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T
1
NL
Nguyễn Lê Phước Thịnh
CTVHS
9 tháng 3
Ta có: \(\frac{x^2+y^2}{x^2-2xy+y^2}-\frac{2}{xy}\)
\(=\frac{x^2+y^2}{\left(x-y\right)^2}-\frac{2}{xy}\)
\(=\frac{xy\left(x^2+y^2\right)-2\left(x-y\right)^2}{xy\left(x-y\right)^2}=\frac{x^3y+xy^3-2x^2+4xy-2y^2}{xy\left(x-y\right)^2}\)
TA có: \(\left(\frac{x^2+y^2}{x^2-2xy+y^2}-\frac{2}{xy}\right):\left(\frac{1}{x}-\frac{1}{y}\right)^2\)
\(=\frac{x^3y+xy^3-2x^2+4xy-2y^2}{xy\left(x-y\right)^2}:\left(\frac{y-x}{xy}\right)^2\)
\(=\frac{x^3y+xy^3-2x^2+4xy-2y^2}{xy\left(x-y\right)^2}:\frac{\left(x-y\right)^2}{x^2y^2}\)
\(=\frac{x^3y+xy^3-2x^2+4xy-2y^2}{xy\left(x-y\right)^2}\cdot\frac{x^2y^2}{\left(x-y\right)^2}=\frac{\left(x^3y+xy^3-2x^2+4xy-2y^2\right)\cdot xy}{\left(x-y\right)^4}\)
T
1
NL
Nguyễn Lê Phước Thịnh
CTVHS
9 tháng 3
Ta có: \(\frac{x^2+y^2}{x^2-2xy+y^2}-\frac{2}{xy}\)
\(=\frac{x^2+y^2}{\left(x-y\right)^2}-\frac{2}{xy}\)
\(=\frac{xy\left(x^2+y^2\right)-2\left(x-y\right)^2}{xy\left(x-y\right)^2}=\frac{x^3y+xy^3-2x^2+4xy-2y^2}{xy\left(x-y\right)^2}\)
TA có: \(\left(\frac{x^2+y^2}{x^2-2xy+y^2}-\frac{2}{xy}\right):\left(\frac{1}{x}-\frac{1}{y}\right)^2\)
\(=\frac{x^3y+xy^3-2x^2+4xy-2y^2}{xy\left(x-y\right)^2}:\left(\frac{y-x}{xy}\right)^2\)
\(=\frac{x^3y+xy^3-2x^2+4xy-2y^2}{xy\left(x-y\right)^2}:\frac{\left(x-y\right)^2}{x^2y^2}\)
\(=\frac{x^3y+xy^3-2x^2+4xy-2y^2}{xy\left(x-y\right)^2}\cdot\frac{x^2y^2}{\left(x-y\right)^2}=\frac{\left(x^3y+xy^3-2x^2+4xy-2y^2\right)\cdot xy}{\left(x-y\right)^4}\)
T
2
19 tháng 12 2021
\(\dfrac{x^3-8}{x+7}\left(\dfrac{3}{x-2}-\dfrac{2}{x^2+2x+4}\right)\\ =\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{x+7}\left(\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}-\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\right)\\ =\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{x+7}\left(\dfrac{3x^2+6x+12}{\left(x-2\right)\left(x^2+2x+4\right)}-\dfrac{2x-4}{\left(x-2\right)\left(x^2+2x+4\right)}\right)\)
\(=\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{x+7}.\dfrac{3x^2+6x+12-2x+4}{\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\dfrac{3x^2+4x+16}{x+7}\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
19 tháng 12 2021
\(=\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{x+7}\cdot\dfrac{3x^2+6x+12-2x+4}{\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\dfrac{3x^2+4x+16}{x+7}\)
30 tháng 8 2021
\(\left(3x-1\right)\left(2x+7\right)-\left(12x^3+8x^2-14x\right):2x\)
\(=6x^2+19x-7-2x\left(6x^2+4x-7\right):2x=6x^2+19x-7-\left(6x^2-4x+7\right)=15x\)
KK
30 tháng 8 2021
(3x - 1)(2x + 7) - (12x3 + 8x2 - 14x) : 2x
= 6x2 + 21x - 2x - 7 - (6x2 + 4x - 7)
= 6x2 + 21x - 2x - 7 - 6x2 - 4x + 7
= 6x2 - 6x2 + 21x - 2x - 4x - 7 + 7
= 5x
Y
25 tháng 4 2022
\(49,=\dfrac{38}{5}-\left(\dfrac{19}{7}+\dfrac{28}{5}\right)\)
\(=\dfrac{38}{5}-\dfrac{19}{7}-\dfrac{28}{5}\)
\(=\left(\dfrac{38}{5}-\dfrac{28}{5}\right)-\dfrac{19}{7}\)
\(=2-\dfrac{19}{7}=-\dfrac{5}{7}\)
\(50,=\dfrac{25}{81}.\dfrac{15}{22}=\dfrac{125}{594}\)
25 tháng 4 2022
mik xinloi ạ câu 50 mik viết sai đề nên mong bạn giải lại giúp ạ <3
a: \(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)
\(=\frac{2x}{x\left(x+2y\right)}+\frac{y}{y\left(x-2y\right)}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2}{x+2y}+\frac{1}{x-2y}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2\left(x-2y\right)+x+2y+4}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x-4y+x+2y+4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{3x-2y+4}{\left(x-2y\right)\left(x+2y\right)}\)
b:Sửa đề: \(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\)
\(=\frac{x^2}{x\left(x^2-4\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\)
\(=\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2}{x-2}+\frac{1}{x+2}=\frac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x-2-2x-4}{\left(x-2\right)\left(x+2\right)}=\frac{-6}{\left(x-2\right)\left(x+2\right)}\)
c: \(\frac{x^2+2}{x^3-1}+\frac{x}{x^2+x+1}+\frac{1}{1-x}\)
\(=\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x}{x^2+x+1}-\frac{1}{x-1}\)
\(=\frac{x^2+2+x\left(x-1\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{x^2+2+x^2-x-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x^2-2x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x-1}{x^2+x+1}\)