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a,A=\(\frac{1}{2}.\left(\frac{2.2}{1.3}.\frac{3.3}{2.4}......\frac{2016.2016}{2015.2017}\right)=\frac{1}{2}.\left(\frac{2.3.4...2016}{1.2....2015}.\frac{2.3.4...2016}{3.4....2017}\right)=\frac{1}{2}.\left(\frac{2016.2}{2017}\right)=\frac{4032}{4034}=\frac{2016}{2017}\)
Hok tốt
\(\left|x\right|=\frac{1}{2}\Rightarrow x=\orbr{\begin{cases}\frac{1}{2}\\-\frac{1}{2}\end{cases}}\)
TH1:\(x=\frac{1}{2}\)
\(\Rightarrow\frac{1}{2}-\frac{3}{2}+5=4\)
TH2:\(x=\frac{-1}{2}\)
\(\Rightarrow\frac{1}{2}+\frac{3}{2}+5=7\)
Vậy
Nhận xét :
\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)
Vì \(x\ge0\) nên pt a) tương đương với : \(100x+\frac{1+2+3+...+100}{101}=101x\)
\(\Leftrightarrow x=\frac{100.101}{2.101}=50\)
b)
Tương tự câu a) , phương trình tương đương với :
\(49x+\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{...1}{97.99}=50x\)
\(\Rightarrow x=\frac{97}{195}\)
a: \(\Leftrightarrow\dfrac{x-214}{86}-1+\dfrac{x-132}{84}-2+\dfrac{x-54}{82}-3=0\)
=>x-300=0
hay x=300
a, \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{18}{90}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{18}\)
⇒ x + 1 = 18
⇒ x = 17
Vậy x = 17
b, \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)
⇒ \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{49.3}{148}\)
⇒ \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(1-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=1-\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=\frac{1}{148}\)
⇒ x + 3 = 148
⇒ x = 145
Vậy x = 145
b \(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{x\cdot\left(x+1\right)}=\frac{19}{100}\)
=>\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{19}{100}\)
=>\(\frac{1}{5}-\frac{1}{x+1}\)\(=\frac{19}{100}\)
=>\(\frac{1}{x+1}=\frac{1}{5}-\frac{19}{100}\)
=>\(\frac{1}{x+1}=\frac{1}{100}\)
=> x+1 =100
=>x=99
b) \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{19}{100}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{19}{100}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{19}{100}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{5}-\frac{19}{100}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{100}\)
\(\Rightarrow x+1=100\)
\(\Rightarrow x=99\)
c) \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}=\frac{49}{99}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{49}{99}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{49}{99}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{49}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{50}{99}\)
\(\Rightarrow50.\left(x+2\right)=99\)
\(\Rightarrow x+2=\frac{99}{50}\)
\(\Rightarrow x=-\frac{1}{99}\)
d) Ta có : 6 = 1.6 = 2.3 = (-2) . (-3)
Lâp bảng xét 6 trường hợp:
Vậy các cặp (x,y) \(\inℤ\)thỏa mãn là : (0;4) ; (1; 4) ; (-2 ; 0)
e) \(x^2-3xy+3y-x=1\)
\(\Rightarrow x\left(x-3y\right)+3y-x=1\)
\(\Rightarrow x\left(x-3y\right)-\left(x-3y\right)=1\)
\(\Rightarrow\left(x-3y\right)\left(x-1\right)=1\)
Lại có : 1 = 1.1 = (-1) . (-1)
Lập bảng xét các trường hợp :
Vậy các cặp(x,y) thỏa mãn là : \(\left(2;\frac{1}{3}\right);\left(0;\frac{1}{3}\right)\)