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À khác cái dấu nhưng đề phải là giải phương trình chứ
Đặt 2017-x=a => x-2018=-a-1 phương trình trở thành:
\(\frac{a^2+a\left(-a-1\right)+\left(a-1\right)^2}{a^2-a\left(-a-1\right)+\left(a-1\right)^2}=\frac{19}{49}\)
\(\Leftrightarrow\frac{a^2+a+1}{3a^2+3a+1}=\frac{19}{49}\)
\(\Leftrightarrow49\left(a^2+a+1\right)=19\left(3a^2+3a+1\right)\)
\(\Leftrightarrow49a^2+49a+49=57a^2+57a+19\)
\(\Leftrightarrow8a^2+8a-30=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=\frac{3}{2}\\a=-\frac{5}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=2015,5\\x=2019,5\end{cases}}}\)
Vậy......................
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+...+\dfrac{1}{\left(x+2017\right)\left(x+2018\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{\left(x+1\right)}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+...+\dfrac{1}{x+2017}-\dfrac{1}{x+2018}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+2018}\)
\(=\dfrac{2018}{x\left(x+2018\right)}\)
Dạng này mình làm suốt rồi, bạn cứ yên tâm.
Đặt \(2x^2+x-2018=a;x^2-5x-2017=b\) ta có :
\(a^2+4b^2=4ab\)
\(\Leftrightarrow\)\(a^2-4ab+4b^2=0\)
\(\Leftrightarrow\)\(\left(a-2b\right)^2=0\)
\(\Leftrightarrow\)\(a-2b=0\)
\(\Leftrightarrow\)\(2x^2+x-2018-2\left(x^2-5x-2017\right)=0\)
\(\Leftrightarrow\)\(2x^2+x-2018-2x^2+10x+4034=0\)
\(\Leftrightarrow\)\(11x+2016=0\)
\(\Leftrightarrow\)\(x=\frac{-2016}{11}\)
Vậy \(x=\frac{-2016}{11}\)
Chúc bạn học tốt ~
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+....+\dfrac{1}{\left(x+2017\right)\left(x+2018\right)}\\ =\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+2017}-\dfrac{1}{x+2018}\\ =\dfrac{1}{x}-\dfrac{1}{x+2018}\\ =\dfrac{2018}{x\left(x+2018\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+2016}-\dfrac{1}{x+2017}+\dfrac{1}{x+2017}-\dfrac{1}{x+2018}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+2018}\)
\(=\dfrac{2018}{x\left(x+2018\right)}\)
\(A=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+2017}-\dfrac{1}{x+2018}\)
\(A=\dfrac{1}{x}-\dfrac{1}{x+2018}=\dfrac{2018}{x\left(x+2018\right)}\)
\(B=\dfrac{1}{4}\left(\dfrac{1}{x\left(x+2\right)}-\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}-\dfrac{1}{\left(x+4\right)\left(x+6\right)}+...+\dfrac{1}{\left(x+96\right)\left(x+98\right)}-\dfrac{1}{\left(x+98\right)\left(x+100\right)}\right)\)
\(B=\dfrac{1}{4}\left(\dfrac{1}{x\left(x+2\right)}-\dfrac{1}{\left(x+98\right)\left(x+100\right)}\right)=\dfrac{1}{4}\left(\dfrac{x^2+198x+9800-x^2-2x}{x\left(x+2\right)\left(x+98\right)\left(x+100\right)}\right)\)
\(B=\dfrac{196x+9800}{4x\left(x+2\right)\left(x+98\right)\left(x+100\right)}\)
Đặt x - 2017 = a
Phương trình trên tương đương:
\(\dfrac{\left(-a\right)^2-\left(-a\right)\left(a-1\right)+\left(a-1\right)^2}{\left(-a\right)^2+\left(-a\right)\left(a-1\right)+\left(a-1\right)^2}=\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{a^2+a^2-a+a^2-2a+1}{a^2-a^2+a+a^2-2a+1}=\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3a^2-3a+1}{a^2-a+1}=\dfrac{5}{3}\)
\(\Leftrightarrow9x^2-9x+3=5x^2-5x+5\)
\(\Leftrightarrow4x^2-4x-2=0\)
\(\Leftrightarrow\left(x-\dfrac{1+\sqrt{3}}{2}\right)\left(x-\dfrac{1-\sqrt{3}}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1+\sqrt{3}}{2}\\\dfrac{1-\sqrt{3}}{2}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình: \(S=\left\{\dfrac{1+\sqrt{3}}{2};\dfrac{1-\sqrt{3}}{2}\right\}\)