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a) ta có: \(\frac{x+13}{2006}+\frac{x+2006}{13}+\frac{x+1}{2018}+3=0\)
\(\Rightarrow\frac{x+13}{2006}+1+\frac{x+2006}{13}+1+\frac{x+1}{2018}+1=0\)
\(\Rightarrow\frac{x+2019}{2006}+\frac{x+2019}{13}+\frac{x+2019}{2018}=0\)
\(\Rightarrow\left(x+2019\right)\left(\frac{1}{2006}+\frac{1}{13}+\frac{1}{2018}\right)=0\)
mà \(\frac{1}{2006}+\frac{1}{13}+\frac{1}{2018}>0\)
\(\Rightarrow x+2019=0\)
\(\Rightarrow x=-2019\)
b) \(\frac{4}{\left(x+3\right)\left(x+7\right)}+\frac{3}{\left(x+7\right)\left(x+10\right)}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)
\(\Rightarrow\frac{\left(x+7\right)-\left(x+3\right)}{\left(x+3\right)\left(x+7\right)}+\frac{\left(x+10\right)-\left(x+7\right)}{\left(x+7\right)\left(x+10\right)}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)
\(\Rightarrow\frac{1}{x+3}-\frac{1}{x+7}+\frac{1}{x+7}-\frac{1}{x+10}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)
\(\Rightarrow\frac{1}{x+3}-\frac{1}{x+10}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)
\(\Rightarrow\frac{7}{\left(x+3\right)\left(x+10\right)}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)
\(\Rightarrow x=7\)
Bài giải
a, \(\frac{x+5}{2017}-\frac{x+5}{2018}+\frac{x+5}{2019}-\frac{x+5}{2020}=0\)
\(\left(x+5\right)\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}\right)=0\)
Do \(\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}\right)\ne0\)
\(\Rightarrow\text{ }x+5=0\)
\(x=0-5\)
\(=-5\)
1 \(=\)\(\frac{46656}{216}\)\(=\)216
2\(=\)\(\frac{64}{1024}\)\(=\)\(\frac{1}{16}\)
3 \(=\)\(\frac{900}{-27000}\)\(=\)\(\frac{-1}{30}\)
4 \(=\)\(\frac{225}{-3375}\)\(=\)\(\frac{-1}{15}\)
Ta có :\(\frac{6^8.2^4-4^5.18^4}{27^3.8^4-3^9.2^{13}}=\frac{\left(2.3\right)^8.2^4-\left(2^2\right)^5.\left(3^2.2\right)^4}{\left(3^3\right)^3.\left(2^3\right)^4-3^9.2^{13}}=\frac{2^{12}.3^8-2^{14}.3^8}{3^9.2^{12}-3^9.2^{13}}=\frac{3^8.2^{12}.\left(2^2-1\right)}{3^9.2^{12}.\left(1-2\right)}\)
\(=\frac{3^9.2^{12}}{-3^9.2^{12}}=-1\)
\(\frac{6^8\cdot2^2-4^5\cdot18^4}{27^3\cdot8^4-3^9\cdot2^{13}}\)
\(=\frac{\left(2.3\right)^8.2^4-\left(2^2\right)^5.\left(3^2.2\right)^4}{\left(3^3\right)^3.\left(2^3\right)^4-3^9.2^{13}}\)
\(=\frac{2^{12}.3^8-2^{14}.3^8}{3^9.2^{12}-3^9.2^{14}}\)
\(=\frac{3^8.2^{12}.\left(2^2-1\right)}{3^9.2^{12}.\left(1-2\right)}\)
\(=\frac{3^9.2^{12}}{-3^9.2^{12}}=-1\)
(a+1)2+(b-2)2=4
=> (a+1)2+(b-2)2=22+02=02+22
TH1: a+1=2 => a=2-1=1
b-2=0 => b=0+2=2
TH2: a+1=0 => a=0-1=-1
b-2=2 => b=2+2=4
Vậy có 2 cặp số nguyên (a;b) thỏa mãn là (1; 2) và (-1; 4).
b)
\(F=\frac{17^{13}.2^{13}}{34^{12}}=\frac{34^{13}}{34^{12}}=34^1=34.\)
Chúc bạn học tốt!