Bài 1 (1 điểm): Thực hiện phép tính
a) $\dfrac{2y-1}{y}-\dfrac{2x+1}{x}$;
b) $\dfrac{2x}{3} \, : \, \dfrac{5}{6x^2}$.
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17 tháng 2 2021
ĐKXĐ: \(a\ne1\)
a. \(\dfrac{3a^2-a+3}{a^3-1}+\dfrac{1-a}{a^2+a+1}+\dfrac{2}{1-a}\)
\(=\dfrac{3a^2-a+3+\left(1-a\right).\left(a-1\right)-2.\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{3a^2-a+3-a^2+2a-1-2a^2-2a-2}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{-a+1}{\left(a-1\right).\left(a^2+a+1\right)}\)
\(=-\dfrac{1}{a^2+a+1}\)
17 tháng 2 2021
a) Ta có: \(\dfrac{3a^2-a+3}{a^3-1}+\dfrac{1-a}{a^2+a+1}+\dfrac{2}{1-a}\)
\(=\dfrac{3a^2-a+3}{\left(a-1\right)\left(a^2+a+1\right)}-\dfrac{\left(a-1\right)^2}{\left(a-1\right)\left(a^2+a+1\right)}-\dfrac{2\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{3a^2-a+3-\left(a^2-2a+1\right)-2a^2-2a-2}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{a^2-3a+1-a^2+2a-1}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{-a}{\left(a-1\right)\left(a^2+a+1\right)}\)
b) Ta có: \(x-\dfrac{xy}{x+y}-\dfrac{x^3}{x^2y^2}\)
\(=x-\dfrac{xy}{x+y}-\dfrac{x}{y^2}\)
\(=\dfrac{xy^2\cdot\left(x+y\right)}{y^2\cdot\left(x+y\right)}+\dfrac{y^2\cdot xy}{y^2\cdot\left(x+y\right)}-\dfrac{x\cdot\left(x+y\right)}{y^2\cdot\left(x+y\right)}\)
\(=\dfrac{x^2y^2+xy^3+xy^3-x^2-xy}{y^2\cdot\left(x+y\right)}\)
\(=\dfrac{x^2y^2+2xy^3-x^2-xy}{y^2\cdot\left(x+y\right)}\)
22 tháng 6 2023
a: =5x^3-5x^2y+5x-2x^2y+2xy^2-2y
=5x^3-7x^2y+2xy^2+5x-2y
b: =(x^2-1)(x+2)
=x^3+2x^2-x-2
c: =1/2x^2y^2(4x^2-y^2)
=2x^4y^2-1/2x^2y^4
d: =(x^2-1/4)(4x-1)
=4x^3-x^2-x+1/4
e: =x^2-2x-35+(2x+1)(x-3)
=x^2-2x-35+2x^2-6x+x-3
=3x^2-7x-38
21 tháng 2 2021
a) Ta có: \(\dfrac{1-x}{x^2-2x+1}+\dfrac{x+1}{x-1}\)
\(=\dfrac{1-x}{\left(x-1\right)^2}-\dfrac{x+1}{1-x}\)
\(=\dfrac{1-x}{\left(1-x\right)^2}-\dfrac{x+1}{1-x}\)
\(=\dfrac{1-x-1}{1-x}=\dfrac{-x}{1-x}=\dfrac{x}{x-1}\)
b) Ta có: \(\dfrac{2x}{3y^4z}\cdot\left(-\dfrac{4y^2z}{5x}\right)\cdot\left(-\dfrac{15y^3}{8xz}\right)\)
\(=\dfrac{2x\cdot4y^2z\cdot15y^3}{3y^4z\cdot5x\cdot8xz}\)
\(=\dfrac{120xy^5z}{120x^2y^4z^2}=\dfrac{y}{xz}\)
LN
1
AH
Akai Haruma
Giáo viên
27 tháng 12 2021
Lời giải:
\(\frac{2}{x^2y}-\frac{4xy^2-1}{2x^3y^3}=\frac{4xy^2}{2x^3y^3}-\frac{4xy^2-1}{2x^3y^3}=\frac{4xy^2-(4xy^2-1)}{2x^3y^3}=\frac{1}{2x^3y^3}\)
30 tháng 5 2023
a: =-1/5x^5y^2
b: =-9/7xy^3
c: =7/12xy^2z
d: =2x^4
e: =3/4x^5y
f: =11x^2y^5+x^6
6 tháng 1 2021
B1: a)\(xy\left(3x-2y\right)-2xy^2=3x^2y-2y^2x-2xy^2=3x^2y-4xy^2\)
b) \(\left(x^2+4x+4\right):\left(x+2\right)=\left(x+2\right)^2:\left(x+2\right)=\left(x+2\right)\)
\(\dfrac{2\left(x-1\right)}{x^2}.\dfrac{x}{\left(x-1\right)}=\dfrac{2\left(x-1\right)x}{x^2\left(x-1\right)}=\dfrac{2}{x}\)
B2:
a)\(2x^2-4x+2=2\left(x^2-2x+1\right)=2\left(x-1\right)^2\)
b)\(x^2-y^2+3x-3y=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
Mấy bài này là mấy bài rất rất rất cơ bản, học sinh TB cũng phải tự làm được, mấy bài kiểu này đừng nên đăng lên hỏi nha:vv
14 tháng 5 2022
a: \(=\dfrac{2x-2x+y}{2\left(2x-y\right)}=\dfrac{y}{2\left(2x-y\right)}\)
b: \(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{6x+2-x^2-x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2+5x+2}{2\left(x-1\right)\left(x+1\right)}\)
c: \(=\dfrac{1}{x+2}+\dfrac{x+8}{3x\left(x+2\right)}\)
\(=\dfrac{3x+x+8}{3x\left(x+2\right)}=\dfrac{4x+8}{3x\left(x+2\right)}=\dfrac{4}{3x}\)
d: \(=\dfrac{4x+6-2x^2+3x+2x+1}{\left(2x-3\right)\left(2x+3\right)}\)
\(=\dfrac{-2x^2+9x+7}{\left(2x-3\right)\left(2x+3\right)}\)
câu a)
\(\dfrac{2y-1}{y}-\dfrac{2x+1}{x}\\ =\dfrac{2xy-x}{xy}-\dfrac{2xy+y}{xy}\\ =\dfrac{2xy-x-2xy-y}{xy}\\ =\dfrac{-x-y}{xy}\)
câu b)
\(\dfrac{2x}{3}:\dfrac{5}{6x^2}\\ =\dfrac{2x}{3}\cdot\dfrac{6x^2}{5}\\ =\dfrac{2x\cdot6x^2}{3\cdot5}\\ =\dfrac{12x^3}{15}=\dfrac{4x^3}{5}\)