Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
a) \(x^4-x^2y^2=x^2(x^2-y^2)=x^2(x-y)(x+y)\)
b) \(6x^2y+3xy+9xy^2=3xy(2x+1+3y)\)
c) \(4x^2-20xy+25y^2=(2x)^2-2.2x.5y+(5y)^2\)
\(=(2x-5y)^2\)
d) \(x^3+6x^2+12x+8\)
\(=x^3+3x^2.2+3x.2^2+2^3=(x+2)^3\)
e) \(64x^3+y^3=(4x)^3+y^3\)
\(=(4x+y)(16x^2-4xy+y^2)\)
f) \(\frac{125}{216}a^3-8b^3=(\frac{5}{6}a)^3-(2b)^3\)
\(=(\frac{5}{6}a-2b)[(\frac{5}{6}a)^2+\frac{5}{6}a.2b+(2b)^2]\)
\(=(\frac{5}{6}a-2b)(\frac{25a^2}{36}+\frac{5}{3}ab+4b^2)\)
g) \(5xy-5x+10y-10\)
\(=5x(y-1)+10(y-1)\)
\(=(5x+10)(y-1)=5(x+2)(y-1)\)
h) \(x^2+2xy-z^2+y^2\)
\(=(x^2+2xy+y^2)-z^2=(x+y)^2-z^2\)
\(=(x+y-z)(x+y+z)\)
\(e,-5x+x^2-14\)
\(=x^2+2x-7x-14\)
\(=x\left(x+2\right)-7\left(x+2\right)\)
\(=\left(x+2\right)\left(x-7\right)\)
\(f,x^3+8+6x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+2x+4\right)+6x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+8x+4\right)\)
\(g,15x^2-7xy-2y^2\)
\(=15x^2+3xy-10xy-2y^2\)
\(=3\left(5x+y\right)-2y\left(5x+y\right)\)
\(=\left(5x+y\right)\left(3-2y\right)\)
\(h,3x^2-16x+5\)
\(=3x^2-x-15x+5\)
\(=x\left(3x-1\right)+5\left(3x-1\right)\)
\(=\left(3x-1\right)\left(x+5\right)\)
\(a,x^3+2x^2y+xy^2=x\left(x^2+2xy+y^2\right)\)
\(=x\left(x+y\right)^2\)
\(b,4x^2-9y^2+4x-6y\)
\(=4x^2+4x+1-\left(9y^2+6y+1\right)\)
\(=\left(2x+1\right)^2-\left(3y+1\right)^2\)
\(=\left(2x-3y\right)\left(2x+3y+2\right)\)
\(c,-x^2+5x+2xy-5y-y^2\)
\(=-\left(x^2-2xy+y^2\right)+5\left(x-y\right)\)
\(=-\left(x-y\right)^2+5\left(x-y\right)\)
\(=\left(x-y\right)\left(y-x+5\right)\)
\(d,x^2+4x-12\)
\(=x^2-2x+6x-12\)
\(=x\left(x-2\right)+6\left(x-2\right)\)
\(=\left(x-2\right)\left(x+6\right)\)
Câu 2 nha
\(a,x^4+2x^3+x^2\)
\(=x^2\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)^2\)
\(c,x^2-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
\(1\hept{\begin{cases}6x^2-8x+3x-4\\2x\left(3x-4\right)+\left(3x-4\right)\\\left(3x-4\right)\left(2x+1\right)\end{cases}}\)
\(2\hept{\begin{cases}7x^2-7xy-5x+5y+6xy\\7x\left(x-y\right)-5\left(x-y\right)+\frac{6xy\left(x-y\right)}{\left(x-y\right)}\\\left(x-y\right)\left(7x-5+\frac{6xy}{\left(x-y\right)}\right)\end{cases}}\)
\(3\hept{\begin{cases}5x\left(x-y\right)-15\left(x-y\right)\\\left(x-y\right)\left(5x-15\right)\end{cases}}\)
\(4,,2x^2+x=x\left(2x+1\right)\)
\(5\hept{\begin{cases}x^3-4x-3x^2+12\\x\left(x^2-4\right)-3\left(x^2-4\right)\\\left(x+2\right)\left(x-2\right)\left(x-3\right)\end{cases}}\)
\(6\hept{\begin{cases}2x+2y+x^2-y^2\\2\left(x+y\right)+\left(x+y\right)\left(x-y\right)\\\left(x+y\right)\left(2+x-y\right)\end{cases}}\)
\(7\hept{\begin{cases}\left(x^2y-2xy\right)-\left(xy-2y\right)+\left(xy-y\right)\\xy\left(x-2\right)-y\left(x-2\right)+y\left(x-1\right)\\y\left(X-2\right)\left(x-1\right)+y\left(x-1\right)\end{cases}}\Leftrightarrow y\left(x-1\right)\left(x-2+1\right)\)
\(8\hept{\begin{cases}x\left(2-y\right)+z\left(2-y\right)\\\left(2-y\right)\left(x+1\right)\end{cases}}\)
a, \(=12x^5+9x^3y^2-6x^2y^3-20x^4y-15x^2y^3-10xy^4-24x^3y^2-18xy^4+12y^5\)
(tự rút gọn cái :P)
b, \(8x^3+4x^2y-2xy^2-y^3\)
\(=4x^2\left(2x+y\right)-y^2\left(2x+y\right)=\left(2x+y\right)^2\left(2x-y\right)\)
\(4x^2y^2-4x^2-4xy-y^2=4x^2y^2-\left(2x+y\right)^2\)
\(=\left(2x+y+2xy\right)\left(2xy-2x+y\right)\)
Mấy cái còn lại nhân tung ra là được mà :))))
k) \(x^3-x+3x^2+3xt^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)
h) \(a^3-a^2x-ay+xy\)
\(=a^2\left(a-x\right)-y\left(a-x\right)\)
\(=\left(a^2-y\right)\left(a-x\right)\)
Bài nhiều quá... nhìn mik nổi gai ốc lun...oh my god sao mà nhiều vậy nè .
Mik định giải giúp bạn nhưng bây h mik hoảng quá ... nhiều vậy chắc mik chết mất... ToT ... >.< =)))
mk viết đáp án, ko biết biến đổi ib mk
a) \(x^3+3x^2y-9xy^2+5y^3=\left(x+5y\right)\left(x-y\right)^2\)
b) \(x^4+x^3+6x^2+5x+5=\left(x^2+5\right)\left(x^2+x+1\right)\)
c) \(x^4-2x^3-12x^2+12x+36=\left(x^2-6\right)\left(x^2-2x-6\right)\)
d) \(x^8y^8+x^4y^4+1=\left(x^2y^2-xy+1\right)\left(x^2y^2+xy+1\right)\left(x^4y^4-x^2y^2+1\right)\)
a) \(x^4-x^2y^2\) \(=x^2(x^2-y^2)\)
\(=x^2\left(x-y\right)\left(x+y\right)\)
b) \(6x^2y+3xy+9xy^2=3xy\left(2x+1+3y\right)\)
c) \(4x^2-20xy+25y^2=\left(2x\right)^2-2.2x.5y+\left(5y\right)^2\)
\(=\left(2x-5y\right)^2\)
d) \(x^3+6x^2+12x+8=x^3+3x^2.2+3x.2^2+2^3\)
\(=\left(x+2\right)^3\)
e) \(64x^3+y^3=\left(4x\right)^3+y^3\)
\(=\left(4x+y\right)\left(16x^2-4xy+y^2\right)\)
f) \(\dfrac{125}{216}a^3-8b^3=\left(\dfrac{5}{6}a\right)^3-\left(2b\right)^3\)
\(=\left(\dfrac{5}{6}a-2b\right)\left[\left(\dfrac{5}{6}a^2+\dfrac{5}{6}a.2b-\left(2b\right)^2\right)\right]\)
\(=\left(\dfrac{5}{6}a-2b\right)\left(\dfrac{25}{36}a^2+\dfrac{5}{6}ab-4b^2\right)\)
g) \(5xy-5x+10y-10=\left(5xy-5x\right)+\left(10y-10\right)\)
\(=5x\left(y-1\right)+10y\left(y-1\right)\)
\(=\left(y-1\right)\left(5x+10y\right)\)
h) \(x^2+2xy-z^2+y^2=\left(x^2+2xy+y^2\right)-z^2\)
\(=\left(x+y\right)^2-z^2\)
\(=\left(x+y-z\right)\left(x+y+z\right)\)
Hình như đề câu h bạn viết sai hay sao á,Mình nghĩ chắc là \(x^2\) thôi,vì \(x^3\) mình tính không ra :v Mong bạn xem lại đề có đúng như mình ns không
Học Tốt !!
câu nào vậy