\(1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}=\frac{k^2.\left(k+1\right)^2+\left(k+1\right)^2+k^2}{k^2\left(k+1\right)^2}\)

\(=\frac{k^2\left(k+1\right)^2+2k\left(k+1\right)+1}{k^2\left(k+1\right)^2}=\frac{\left(k\left(k+1\right)+1\right)^2}{k^2\left(k+1\right)^2}\)

=> \(\sqrt{1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}}=\frac{k\left(k+1\right)+1}{k\left(k+1\right)}=1+\frac{1}{k\left(k+1\right)}=1+\frac{1}{k}-\frac{1}{k+1}\)

=> \(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+....+\sqrt{1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}}\)

\(=1+\frac{1}{1}-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{k}-\frac{1}{k+1}\)

\(=k+1-\frac{1}{k+1}\)

=> \(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+....+\sqrt{1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}}=\frac{2017^2-1}{2017}\)

<=> \(k+1-\frac{1}{k+1}=2017-\frac{1}{2017}\)

\(\Leftrightarrow\left(k+1-2017\right)-\left(\frac{1}{k+1}-\frac{1}{2017}\right)=0\)

\(\Leftrightarrow\left(k-2016\right)\left(1+\frac{1}{2017.\left(k+1\right)}\right)=0\)

<=> k=2016

Câu hỏi của phạm trung hiếu - Toán lớp 9 - Học toán với OnlineMath

Em tham khảo nhé!

\(a,\frac{6-3\sqrt{6}}{6}=\frac{3\left(2-\sqrt{6}\right)}{6}=\frac{2-\sqrt{6}}{2}\)

\(b,\frac{a-b}{\sqrt{a}-\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}=\sqrt{a}+\sqrt{b}\)

\(b,ĐKXĐ:x>0\)

\(D=2011\sqrt{x}-2+\frac{1}{\sqrt{x}}\)\(=2011\sqrt{x}+\frac{1}{\sqrt{x}}-2\)

Áp dụng bđt Cauchy cho 2 số dương \(2011\sqrt{x}\)và\(\frac{1}{\sqrt{x}}\)ta được:

\(2011\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{2011\sqrt{x}.\frac{1}{\sqrt{x}}}\)

\(\Leftrightarrow2011\sqrt{x}+\frac{1}{\sqrt{x}}-2\ge2\sqrt{2011}-2\)

\(\Leftrightarrow D\ge2\sqrt{2011}-2\)

Dấu "=" xảy ra \(\Leftrightarrow2011\sqrt{x}=\frac{1}{\sqrt{x}}\Leftrightarrow x=\frac{1}{2011}\left(TMĐK\right)\)

^{\(a,\sqrt{2x-1}=2\)}

\(\Rightarrow2x-1=4\)

\(\Rightarrow2x=5\)

\(\Rightarrow x=\frac{5}{2}\)

\(b,\sqrt{2x-1}=x+1\)

\(\Rightarrow2x-1=\left(x+1\right)^2\)

\(\Rightarrow2x-1=x^2+2x+1\)

\(\Rightarrow x^2+2x-2x=-1-1\)

\(\Rightarrow x^2=-2VN\)

Ta có :A = x² + 4y² - 4x + 32y + 2078 = (x² - 4x  + 4) + (4y² + 32y + 64) + 2010 = (x - 2)² + (2y + 8)² + 2010

Ta luôn có: (x - 2)² \(\ge\)0 \(\forall\)x

        (2y + 8)² \(\ge\)0 \(\forall\)y

=> (x - 2)² + (2y + 8)² + 2010 \(\ge\)2010

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-1=0\\2y+8=0\end{cases}}\) <=> \(\hept{\begin{cases}x=1\\2y=-8\end{cases}}\) <=> \(\hept{\begin{cases}x=1\\y=-4\end{cases}}\)

Vậy Min của A = 2010 tại x =  1 và y = -4

sửa đề B = 3x² + y² + 4x - y

Ta có B = \(3\left(x+\frac{2}{3}\right)^2+\left(y-\frac{1}{2}\right)^2-\frac{19}{12}\ge\frac{-19}{12}\)

Vậy GTNN của B là \(\frac{-19}{12}\)khi \(x=\frac{-2}{3};y=\frac{1}{2}\)

Áp dụng bđt sau \(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\)

Có: \(A=\left(x+2\right)^4+\left(x-4\right)^4\)

           \(=\left(x+2\right)^4+\left(4-x\right)^4\)

          \(\ge\frac{\left[\left(x+2\right)^2+\left(4-x\right)^2\right]^2}{2}\ge\frac{\left[\frac{\left(x+2+4-x\right)^2}{2}\right]^2}{2}\)

                                                                       \(=\frac{\left(\frac{6^2}{2}\right)^2}{2}=162\)

Dấu "=" xảy ra <=> x + 2 = 4 - x

                        <=> 2x = 2

                        <=> x = 1 

tìm Min là sao???

bn kham khảo ở 

chuyên đề cực trị GTLN và GTNN , rất chi tiết và đầy đủ

vào thống kê của mk nhé 

hc tốt

ĐKXĐ: \(x\ge0;x\ne1\)

\(P=\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}}{\sqrt{x}-1}\right):\frac{2}{\sqrt{x}+1}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{2}{\sqrt{x}+1}\)

\(=\frac{x-\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{2}{\sqrt{x}+1}\)

\(=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}+1}{2}\)

\(=\frac{-\sqrt{x}}{\sqrt{x}-1}\)

Để p = -2 \(\Rightarrow\frac{-\sqrt{x}}{\sqrt{x}-1}=-2\)

\(\frac{-\sqrt{x}}{\sqrt{x}-1}=-2\)

\(\Rightarrow-\sqrt{x}=-2\left(\sqrt{x}-1\right)\)

\(\Rightarrow-\sqrt{x}=-2\sqrt{x}+2\)

\(\Rightarrow-\sqrt{x}+2\sqrt{x}=2\)

\(\Rightarrow\sqrt{x}=2\)

\(\Rightarrow x=4\)