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\(B=\sqrt{1+2017^2+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
Đặt B = 2017 => B + 1 = 2018
Khi B bằng:
\(B=\sqrt{1+B^2+\frac{B}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\sqrt{\frac{\left(B+1\right)^2+B^2\left(B+1\right)^2+B^2}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\sqrt{\frac{B^2\left(B+1\right)^2+2B\left(B+1\right)^2+B^2}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\sqrt{\frac{\left[B\left(B+1\right)+1\right]^2}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\frac{B^2+B+1}{B+1}+\frac{B}{B+1}\left(\text{vi}:a>0\right)\)
\(B=\frac{B^2+2B+1}{B+1}\)
\(B=\frac{\left(B+1\right)^2}{B+1}\)
\(B=B+1\left(\text{vi}:a>0\Rightarrow B+1>0\right)\)
\(B=2017+1\left(\text{vi}:B=2017\right)\)
\(\Rightarrow B=2018\)
Bài 1:
Đặt \(\underbrace{111....1}_{1009}=t\Rightarrow 9t+1=10^{1009}\)
Ta có:
\(a+b+1=\underbrace{11...11}_{1009}.10^{1009}+\underbrace{11...1}_{1009}+4.\underbrace{11....1}_{1009}+1\)
\(=t(9t+1)+t+4.t+1=9t^2+6t+1=(3t+1)^2\) là scp.
Ta có đpcm.
Bài 2:
Đặt \(\underbrace{111....1}_{n}=t\Rightarrow 9t+1=10^n\)
Ta có:
\(a+b+c+8=\underbrace{111..11}_{n}.10^n+\underbrace{111....1}_{n}+\underbrace{11...1}_{n}.10+1+6.\underbrace{111...1}_{n}+8\)
\(t(9t+1)+t+10t+1+6t+8=9t^2+18t+9\)
\(=(3t+3)^2\) là scp.
Ta có đpcm.