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Ta có : \(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}\)
Mà \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};...;\frac{1}{8^2}<\frac{1}{7.8}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}=1-\frac{1}{8}<1\)
Vậy B < 1
Ta có : \(\frac{1}{2^2}<\frac{1}{1.2}\)
\(\frac{1}{3^2}<\frac{1}{2.3}\)
...
\(\frac{1}{n^2}<\frac{1}{\left(n-1\right)n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right).n}=1-\frac{1}{n}<1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}<1\)
đk: \(\begin{cases}x+2\ne0\\4-x>0\\6+x>0\end{cases}\)
ta có \(3\log_{\frac{1}{4}}\left(x+2\right)-3=3\log_{\frac{1}{4}}\left(4-x\right)+3\log_{\frac{1}{4}}\left(6+x\right)\) suy ra \(\log_{\frac{1}{4}}\left(x+2\right)-\log_{\frac{1}{4}}\frac{1}{4}=\log_{\frac{1}{4}}\left(4-x\right)\left(6+x\right)\) suy ra \(\log_{\frac{1}{4}}\left(x+2\right).\frac{1}{4}=\log_{\frac{1}{4}}\left(4-x\right)\left(6+x\right)\) suy ra \(\frac{x+2}{4}=\left(4-x\right)\left(6+x\right)\)
giải pt tìm ra x
đối chiếu với đk của bài ta suy ra đc nghiệm của pt
a) ĐK: \(x\ge0,x\ne1,x\ne\frac{1}{4}\)
\(A=1+\left(\frac{2x+\sqrt{x}-1}{1-x}-\frac{2x\sqrt{x}-\sqrt{x}+x}{1-x\sqrt{x}}\right)\frac{x-\sqrt{x}}{2\sqrt{x}-1}\)
\(A=1+\left[\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(1-\sqrt{x}\right)}-\frac{\sqrt{x}\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}\right]\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}-1}\)
\(A=1+\left[\frac{2\sqrt{x}-1}{1-\sqrt{x}}-\frac{\sqrt{x}\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}\right]\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}-1}\)
\(A=1-\sqrt{x}+\frac{x\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\)
\(A=\frac{x+1}{x+\sqrt{x}+1}\)
Để \(A=\frac{6-\sqrt{6}}{5}\Rightarrow\frac{x+1}{x+\sqrt{x}+1}=\frac{6-\sqrt{6}}{5}\)
\(\Rightarrow5x+5=\left(6-\sqrt{6}\right)x+\left(6-\sqrt{6}\right)\sqrt{x}+6-\sqrt{6}\)
\(\Rightarrow\left(1-\sqrt{6}\right)x+\left(6-\sqrt{6}\right)\sqrt{x}+1-\sqrt{6}=0\)
\(\Rightarrow x-\sqrt{6}.\sqrt{x}+1=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{\sqrt{2}+\sqrt{6}}{2}\\\sqrt{x}=\frac{-\sqrt{2}+\sqrt{6}}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{cases}}\left(tmđk\right)\)
b) Xét \(A-\frac{2}{3}=\frac{x+1}{x+\sqrt{x}+1}-\frac{2}{3}=\frac{3x+3-2x-2\sqrt{x}-2}{3\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x-2\sqrt{x}+1}{3\left(x+\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}\)
Do \(x\ge0,x\ne1,x\ne\frac{1}{4}\Rightarrow\left(\sqrt{x}-1\right)^2>0\)
Lại có \(x+\sqrt{x}+1=\left(\sqrt{x}+\frac{1}{2}\right)+\frac{3}{4}>0\)
Nên \(A-\frac{2}{3}>0\Rightarrow A>\frac{2}{3}\).
\(\frac{1}{3}+\frac{1}{2.3}\left(1+2\right)+\frac{1}{3.3}\left(1+2+3\right)+...+\frac{1}{3.2015}\left(1+2+3+...+2015\right)=\frac{1}{3}\left[\frac{2}{2}+\frac{1}{2}\left(\frac{2.3}{2}\right)+\frac{1}{3}\left(\frac{3.4}{2}\right)+...+\frac{1}{2015}\left(\frac{2016.2015}{2}\right)\right]=\frac{1}{3}.\frac{1}{2}\left(2+3+4+....+2016\right)=\frac{1}{6}\left(\frac{2016.2017}{2}-1\right)\)
bit lm bài này k giup tui