)
=> \(44x-4-36+6x=0\)
=> \(\)\(50x=40\)
=> \(x=\frac{4}{5}\)
c, Ta có : \(\frac{5\left(1-2x\right)}{3}+\frac{x}{2}=\frac{3\left(x-5\right)}{4}-2\)
=> \(\frac{20\left(1-2x\right)}{12}+\frac{6x}{12}=\frac{9\left(x-5\right)}{12}-\frac{24}{12}\)
=> \(20\left(1-2x\right)+6x=9\left(x-5\right)-24\)
=> \(20-40x+6x-9x+45+24=0\)
=> \(43x=89\)
=> \(x=\frac{89}{43}\)
Trong app này có cả bộ đề thi + thi thử bạn thử xem nha! https://giaingay.com.vn/downapp.html
a: ĐK của A là x<>-3; x<>2
ĐKXĐ của B là x<>3
DKXĐ của C là x<>0; x<>4/3
ĐKXĐ của D là x<>-2
ĐKXĐ của E là x<>2; x<>-2
ĐKXĐ của F là x<>2
b,c:
\(A=\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{2}{x-2}\)
Để A=0 thì 2=0(loại)
\(B=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x-3\right)}=\dfrac{x+3}{x-3}\)
Để B=0 thì x+3=0
=>x=-3
\(C=\dfrac{\left(3x-4\right)\left(3x+4\right)}{x\left(3x-4\right)}=\dfrac{3x+4}{x}\)
Để C=0 thì 3x+4=0
=>x=-4/3
\(D=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}=\dfrac{x+2}{2}\)
Để D=0 thì x+2=0
=>x=-2(loại)
\(E=\dfrac{x\left(2-x\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{-x}{x+2}\)
Để E=0 thì x=0
\(F=\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}\)
Để F=0 thì 3=0(loại)
/ (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 ....