Tìm GTNN của:
A=x2+y2+xy-x+y+1
Biết xy=11 và x2y+xy2+x+y=2010. Tính x2+y2.
(x+1)/x2+2x-3 và (-2x)/x2+7x+10
x-y/x2+xy vÀ 2x-3y/xy2
x-2y/2 và x2+y2/2x-2xy
x+2y/x2y+xy2 và x-yy/x2+2xy+y2
a: \(\dfrac{\left(x+1\right)}{x^2+2x-3}=\dfrac{\left(x+1\right)}{\left(x+3\right)\cdot\left(x-1\right)}=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+5\right)}{\left(x+3\right)\left(x-1\right)\left(x+2\right)\left(x+5\right)}\)
\(\dfrac{-2x}{x^2+7x+10}=\dfrac{-2x}{\left(x+2\right)\left(x+5\right)}=\dfrac{-2x\left(x+3\right)\left(x-1\right)}{\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x-1\right)}\)
b: \(\dfrac{x-y}{x^2+xy}=\dfrac{x-y}{x\left(x+y\right)}=\dfrac{y^2\left(x-y\right)}{xy^2\left(x+y\right)}\)
\(\dfrac{2x-3y}{xy^2}=\dfrac{\left(2x-3y\right)\left(x+y\right)}{xy^2\left(x+y\right)}\)
c: \(\dfrac{x-2y}{2}=\dfrac{\left(x-2y\right)\left(x-xy\right)}{2\left(x-xy\right)}\)
\(\dfrac{x^2+y^2}{2x-2xy}=\dfrac{x^2+y^2}{2\left(x-xy\right)}\)
Phân tích tử và mẫu thành nhân tử rồi rút gọn phân thức:
a) x2 + xy +x + y / x2 - xy + x - y
b) x2 - 6x+ 9 / 3x2 - 9x
c) y2 - x2 / x2y - xy2
\(a,=\dfrac{\left(x+1\right)\left(x+y\right)}{\left(x-y\right)\left(x+1\right)}=\dfrac{x+y}{x-y}\\ b,=\dfrac{\left(x-3\right)^2}{3x\left(x-3\right)}=\dfrac{x-3}{3x}\\ c,=\dfrac{\left(y-x\right)\left(y+x\right)}{xy\left(x-y\right)}=\dfrac{-x-y}{xy}\)
Lời giải:
a.
\(\frac{x^2+xy+x+y}{x^2-xy+x-y}=\frac{x(x+y)+(x+y)}{x(x+1)-y(x+1)}=\frac{(x+y)(x+1)}{(x+1)(x-y)}=\frac{x+y}{x-y}\)
b.
\(\frac{x^2-6x+9}{3x^2-9x}=\frac{(x-3)^2}{3x(x-3)}=\frac{x-3}{3x}\)
c.
\(\frac{y^2-x^2}{x^2y-xy^2}=\frac{(y-x)(y+x)}{-xy(y-x)}=\frac{x+y}{-xy}\)
a) 5x-5y+ax-ay b) ax+ay+bx+by c) x2+x+ax+a
d) x2y+xy2+xy2-3x-3y e) x2y+xy-x-1 f) x2+2x-2x-4
g) x2+6x-y2+9 h) x2-y2+10x+25 i) x2-8x-24y2+16
\(a,=5\left(x-y\right)+a\left(x-y\right)=\left(5+a\right)\left(x-y\right)\\ b,=a\left(x+y\right)+b\left(x+y\right)=\left(a+b\right)\left(x+y\right)\\ c,=x\left(x+1\right)+a\left(x+1\right)=\left(x+a\right)\left(x+1\right)\\ d,Sửa:x^2y+xy^2-3x-3y=xy\left(x+y\right)-3\left(x+y\right)=\left(xy-3\right)\left(x+y\right)\\ e,=xy\left(x+1\right)-\left(x+1\right)=\left(xy-1\right)\left(x+1\right)\\ f,=x^2-4=\left(x-2\right)\left(x+2\right)\\ g,=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\\ h,=\left(x+5\right)^2-y^2=\left(x-y+5\right)\left(x+y+5\right)\\ i,=\left(x-4\right)^2-24y^2=\left(x-2\sqrt{6}y-4\right)\left(x+2\sqrt{6}y+4\right)\)
Bài 2:Phân tích đa thức thành nhân tử chung
a, 4(2-x)2+xy-2y
b, x(x-y)3-y(y-x)2-y2(x-y)
c, x2y-xy2-3x+3y
d, x(x+y)2-y(x+y2)+xy-x2
a) \(4\left(2-x\right)^2+xy-2y\)
\(=4\left(x-2\right)^2+\left(xy-2y\right)\)
\(=4\left(x-2\right)\left(x-2\right)+y\left(x-2\right)\)
\(=\left(x-2\right)\left(4x-8\right)+y\left(x-2\right)\)
\(=\left(x-2\right)\left(4x-8+x-2\right)\)
\(=\left(x-2\right)\left(5x-10\right)\)
\(=5\left(x-2\right)^2\)
a, \(=4\left(x-2\right)^2+y\left(x-2\right)=\left(x-2\right)\left(4x-8+y\right)\)
b, \(=x\left(x-y\right)^3-y\left(x-y\right)^2-y^2\left(x-y\right)=\left(x-y\right)\left[x\left(x-y\right)^2-y\left(x-y\right)-y^2\right]=\left(x-y\right)\left[x\left(x^2-2xy+y^2\right)-xy+y^2-y^2\right]=\left(x-y\right)\left(x^3-2x^2y+xy^2-xy\right)=x\left(x-y\right)\left(x^2-2xy+y^2-y\right)\)
c, \(=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\)
d, không phân tích được
c, x2y - xy2 - 3x + 3y
= xy(x-y) - 3(x-y)
= (x-y)(x-3)
Bài 1. Làm tính nhân:
a) 3x(5x2 - 2x - 1);
b) (x2 - 2xy + 3)(-xy);
c) x2y(2x3 - xy2 - 1);
d) x(1,4x - 3,5y);
e) xy(x2 - xy + y2);
f)(1 + 2x - x2)5x;
g) (x2y - xy + xy2 + y3). 3xy2;
h) x2y(15x - 0,9y + 6);
a) \(3x\left(5x^2-2x-1\right)\)
\(=3x.5x^2-3x.2x+3x.\left(-1\right)\)
\(=15x^3-6x^2-3x\)
b) \(\left(x^3-2xy+3\right)\left(-xy\right)\)
\(=\left(-xy\right).\left(x^2+2xy-3\right)\)
\(=\left(-xy\right).x^2+\left(-xy\right).2xy+\left(-xy\right).\left(-3\right)\)
\(=x^3y-2x^2y^2+3xy\)
mấy câu sau vt lại đè
Phân tích đa thức thành nhân tử:
a) 4 ( 2 - x ) 2 + xy - 2y;
b) x ( x - y ) 3 - y ( y - x ) 2 - y 2 (x - y);
c) x 2 y - xy 2 - 3x + 3y;
d) x ( x + y ) 2 - y ( x + y ) 2 + xy - x 2
Chứng minh rằng: Giá trị của mỗi đa thức sau là hằng số. Cho x - y = 1.
a, P = x2 - xy - x + xy2 - y3 - y2 + 5.
b, Q = x3 - x2y + xy2 - y3 - y2 + 5x - 5y - 2017.
Chứng minh rằng: Giá trị của mỗi đa thức sau là hằng số. Cho x - y = 1.
a, P = x2 - xy - x + xy2 - y3 - y2 + 5.
b, Q = x3 - x2y + xy2 - y3 - y2 + 5x - 5y - 2017.
Viết các đa thức sau dưới dạng tổng của các đơn thức rồi thu gọn các đơn thức đồng dạng(nếu có)và tìm bậc của những đa thức đó với tập hợp các biến.
a) (x2 - y2) (x2 + y2) - 3xy2(x + y) + 5x2y2 + x2y(x - y)
b) 3x(x2y + xy2) - 7xy(x2 - y2) - x(3y2 - 2xy2 - 5y - 1)
6). – x2 y(xy2 – 1/2 xy + 3/4 x2 y2 )
7). (3xy – x2 + y). 2/3 x2 y
8). (4x3 – 5xy + 2x)( – 1/2 xy)
9). 2x2 (x2 + 3x + 1/2 )
10). – 3/2 x4 y2 (6x4 − 10/9 x2 y3 – y5 )
11). 2 3 x3 (x + x2 – 3/4 x5 )
12). 2xy2 (xy + 3x2 y – 2/3 xy3 )
13). 3x(2x3 – 1/3 x2 – 4x)
14). 3/5 x3 y5 (7x4 + 5x2 y − 10/21 x4 y3 –y4 )
6: \(-x^2y\left(xy^2-\dfrac{1}{2}xy+\dfrac{3}{4}x^2y^2\right)\)
\(=-x^3y^3+\dfrac{1}{2}x^3y^2-\dfrac{3}{4}x^4y^3\)
7: \(\dfrac{2}{3}x^2y\cdot\left(3xy-x^2+y\right)\)
\(=2x^3y^2-\dfrac{2}{3}x^4y+\dfrac{2}{3}x^2y^2\)
8: \(-\dfrac{1}{2}xy\left(4x^3-5xy+2x\right)\)
\(=-2x^4y+\dfrac{5}{2}x^2y^2-x^2y\)
9: \(2x^2\left(x^2+3x+\dfrac{1}{2}\right)=2x^4+6x^3+x^2\)
10: \(-\dfrac{3}{2}x^4y^2\left(6x^4-\dfrac{10}{9}x^2y^3-y^5\right)\)
\(=-9x^8y^2+\dfrac{5}{3}x^6y^5+\dfrac{3}{2}x^4y^7\)
11: \(\dfrac{2}{3}x^3\left(x+x^2-\dfrac{3}{4}x^5\right)=\dfrac{2}{3}x^3+\dfrac{2}{3}x^5-\dfrac{1}{2}x^8\)
12: \(2xy^2\left(xy+3x^2y-\dfrac{2}{3}xy^3\right)=2x^2y^3+6x^3y^3-\dfrac{4}{3}x^2y^5\)
13: \(3x\left(2x^3-\dfrac{1}{3}x^2-4x\right)=6x^4-x^3-12x^2\)