Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a. \(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)\)
\(\Rightarrow x^5+x^4y+x^3y^2+x^2y^3+y^5-yx^4-x^3y^2-x^2y^3-xy^4-y^5=VP\)
\(\Rightarrow dpcm\)
b. \(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\)
\(\Rightarrow x^5-x^4y+x^3y^2-x^2y^3+xy^4+yx^4-x^3y^2-xy^4+y^5=VP\)
\(\Rightarrow dpcm\)
c.d làm tương tự
Bài làm
a) Biến đổi vế trái, ta được:
\(VT=\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)\)
\(=x^5+x^4y+x^3y^2+x^2y^3+xy^4-x^4y-x^3y^2-x^2y^3-xy^4-y^5\)
\(=\left(x^5-y^5\right)+\left(x^4y-x^4y\right)+\left(x^3y^2-x^3y^2\right)+\left(x^2y^3-x^2y^3\right)+\left(xy^4-xy^4\right)\)
\(=x^5-y^5=VP\left(đpcm\right)\)
b) Biến đổi vế trái, ta có:
\(VT=\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\)
\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5\)
\(=\left(x^5+y^5\right)+\left(-x^4y+x^4y\right)+\left(x^3y^2-x^3y^2\right)+\left(-x^2y^3+x^2y^3\right)+\left(xy^4-xy^4\right)\)
\(=x^5+y^5=VP\left(đpcm\right)\)
c) Biến đổi vế trái, ta có:
\(VT=\left(a+b\right)\left(a^3-a^2b+ab^2-b^3\right)\)
\(=a^4-a^3b+a^2b^2-ab^3+a^3b-a^2b^2+ab^3-b^4\)
\(=\left(a^4-b^4\right)+\left(-a^3b+a^3b\right)+\left(a^2b^2-a^2b^2\right)+\left(-ab^3+ab^3\right)\)
\(=a^4-b^4=VP\left(đpcm\right)\)
d) Đây là hằng đẳng thức, như vế phải hình như bạn viết bị sai, mik sửa là vế phải nha.
\(\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3\)
Biến đổi vế trái, ta có:
\(VT=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=a^3-a^2b+ab^2+a^2b-ab^2+b^3\)
\(=\left(a^3+b^3\right)+\left(-a^2b+a^2b\right)+\left(ab^2-ab^2\right)\)
\(=a^3+b^3=VP\left(đpcm\right)\)
Ta có:
\(\left(a+b+c\right)^2=\left(a+b\right)^2+2\left(a+b\right)c+c^2\)
\(=a^2+2ab+b^2+2ac+2bc+c^2\)
\(=a^2+b^2+c^2+2\left(ab+bc+ca\right)\) \(\Rightarrowđpcm\)
Bạn làm bài kiểm tra hả sao nhiều bài tek. Mk làm mất khá nhiều tg luôn đó 
Có một số câu thì mình không làm được. Mong bạn thông cảm!!!
\(a,3x^3y^3-15x^2y^2=3x^2y^2\left(xy-5\right)\)
\(b,5x^3y^2-25x^2y^3+40xy^4\)
\(=5xy^2\left(x^2-5xy+8y^2\right)\)
\(c,-4x^3y^2+6x^2y^2-8x^4y^3\)
\(=-2x^2y^2\left(2x-3+4x^2y\right)\)
\(d,a^3x^2y-\frac{5}{2}a^3x^4+\frac{2}{3}a^4x^2y\)
\(=a^3x^2\left(y-\frac{5}{2}x^2+\frac{2}{3}ay\right)\)
\(e,a\left(x+1\right)-b\left(x+1\right)=\left(x+1\right)\left(a-b\right)\)
\(f,2x\left(x-5y\right)+8y\left(5y-x\right)\)
\(=2x\left(x-5y\right)-8y\left(x-5y\right)=\left(x-5y\right)\left(2x-8y\right)\)
\(g,a\left(x^2+1\right)+b\left(-1-x^2\right)-c\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(a-b-c\right)\)
\(h,9\left(x-y\right)^2-27\left(y-x\right)^3\)
\(=9\left(x-y\right)^2+27\left(x-y\right)^3\)
\(=9\left(x-y\right)^2\left(1+3x-3y\right)\)
a,3x3y3−15x2y2=3x2y2(xy−5)a,3x3y3−15x2y2=3x2y2(xy−5)
b,5x3y2−25x2y3+40xy4b,5x3y2−25x2y3+40xy4
=5xy2(x2−5xy+8y2)=5xy2(x2−5xy+8y2)
c,−4x3y2+6x2y2−8x4y3c,−4x3y2+6x2y2−8x4y3
=−2x2y2(2x−3+4x2y)=−2x2y2(2x−3+4x2y)
d,a3x2y−52a3x4+23a4x2yd,a3x2y−52a3x4+23a4x2y
=a3x2(y−52x2+23ay)=a3x2(y−52x2+23ay)
e,a(x+1)−b(x+1)=(x+1)(a−b)e,a(x+1)−b(x+1)=(x+1)(a−b)
f,2x(x−5y)+8y(5y−x)f,2x(x−5y)+8y(5y−x)
=2x(x−5y)−8y(x−5y)=(x−5y)(2x−8y)=2x(x−5y)−8y(x−5y)=(x−5y)(2x−8y)
g,a(x2+1)+b(−1−x2)−c(x2+1)g,a(x2+1)+b(−1−x2)−c(x2+1)
=(x2+1)(a−b−c)=(x2+1)(a−b−c)
h,9(x−y)2−27(y−x)3h,9(x−y)2−27(y−x)3
=9(x−y)2+27(x−y)3
Bài 2:
a)x3+2x2+x
=x(x2+2x+12)
=x(x+1)2
b)xy+y2-x-y
=(xy-x)+(y2-y)
=x(y-1)+y(y-1)
=(y-1)(x+y)
Bai 1:
a) = 2x^3 + 14x^2 - 2x^3 - x^2 + 9x - 12
= 13x^2 + 9x - 12
b) = x^2 - 2x + 1 - x^2 + 4x - 4x + 16
= -2x + 17
Bài làm :
Bài 1 :
\(a,-2x^3y.\left(2x^2-3y+5y^2\right)\)
\(=-4x^5y+6x^3y^2-10x^3y^3\)
\(b,\left(x+1\right)\left(x^2-x+1\right)\)
\(=x^3-x^2+x+x^2-x+1\)
\(=x^3+1\)
\(c,\left(2x-1\right).\left(3x+2\right).\left(3-x\right)\)
\(=\left[\left(2x-1\right)\left(3x+2\right)\right]\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=18x^2-6x^3+12x-4x^2-9x+3x^2-6+2x\)
\(=-6x^3+\left(18x^2-4x^2+3x^2\right)+\left(12x-9x+2x\right)-6\)
\(=-6x^3+17x^2+5x-6\)
Bài 2 :
\(\left(a+b\right).\left(a^3-a^2b+ab^2-b^3\right)\)
\(=a^4-a^3b+a^2b^2-ab^3+ba^3-a^2b^2+ab^3-b^4\)
\(=a^4+\left(-a^3b+ba^3\right)+\left(a^2b^2-a^2b^2\right)+\left(-ab^3+ab^3\right)-b^4\)
\(=a^4-b^4\)
=> đpcm
Học tốt nha
e)
$x^3+6x^2+12x+8=x^3+3.2.x^2+3.2^2.x+2^3=(x+2)^3$
f)
$a^3-2a^2-ab^2+2b^2=(a^3-ab^2)-(2a^2-2b^2)$
$=a(a^2-b^2)-2(a^2-b^2)=(a^2-b^2)(a-2)=(a-b)(a+b)(a-2)$
g)
$2a^2x-2a^2-2abx+4ab-2b^2=(2a^2x-2abx)-(2a^2-4ab+2b^2)$
$=2ax(a-b)-2(a-b)^2=2(a-b)(ax-a+b)$
h)
\(x^2-2xy+y^2-25=(x-y)^2-25=(x-y)^2-5^2=(x-y+5)(x-y-5)\)
a)
$4x^2-40x^4+100x^3=4x^2(1-10x^2+25x)$
b)
\(3xy(x-5)-7x+35=3xy(x-5)-7(x-5)\)
\(=(x-5)(3xy-7)\)
c)
\(a^2-am-b^2-bm=(a^2-b^2)-(am+bm)=(a-b)(a+b)-m(a+b)\)
\(=(a+b)(a-b-m)\)
d)
\(x^3-4x-x^2y+4y=(x^3-x^2y)-(4x-4y)\)
\(=x^2(x-y)-4(x-y)=(x^2-4)(x-y)=(x-2)(x+2)(x-y)\)
Bài 2
\(a,x^3+2x^2+x\)
\(=x.\left(x^2+2x+1\right)\)
\(b,xy+y^2-x-y\)
\(=y.\left(x+y\right)-\left(x+y\right)\)
\(=\left(y-1\right).\left(x+y\right)\)
bài 3
\(a,3x.\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\x^2=4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=2,x=-2\end{cases}}\)
vậy x=0,x=2 hay x=-2
\(b,xy+y^2-x-y=0\)
\(y.\left(x+y\right)-\left(x+y\right)=0\)
\(\left(y-1\right).\left(x+y\right)=0\)
\(\Rightarrow\orbr{\begin{cases}y-1=0\\x+y=0\end{cases}\Rightarrow\orbr{\begin{cases}y=1\\x=-1\end{cases}}}\)
vậy x=-1, y=1