Quy đồng mẫu thức a) 74x+1674�+16 và 11x2−1611�2−16b) 6(x+3)36(�+3)3 và x−32x(x+3)2�−32�(�+3)2 c) −61−x−61−� ; 3xx2+x+13��2+�+1 và x2−3x+5x3−1�2−3�+5�3−1 d)175x175� ; 24x−2y24�−2� và...
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Quy đồng mẫu thức a) và b) và c) ; và d) ; và
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NT
0
7 tháng 12 2021
Bài 2:
a: \(\dfrac{1}{2x^3y}=\dfrac{6yz^3}{12x^3y^2z^3}\)
\(\dfrac{2}{3xy^2z^3}=\dfrac{2\cdot4x^2}{12x^3y^2z^3}=\dfrac{8x^2}{12x^3y^2z^3}\)
11 tháng 10 2023
a: \(\dfrac{3}{x-1}=\dfrac{3\cdot9}{9\cdot\left(x-1\right)}=\dfrac{27}{9\left(x-1\right)}\)
\(\dfrac{4}{3x-3}=\dfrac{12}{9x-9}=\dfrac{12}{9\left(x-1\right)}\)
\(\dfrac{10}{9-9x}=\dfrac{-10}{9x-9}=-\dfrac{10}{9\left(x-1\right)}\)
b: \(\dfrac{3}{2\left(x-3\right)}=\dfrac{3x-9}{2\left(x-3\right)^2}\)
\(\dfrac{3x-2}{x^2-6x+9}=\dfrac{6x-4}{2\left(x-3\right)^2}\)
c: \(\dfrac{3}{x^2+2x+1}=\dfrac{3}{\left(x+1\right)^2}=\dfrac{3x}{x\left(x+1\right)^2}\)
\(-\dfrac{2}{x^2+x}=\dfrac{-2}{x\left(x+1\right)}=\dfrac{-2\left(x+1\right)}{x\left(x+1\right)^2}\)
28 tháng 7 2023
a: 1/x^2y=1/x^2y
3/xy=3x/x^2y
b: \(\dfrac{x}{x^2+2xy+y^2}=\dfrac{x}{\left(x+y\right)^2}\)
\(\dfrac{2x}{x^2+xy}=\dfrac{2}{x+y}=\dfrac{2x+2y}{\left(x+y\right)^2}\)
T
10 tháng 5 2020
c1
a,3/15 = 3:3/15:3 = 15
33/44 = 33:11/44:11 = 34
2/8 = 2:2/8:2 = 1/4
b,9/12 =9:3/12:3 = 34
24/36 =24:12/36:12 = 23
3/8 = 3:1/8:1 = 3/8
c2
a) =12x(4+6)/24
= 12x10/24
=120/24
=5
b,16x8-16x2/12x4
=16x(8-2)/48
=16x6/48
=2
c3
5/8=45/72
20/15=4/3=96/72
24/32=3/4=54/72
15/18=5/6=60/72
77/99=7/9=56/72
c4
2/3=2/3
12/15=4/5
24/18=4/3
16/48=1/3
75/100=3/4
30/45=2/3
12/36=1/3
20/15=4/3
các phân số lớn hơn 1 luôn có mẫu số bé hơn tử số
vậy các số lớn hơn 1 là 24/18,20/15
k mk nha thank mọi ng'
3
10 tháng 5 2020
a, \(\frac{3}{15}=\frac{1}{5}=\frac{4}{20}\); \(\frac{33}{44}=\frac{3}{4}=\frac{15}{20}\); \(\frac{2}{8}=\frac{1}{4}=\frac{5}{20}\)
b, \(\frac{9}{12}=\frac{3}{4}=\frac{18}{24}\); \(\frac{24}{36}=\frac{2}{3}=\frac{16}{24}\); \(\frac{3}{8}=\frac{9}{24}\)
Bài 2 :
a,\(\frac{12x4+12x6}{24}=\frac{12x\left(4+6\right)}{24}=\frac{1x10}{2}=\frac{10}{2}=\frac{5}{1}\)
b, \(\frac{16x8-16x2}{12}=\frac{16x\left(8-2\right)}{12}=\frac{8x6}{6}=\frac{8}{1}\)
a: ĐKXĐ: \(x\notin\left\{4;-4\right\}\)
\(\dfrac{7}{4x+16}=\dfrac{7}{4\left(x+4\right)}=\dfrac{7\left(x-4\right)}{4\left(x+4\right)\left(x-4\right)}\)
\(\dfrac{11}{x^2-16}=\dfrac{11\cdot4}{4\left(x^2-16\right)}=\dfrac{44}{4\left(x-4\right)\left(x+4\right)}\)
b: \(\dfrac{6}{x\left(x+3\right)^2};\dfrac{x-3}{2x\left(x+3\right)^2}\)
ĐKXĐ: \(x\notin\left\{0;-3\right\}\)
\(\dfrac{6}{x\left(x+3\right)^2}=\dfrac{6\cdot2}{2x\left(x+3\right)^2}=\dfrac{12}{2x\left(x+3\right)^2}\)
\(\dfrac{x-3}{2x\left(x+3\right)^2}=\dfrac{x-3}{2x\left(x+3\right)^2}\)
c: \(\dfrac{-6}{1-x};\dfrac{3x}{x^2+x+1};\dfrac{x^2-3x+5}{x^3-1}\)
ĐKXĐ: \(x\ne1\)
\(-\dfrac{6}{1-x}=\dfrac{6}{x-1}=\dfrac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{6x^2+6x+6}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\dfrac{3x}{x^2+x+1}=\dfrac{3x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{3x^2-3x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\dfrac{x^2-3x+5}{x^3-1}=\dfrac{x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
d: \(\dfrac{17}{5x};\dfrac{24}{x-2y};\dfrac{x-y}{8y^2-2x^2}\)
ĐKXĐ: \(x\ne0;x\ne\pm2y\)
\(\dfrac{17}{5x}=\dfrac{17\cdot2\left(x-2y\right)\left(x+2y\right)}{5x\cdot2\cdot\left(x-2y\right)\left(x+2y\right)}=\dfrac{34\left(x^2-4y^2\right)}{10x\left(x-2y\right)\left(x+2y\right)}\)
\(\dfrac{24}{x-2y}=\dfrac{24\cdot10x\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}=\dfrac{240x\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}\)
\(\dfrac{x-y}{8y^2-2x^2}=\dfrac{-\left(x-y\right)}{2x^2-8y^2}=\dfrac{-\left(x-y\right)}{2\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{-5x\left(x-y\right)}{10x\left(x-2y\right)\left(x+2y\right)}=\dfrac{-5x^2+5xy}{10x\left(x-2y\right)\left(x+2y\right)}\)