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Đậu Thị Thanh Huyền
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loveyoongi03
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Lê Việt
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Edogawa Conan
14 tháng 7 2017 lúc 20:27

Ta có:

\(\frac{2^{19}.27^9+15.4^9.9^4}{6^9.2^{12}+12^{10}}=\frac{2^{19}.\left(3^3\right)^9+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{12}+\left(2^2.3\right)^{10}}\)

\(=\frac{2^{19}.3^{27}+3.5.2^{18}.3^8}{2^9.3^9.2^{12}+2^{20}.3^{10}}=\frac{2^{19}.3^{27}+3^9.2^{18}.5}{2^{21}.3^9+2^{20}.3^{10}}=\frac{2^{18}.3^9.\left(2.3^{18}+5\right)}{2^{20}.3^9.\left(2+3\right)}\)

\(=\frac{1.1.\left(2.3^{18}+5\right)}{2^2.1.5}=\frac{2.3^{18}+5}{20}\)

nguyen thi man
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Ngô Thị Yến Nhi
19 tháng 2 2017 lúc 16:56

1.

a) \(\frac{16}{24}-\frac{1}{3}=\frac{16}{24}-\frac{8}{24}=\)\(\frac{8}{24}=\frac{1}{3}\)

b) \(\frac{4}{5}-\frac{12}{60}=\frac{48}{60}-\frac{12}{60}=\frac{36}{60}=\frac{9}{15}\)

3.

a)\(\frac{17}{6}-\frac{2}{6}=\frac{17-2}{6}=\frac{15}{6}\)

b) \(\frac{16}{15}-\frac{11}{15}=\frac{16-11}{15}=\frac{5}{15}=\frac{1}{3}\)

c) \(\frac{19}{12}-\frac{13}{12}=\frac{19-13}{12}=\frac{6}{12}=\frac{1}{2}\)

Neo Amazon
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gunny
25 tháng 12 2019 lúc 20:56

chịu ko bt *_*

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Cô chủ nhỏ
25 tháng 12 2019 lúc 21:17

ĐKXĐ : x\(\ne\mp2\)

A = \(\frac{x}{x-2}\)+\(\frac{2-x}{x+2}\)+\(\frac{12-10x}{\left(x-2\right)\left(x+2\right)}\)

=\(\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)+\(\frac{\left(2-x\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)+\(\frac{12-10x}{\left(x-2\right)\left(x+2\right)}\)

=\(\frac{x^2+2x-x^2+4x-4+12-10x}{\left(x-2\right)\left(x+2\right)}\)

=\(\frac{8-4x}{\left(x-2\right)\left(x+2\right)}\)

=\(\frac{-4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

=\(\frac{-4}{x+2}\)

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Chu Công Đức
26 tháng 12 2019 lúc 16:23

\(ĐKXĐ:x\ne\pm2\)

\(A=\frac{x}{x-2}+\frac{2-x}{x+2}+\frac{12-10x}{x^2-4}\)\(=\frac{x}{x-2}+\frac{-\left(x-2\right)}{x+2}+\frac{-\left(10x-12\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}+\frac{-10x+12}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2+2x}{\left(x-2\right)\left(x+2\right)}+\frac{-x^2+4x-4}{\left(x-2\right)\left(x+2\right)}+\frac{-10x+12}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2+2x-x^2+4x-4-10x+12}{\left(x-2\right)\left(x+2\right)}=\frac{-4x+8}{\left(x-2\right)\left(x+2\right)}=\frac{-4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{-4}{x+2}\)

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hoang le
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Trần Tuyết Như
10 tháng 6 2015 lúc 10:54

\(P=\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{5}{12}+1-\frac{7}{11}}=\frac{\frac{115}{132}}{\frac{103}{132}}=\frac{115}{132}\cdot\frac{132}{103}=\frac{115}{103}\)

Nguyễn Minh Hoàng
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Lê Quang Phúc
16 tháng 9 2017 lúc 19:49

\(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)

\(\Rightarrow\frac{2^{12}.\left(13+65\right)}{2^{10}.104}+\frac{3^{10}.\left(11+5\right)}{3^9.2^4}\)

\(\Rightarrow\frac{2^{12}.78}{2^{10}.104}+\frac{3^{10}.2^4}{3^9.2^4}\)

\(=\frac{2^2.3}{4}+3\)

\(=3+3=6\)

Sammy Nguyen
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✓ ℍɠŞ_ŦƦùM $₦G ✓
4 tháng 6 2015 lúc 11:28

\(\frac{\frac{3}{41}-\frac{12}{47}+\frac{27}{53}}{\frac{4}{41}-\frac{16}{47}+\frac{36}{53}}=\frac{3.\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}{4.\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}=\frac{3}{4}\)

Love
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JiYoonMinn
17 tháng 7 2018 lúc 17:25

\(A=\frac{x-1}{x+2}-\frac{x+2}{x-2}-\frac{x^2+12}{4-x^2}\)                    ĐKXĐ: \(x\ne\pm2\)

\(=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\frac{x^2+12}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2-2x-x+2-x^2-4x-4+x^2+12}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{x^2-7x+10}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{x^2-2x-5x+10}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{x\left(x-2\right)-5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{\left(x-5\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{x-5}{x+2}\)