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2 tháng 12 2019
a) \(\frac{3x+5}{2\left(x-1\right)}+\frac{4}{x-2}=\frac{\left(3x+5\right)\left(x-2\right)+4\cdot2\left(x-1\right)}{2\left(x-1\right)\left(x-2\right)}=\frac{3x^2-6x+5x-10+8x-8}{2\left(x-1\right)\left(x-2\right)}\)
\(=\frac{3x^2+7x-18}{2\left(x-1\right)\left(x-2\right)}\)
b) \(\frac{2x^2+1}{4x^2-2x}+\frac{3-3x}{1-2x}+\frac{3}{2x}=\frac{2x^2+1+4x\left(3-3x\right)+2\cdot3\left(1-2x\right)}{4x\left(1-2x\right)}=\frac{2x^2+1+12-12x+6-12x}{4x\left(1-2x\right)}\)\(=\frac{2x^2-24x+19}{4x\left(1-2x\right)}\)
Đề này... bạn xem lại đi. Chứ thế này thì dùng máy tính cũng không làm nổi T-T
11 tháng 12 2019
\(a)=\frac{-2\left(x+3\right)}{x\left(1-3x\right)}.\frac{1-3x}{x\left(x+3\right)}\)
\(=\frac{-2}{x^2}\)
\(b)=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x^2}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)
\(=\frac{x^2-3x+3x-9-x^2+9}{x\left(x-3\right)}\)
\(=x\left(x-3\right)\)
\(c)=\frac{x+3}{\left(x-1\right)\left(x+1\right)}-\frac{1}{x\left(x+1\right)}\)
\(=\frac{\left(x+3\right).x}{x\left(x-1\right)\left(x+1\right)}-\frac{1.\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x\left(x+3\right)-\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+3}{x+1}\)
# Sắp ik ngủ nên làm vậy hoi, ko chắc phần kq câu b và c đâu nha
S
14 tháng 12 2018
\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x^2+x}\)
b, \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{y^2-xy-xy+x^2}{\left(xy-x^2\right)\left(y^2-xy\right)}=\frac{x^2+y^2}{xy^3-xyxy-xyxy+x^3y}\)Tu rut gon tiep
c, tt
d, cx r
14 tháng 12 2018
a) \(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}\)
\(=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)
b) \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)
\(=\frac{y}{xy\left(y-x\right)}-\frac{x}{xy\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)
c) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)
\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)
PK
2
T
12 tháng 12 2018
ĐKXĐ: \(x\ne0;x\ne\frac{1}{2}\)
\(\frac{2x^2+1}{4x^2-2x}+\frac{3}{2x}-\frac{3-3x}{2x-1}\)
\(=\frac{2x^2+1}{4x^2-2x}+\frac{3}{2x}-\frac{6x-6x^2}{4x^2-2x}\)
\(=\frac{8x^2-6x+1}{4x^2-2x}+\frac{3}{2x}=\frac{8\left(x-\frac{1}{2}\right)\left(x-\frac{1}{4}\right)}{4x\left(x-\frac{1}{2}\right)}+\frac{3}{2x}\)
\(=\frac{8x-2}{4x}+\frac{3}{2x}=\frac{8x-2}{4x}+\frac{6}{4x}=\frac{8x-2+6}{4x}\)
\(=\frac{8x+4}{4x}=1+\frac{4x+4}{4x}=2+\frac{4}{4x}=2+\frac{1}{x}\)
28 tháng 2 2020
Bài 1:
a)\(-x^2\left(3x^3-2x+\frac{1}{2}\right)=-3x^5+2x^3-\frac{x^2}{2}\)
b) Sửa đề: \(\left(2xy-y^2+x\right)\frac{2}{3}x^3y=\frac{4}{3}x^4y^2-\frac{2}{3}x^3y^3+\frac{2}{3}x^4y\)
Không sửa đề: \(\left(2xy-y2+x\right)\frac{2}{3}x^3y=\frac{4}{3}x^4y^2-\frac{4}{3}x^3y^2+\frac{2}{3}x^4y\)
Bài 2:
a)\(3x\left(12x-5\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-15x-36x^2+27x=30\)
\(\Leftrightarrow12x=30\Leftrightarrow x=\frac{5}{2}\)
b) \(x\left(7-2x\right)+2x\left(x-4\right)=15\)
\(\Leftrightarrow7x-2x^2+2x^2-8x=15\)
\(\Leftrightarrow-x=15\Leftrightarrow x=-15\)
Hok tốt
BN
3 tháng 12 2019
a) \(\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}=\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x\left(x-3\right)}\)
\(=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x.x}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)
\(=\frac{x^2-3x+3x-9-x^2+9}{x\left(x-3\right)}=\frac{0}{x\left(x-3\right)}=0\)
b) \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\)
\(=\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10+8}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\frac{1\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}-\frac{4\left(3x-2\right)}{\left(3x+2\right)\left(3x-2\right)}-\frac{-10x+8}{\left(3x-2\right)\left(3x+2\right)}\)
\(\frac{3x+2-12x+2+10x-8}{\left(3x-2\right)\left(3x+2\right)}=\frac{x-4}{\left(3x-2\right)\left(3+2\right)}\)
BN
3 tháng 12 2019
c) \(\frac{4a^2-3a+5}{a^3-1}-\frac{1-2a}{a^2+a+1}-\frac{6}{a-1}\)
\(=\frac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{2a-1}{a^2+a+1}-\frac{6}{a-1}\)
\(=\frac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{\left(2a-1\right)\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}-\frac{6\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{4a^2-3a+5+2a^2-2a-a+1-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{-12}{\left(a-1\right)\left(a^2+a+1\right)}\)
d) \(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}=\frac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\frac{3y}{x\left(x+3y\right)}=\frac{x\left(x+9y\right)}{x\left(x-3y\right)\left(x+3y\right)}-\frac{3y\left(x-3y\right)}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\frac{x^2+9xy-3xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}=\frac{x^2-6xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}=\frac{\left(x-3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}=\frac{x-3y}{x\left(x+3y\right)}\)
e) \(\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\frac{3x-2}{x^2+2x+1}\)
\(=\frac{3x-2}{\left(x-1\right)^2}-\frac{6}{\left(x-1\right)\left(x+1\right)}-\frac{3x-2}{\left(x+1\right)^2}\)
\(=\frac{\left(3x+2\right)\left(x+1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}-\frac{6\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x-1\right)\left(x+1\right)}-\frac{\left(3x-2\right)\left(x-1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}\)
\(=\frac{3x^3+6x^2+3x+2x^2+4x+2-6x^2+6-3x^3+6x^2-3x+2x^2-4x+2}{\left(x-1\right)^2\left(x+1\right)^2}\)
\(=\frac{8x^2+10}{\left(x-1\right)^2\left(x+1\right)^2}\)
f) \(\frac{5}{a+1}-\frac{10}{a-\left(a^2+1\right)}-\frac{15}{a^3+1}=\frac{5a^2}{a^3+1}+\frac{10}{a^3+1}-\frac{15}{a^3+1}\)
\(=\frac{5a^2+10-15}{a^3+1}=\frac{5a^2-5}{a^3+1}\)
\(\frac{3x^2+3x+3}{4x+4}\): \(\frac{9x^3-9}{2x^2-2}\)= \(\frac{3\left(x^2+x+1\right)}{4\left(x+1\right)}\): \(\frac{9\left(x^3-1\right)}{2\left(x^2-1\right)}\)
= \(\frac{3\left(x^2+x+1\right)}{4\left(x+1\right)}\). \(\frac{2\left(x-1\right)\left(x+1\right)}{9\left(x-1\right)\left(x^2+x+1\right)}\)= \(\frac{1}{6}\)