GTNN: \(\sqrt{x-1-2\sqrt{ }x-2}+\sqrt{x+7-6\sqrt{ }x-2}\)
tính: \(100^2-99^2+98^2-97^2+.....+2^2-1\)
cho B=\(\dfrac{1}{\sqrt{1}+\sqrt{2}}\)+\(\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{98}+\sqrt{99}}+\dfrac{1}{\sqrt{99}+\sqrt{100}}\).số nghiệm của phương trình \(x^3+3Bx^2+27Bx+9B^2=0\) là ?
\(B=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{100}-\sqrt{99}\right)\left(\sqrt{100}+\sqrt{99}\right)}\)
\(=\dfrac{\sqrt{2}-1}{1}+\dfrac{\sqrt{3}-\sqrt{2}}{1}+...+\dfrac{\sqrt{100}-\sqrt{99}}{1}\)
\(=\sqrt{100}-1=9\)
\(x^3+3.9x^2+3.9^2x+9^3=0\)
\(\Leftrightarrow\left(x+9\right)^3=0\)
\(\Leftrightarrow x=-9\)
Bài 1: Tính :
\(C=\sqrt{\frac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\frac{\sqrt{3}+4}{5-2\sqrt{3}}}\)
\(B=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+....+\frac{1}{\sqrt{99}+\sqrt{100}}\)
\(D=\sqrt{1+\sqrt{3+\sqrt{13+4\sqrt{3}}}}+\sqrt{1-\sqrt{3-\sqrt{13-4\sqrt{3}}}}\)
Bài 2 : Cho \(P=\left(\frac{1}{\sqrt{x}-1}+\frac{x-\sqrt{x}+6}{x+\sqrt{x}-2}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{x-\sqrt{x}-2}{x+\sqrt{x}+2}\right)\)
a, Rút gọn P
b, Tìm GTNN
c, Tìm x để \(P.\frac{x-1}{x^2+8x}< -2\)
Cho x=\(\dfrac{\sqrt{2}-\sqrt{1}}{1+2}+\dfrac{\sqrt{3}-\sqrt{2}}{2+3}+...+\dfrac{\sqrt{100}-\sqrt{99}}{99+100}\)
chung minh x<\(\dfrac{1}{2}\)
Lời giải:
Xét số hạng tổng quát:
\(\frac{\sqrt{n+1}-\sqrt{n}}{n+(n+1)}< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n(n+1)}}=\frac{1}{2}(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})\) theo BĐT Cô-si.
Do đó:
\(x< \frac{1}{2}\left[\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\right]=\frac{1}{2}(1-\frac{1}{\sqrt{100}})< \frac{1}{2}\)
Ta có đpcm.
Cho \(x=\dfrac{\sqrt{2}-\sqrt{1}}{1+\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{2+\sqrt{3}}+\dfrac{\sqrt{4}-\sqrt{3}}{3+4}+...+\dfrac{\sqrt{100}-\sqrt{99}}{99+100}\). Chứng minh \(x< \dfrac{1}{2}\)
Bài 1: Tính :
\(C=\sqrt{\frac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\frac{\sqrt{3}+4}{5-2\sqrt{3}}}\)
\(B=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+....+\frac{1}{\sqrt{99}+\sqrt{100}}\)
\(D=\sqrt{1+\sqrt{3+\sqrt{13+4\sqrt{3}}}}+\sqrt{1-\sqrt{3-\sqrt{13-4\sqrt{3}}}}\)
Bài 2 : Cho \(P=\left(\frac{1}{\sqrt{x}-1}+\frac{x-\sqrt{x}+6}{x+\sqrt{x}-2}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{x-\sqrt{x}-2}{x+\sqrt{x}+2}\right)\)
a, Rút gọn P
b, Tìm GTNN
c, Tìm x để \(P.\frac{x-1}{x^2+8x}< -2\)
Tìm GTNN của biểu thức sau :
\(\sqrt{\text{x-1}\text{-2}\sqrt{\text{x-2}}}-\sqrt{\text{x+7}\text{-6}\sqrt{\text{x-2}}}\)
Bài 1: Tìm GTNN: \(A=\sqrt{x-1-2\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}}\)
Bài 2: giải phương trình: \(\frac{1}{\left(x-1\right)^2}+\sqrt{3x+1}=\frac{1}{x^2}+\sqrt{x+2}\)
Cho số thực x. Tìm GTNN của:
A=\(\sqrt{x-1-2\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}}\)
\(A=\sqrt{x-1-2\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}}\)
\(A=\sqrt{x-2-2\sqrt{x-2}+1}+\sqrt{x-2-6\sqrt{x-2}+9}\)
\(A=\sqrt{\left(\sqrt{x-2}-1\right)^2}+\sqrt{\left(\sqrt{x-2}-3\right)^2}\)
\(A=\left|\sqrt{x-2}-1\right|+\left|\sqrt{x-2}-3\right|\)
\(A=\left|\sqrt{x-2}-1\right|+\left|3-\sqrt{x-2}\right|\)
\(A\ge\left|\sqrt{x-2}-1+3-\sqrt{x-2}\right|=\left|2\right|=2\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(\sqrt{x-2}-1\right)\left(3-\sqrt{x-2}\right)\ge0\)
TH1 : \(\hept{\begin{cases}\sqrt{x-2}-1\ge0\\3-\sqrt{x-2}\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge3\\x\le11\end{cases}\Leftrightarrow}3\le x\le11}\)
TH2 : \(\hept{\begin{cases}\sqrt{x-2}-1\le0\\3-\sqrt{x-2}\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le3\\x\ge11\end{cases}}}\) ( loại )
Vậy GTNN của \(A\) là \(2\) khi \(3\le x\le11\)
Chúc bạn học tốt ~
1) Tính tổng \(S=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
2) Giải phương trình sau : \(\left(x^2-x+1\right)^4-\left(x^2+1\right)\left(x^2-x+1\right)^2+x^2=0\)
\(S=\frac{-1+\sqrt{2}}{2-1}+\frac{-\sqrt{2}+\sqrt{3}}{3-2}+...+\frac{-\sqrt{99}+\sqrt{100}}{100-99}\)
\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-....-\sqrt{99}+\sqrt{100}\)
\(=-1+\sqrt{100}\)
\(\hept{\begin{cases}a=\left(x^2-x+1\right)^2\\b=x^2\end{cases}}\)
\(a^2-\left(b+1\right)a+b=0\Leftrightarrow\left(a-1\right)\left(a-b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=1\\a=b\end{cases}\Leftrightarrow}\orbr{\begin{cases}\left(x^2-x+1\right)^2=1\\\left(x^2-x+1\right)^2=x^2\end{cases}}\)(easy)