d,125x3+y6
e, 0,125(a+1)3-1
a) (x2 + 2xy + y2) : (x + y)
b) (125x3 + 1) : (5x + 1)
c) (x2 – 2xy + y2) : (y – x)
a) \(\left(x^2+2xy+y^2\right):\left(x+y\right)=\left(x+y\right)^2:\left(x+y\right)=x+y\)
b) \(=\left[\left(5x+1\right)\left(25x^2-5x+1\right)\right]:\left(5x+1\right)=25x^2-5x+1\)
c) \(=\left(y-x\right)^2:\left(y-x\right)=y-x\)
\(a,=\left(x+y\right)^2:\left(x+y\right)=x+y\\ b,=\left(5x+1\right)\left(25x^2-5x+1\right):\left(5x+1\right)=25x^2-5x+1\\ c,=\left(y-x\right)^2:\left(y-x\right)=y-x\)
1.chia đa thức cho đa thức:
(125x3+1):(5x+1)
\(\text{(125x^3+1):(5x+1)}\)
\(\text{= [(5x)^3+1)]:(5x+1)}\)
\(\text{=(5x+1)(25x^2-5x+1):(5x+1)}\)
\(\text{= 25x^2-5x+1}\)
Viết các đa thức sau thành tích
1, (4x2 + 4x + 1) 2, x2 – 20x + 100
3, y4 – 14y2 + 49 4, 125x3 – 64y3
Mình cần quá trình làm nữa nha, thanks
1: \(4x^2+4x+1=\left(2x+1\right)^2\)
2: \(x^2-20x+100=\left(x-10\right)^2\)
3: \(y^4-14y^2+49=\left(y^2-7\right)^2\)
4: \(125x^3-64y^3=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)
Áp dụng hằng đẳng thức đáng nhớ để thực hiện phép chia: (125x3 + 1) : (5x + 1)
(125x3 + 1) : (5x + 1)
= [(5x)3 + 1] : (5x + 1)
= (5x + 1)[(5x)2 – 5x + 1]] : (5x + 1)
= (5x)2 – 5x + 1
= 25x2 – 5x + 1
1, A= \(\dfrac{-3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
2, B= \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+25\%\right):\dfrac{7}{3}\)
3, C= \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
4, D= \(6\dfrac{5}{12}:2\dfrac{5}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)
\(A=\dfrac{1}{2}-\dfrac{1}{4}\)
\(A=\dfrac{2}{4}-\dfrac{1}{4}\)
\(A=\dfrac{1}{4}\)
\(D=6\dfrac{5}{12}:2\dfrac{5}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(D=\dfrac{77}{12}:\dfrac{13}{4}+\dfrac{45}{4}.\dfrac{2}{15}\)
\(D=\dfrac{77}{39}+\dfrac{3}{2}\)
\(D=\dfrac{271}{78}\)
\(C=\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
\(C=\dfrac{5}{16}:0,125-\left(\dfrac{9}{4}-0,6\right).\dfrac{10}{11}\)
\(C=\dfrac{5}{16}:0,125-\dfrac{33}{20}.\dfrac{10}{11}\)
\(C=\dfrac{5}{2}-\dfrac{3}{2}\)
\(C=1\)
1. Viết dưới dạng
a, M= x2-100
b,N=x3+1000
c,P=125x3-1
d,Q=8x3-27
a) \(M=x^2-100=\left(x-10\right)\left(x+10\right)\)
b) \(N=x^3+1000=\left(x+10\right)\left(x^2-10x+100\right)\)
c) \(125x^3-1=\left(5x-1\right)\left(25x^2+5x+1\right)\)
d) \(8x^3-27=\left(2x-3\right)\left(4x^2+6x+9\right)\)
a, \(M=x^2-100=\left(x-10\right)\left(x+10\right)\)
b, \(N=x^3+1000=\left(x+10\right)\left(x^2-10x+100\right)\)
c, \(P=125x^3-1=\left(5x-1\right)\left(25x^2+5x+1\right)\)
d, \(Q=8x^3-27=\left(2x-3\right)\left(2x+3\right)\)
l) 5xy2- 10xyz + 5xz2
m) x3+ 3x2+ 3x + 1 - 27y3
n) x2- 6xy + 9y2
o) x3+ 6x2y + 12xy2+ 8y3
p) 125x3+ y6
q) x2 + 4xy + 4y2 - 2x - 4y + 1
\(l,=5x\left(y^2-2yz+5z\right)\\ m,=\left(x+1\right)^3-27y^3\\ =\left(x+1-3y\right)\left(x^2+2x+1+3xy+3y+9y^2\right)\\ n,=\left(x-3y\right)^2\\ o,=\left(x+2y\right)^3\\ p,=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\\ q,=\left(x+2y\right)^2-2\left(x-2y\right)+1\\ =\left(x+2y-1\right)^2\)
D=\(2\dfrac{1}{3}-30\%+0,125-6,5\)
Phân tích đa thức thành nhân tử:
A= 10x2+ 20xy + 10y - 90.
B= x3y - 3x2y - 4xy + 12y.
C= 125x3 - 10x2 + 2x - 1.
a: Ta có: \(A=10x^2+20xy+10y^2-90\)
\(=10\left(x^2+2xy+y^2-9\right)\)
\(=10\left(x+y-3\right)\left(x+y+3\right)\)
b: Ta có: \(B=x^3y-3x^2y-4xy+12y\)
\(=x^2y\left(x-3\right)-4y\left(x-3\right)\)
\(=y\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
c: Ta có: \(C=125x^3-10x^2+2x-1\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)-2x\left(5x-1\right)\)
\(=\left(5x-1\right)\left(25x^2+3x+1\right)\)