Rút gọn: P = 219 . 273+ 15 . 49 .94 phần 69 . 210+ 1210
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Những câu hỏi liên quan
13 tháng 11 2023
a: \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
=>\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
=>\(2A+A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2+2^{100}-2^{99}+...+2^2-2\)
=>\(3A=2^{101}-2\)
=>\(A=\dfrac{2^{101}-2}{3}\)
b: Sửa đề: \(A=\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
\(A=\dfrac{2\cdot2^{12}\cdot3^6+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}\)
\(=\dfrac{2^{11}\cdot3^6\left(2^3+3^3\right)}{2^{10}\cdot3^7\left(2^4+5\cdot3\right)}\)
\(=\dfrac{2}{3}\cdot\dfrac{4+27}{16+15}=\dfrac{2}{3}\)
c: \(B=\dfrac{4^5\cdot9^4-2\cdot6^4}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^4\cdot3^4}{2^{10}\cdot3^8+2^8\cdot2^2\cdot5\cdot3^8}\)
\(=\dfrac{2^5\cdot3^4\left(2^5\cdot3^4-1\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{1}{2^5\cdot3^4}\cdot\dfrac{32\cdot81-1}{6}\)
\(=\dfrac{2591}{2^6\cdot3^5}\)
TN
2 tháng 8 2016
a) x+(-30)-(95-40-30+x)
=x+(-30)-95+40+30-x
=(x-x) +[(-30)-95+40+30]
=-55
b) a+(273-a-120)-(270-120)
=a+273-a-120-150
=(a-a)+(273-120-150)
=3
c)b-(294+130)+(94+130)
=b-424+224
=b-(424-224)
=b-200
NH
1
18 tháng 2 2022
\(\dfrac{63}{27}=\dfrac{7}{3}\)
\(\dfrac{3\times4\times6}{12\times8\times5}=\dfrac{3\times1\times1}{1\times1\times5}=\dfrac{3}{5}\)
14 tháng 8 2020
a)\(\frac{20}{-160}=-\frac{1}{8}\)
b)\(\frac{-24}{-72}=\frac{1}{3}\)
c)\(-\frac{270}{450}=-\frac{3}{5}\)
d)\(\frac{3.7.11}{33.14}=\frac{33.7}{33.14}=\frac{7}{14}=\frac{1}{2}\)
e)\(\frac{49+7.49}{-49}=\frac{49.\left(1+7\right)}{-49}=\frac{49.8}{-49}=-8\)
T
14 tháng 8 2020
\(a,\frac{20}{-160}=\frac{-20:20}{160:20}=\frac{-1}{8}\)
\(b,\frac{-24}{-72}=\frac{-24:24}{-72:24}=\frac{-1}{-3}=\frac{1}{3}\)
\(c,\frac{-270}{450}=\frac{-270:90}{450:90}=\frac{-3}{5}\)
\(d,\frac{3.7.11}{33.14}=\frac{3.7.11}{3.11.7.2}=\frac{1}{2}\)
\(e,\frac{49+7.49}{-49}=-\frac{49.\left(7+1\right)}{49}=-8\)
Học tốt
10 tháng 2 2022
49/69 Phân.số.tối.giản
50/ 80 =5/ 8
35/ 40=7/ 8
125/ 1000=1/ 8
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
10 tháng 2 2022
\(\dfrac{49}{69}Phân.số.tối.giản\\ \dfrac{50}{80}=\dfrac{50: 10}{80:10}=\dfrac{5}{8}\\ \dfrac{35}{40}=\dfrac{35:5}{40:5}=\dfrac{7}{8}\\ \dfrac{125}{1000}=\dfrac{125:125}{1000:125}=\dfrac{1}{8}\)
MO
2
AT
13 tháng 7 2021
\(P=\left(\dfrac{x-1}{\sqrt{x}+1}-\dfrac{x-2\sqrt{x}+1}{x-\sqrt{x}}+1\right).\dfrac{1}{x\sqrt{x}+1}\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}-\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}+1\right).\dfrac{1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\left(\sqrt{x}-1-\dfrac{\sqrt{x}-1}{\sqrt{x}}+1\right).\dfrac{1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)+\sqrt{x}}{\sqrt{x}}.\dfrac{1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}.\dfrac{1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
13 tháng 7 2021
Bài 2:
Ta có: \(P=\left(\dfrac{x-1}{\sqrt{x}+1}-\dfrac{x-2\sqrt{x}+1}{x-\sqrt{x}}+1\right)\cdot\dfrac{1}{x\sqrt{x}+1}\)
\(=\left(\sqrt{x}-1-\dfrac{\sqrt{x}-1}{\sqrt{x}}+1\right)\cdot\dfrac{1}{x\sqrt{x}+1}\)
\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{1}{x+\sqrt{x}}\)
Lời giải:
\(P=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\\ =\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\\ =\frac{2^{19}.3^9+2^{18}.3^9.5}{2^{19}.3^9+2^{20}.3^{10}}\\ =\frac{2^{18}.3^9(2+5)}{2^{19}3^9(1+2.3)}\\ =\frac{1}{2}\)