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có 1 nghiệm là 2

Trả lời:

a, \(B=\left(\frac{x+\sqrt{x}-1}{x\sqrt{x}-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\right):\frac{1}{\sqrt{x}-1}\left(ĐK:x>0;x\ne1\right)\)

\(=\left(\frac{x+\sqrt{x}-1}{\left(\sqrt{x}\right)^3-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\right):\frac{1}{\sqrt{x}-1}\)

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

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

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

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

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

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

b, \(B< \frac{1}{3}\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}< \frac{1}{3}\)

\(\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{3}< 0\)

\(\Leftrightarrow\frac{3\sqrt{x}}{3\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{3\left(x+\sqrt{x}+1\right)}< 0\)

\(\Leftrightarrow\frac{3\sqrt{x}-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}< 0\)

\(\Leftrightarrow\frac{-x+2\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}< 0\)

\(\Leftrightarrow\frac{-\left(x-2\sqrt{x}+1\right)}{3\left(x+\sqrt{x}+1\right)}< 0\)

\(\Leftrightarrow\frac{-\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\)

Vì  \(-\left(\sqrt{x}-1\right)^2< 0\) với mọi \(x>0;x\ne1\)

      \(3\left(x+\sqrt{x}+1\right)>0\) với mọi  \(x>0;x\ne1\)

\(\Rightarrow\frac{-\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\)  luôn đúng với mọi \(x>0;x\ne1\)

Vậy \(B< \frac{1}{3}\)

c, \(B=\frac{1}{2\sqrt{x}+1}\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}=\frac{1}{2\sqrt{x}+1}\)

\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)=x+\sqrt{x}+1\)

\(\Leftrightarrow2x+\sqrt{x}=x+\sqrt{x}+1\)

\(\Leftrightarrow x=1\) (tm)

Vậy x = 1 là giá trị cần tìm 

Áp dụng công thức tính diện tích và lập tỉ số ta có:

\(\hept{\begin{cases}\frac{x}{y+3}=\frac{2}{4}=\frac{1}{2}\\\frac{y}{x+2}=\frac{3}{4}\end{cases}}\Leftrightarrow\hept{\begin{cases}2x-y=3\\3x-4y=-6\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{18}{5}\\y=\frac{21}{5}\end{cases}}\)

Vậy phần diện tích cần tìm là \(x+y=\frac{18}{5}+\frac{21}{5}=\frac{39}{5}\)

C/m tổng quát : \(A=\left(a+1\right)\left(a^2+1\right)\left(a^4+1\right)\left(a^8+1\right)...\left(a^{2^n}+1\right)=\frac{a^{2^{n+1}}-1}{a-1}\)

Có : \(A=\frac{\left(a+1\right)\left(a-1\right)}{a-1}.\frac{\left(a^2+1\right)\left(a^2-1\right)}{a^2-1}.\frac{\left(a^4+1\right)\left(a^4-1\right)}{a^4-1}...\frac{\left(a^{2^n}+1\right)\left(a^{2^n}-1\right)}{a^{2^n}-1}\)

\(=\frac{\left(a^2-1\right)\left(a^4-1\right)\left(a^8-1\right)...\left(a^{2^{n+1}}-1\right)}{\left(a-1\right)\left(a^2-1\right)\left(a^4-1\right)...\left(a^{2^n}-1\right)}=\frac{a^{2^{n+1}}-1}{a-1}\)(đpcm)

Với a = 2 ; n = 11 => \(A=2^{4096}-1\)

A=2^4096-1 nha

HT

k cho mình nha

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Answer:

a) Gọi PT đường thẳng AB là \(y=ax+b\)

Vì A thuộc AB

\(\Rightarrow-7=\frac{-2}{3}a+b\left(\text{*}\right)\)

Vì B thuộc AB

\(\Rightarrow1=2a+b\left(\text{*}\text{*}\right)\)

Từ (*) và (**) \(\Rightarrow\hept{\begin{cases}a=3\\b=-5\end{cases}}\Rightarrow y=3x-5\)

b) Hoành độ giao điểm của hai đường thẳng (P) và AB là \(-2x^2=3x-5\)

\(\Leftrightarrow2x^2+3x-5=0\)

\(\Leftrightarrow\hept{\begin{cases}x=-\frac{5}{2}\Rightarrow y=-\frac{25}{2}\Rightarrow B\left(-\frac{5}{2};-\frac{25}{2}\right)\\x=1\Rightarrow y=-2\Rightarrow C\left(1;-2\right)\end{cases}}\)