ko biet

đặt \(A=x\sqrt{6-x}+\left(5-x\right)\sqrt{x+1}\)

\(A=\sqrt{x}\sqrt{x\left(6-x\right)}+\sqrt{5-x}\sqrt{\left(5-x\right)\left(x+1\right)}\)

Áp dụng BĐT bunyakovsky :

\(A^2\le\left(x+5-x\right)\left[x\left(6-x\right)+\left(5-x\right)\left(x+1\right)\right]\)

\(A^2\le5\left(-2x^2+10x+5\right)=5\left[-2\left(x-\frac{5}{2}\right)^2+\frac{35}{2}\right]\)

\(A^2\le\frac{5.35}{2}=\frac{175}{2}=87,5\Leftrightarrow A\le\sqrt{87,5}\)

dấu = xảy ra khi \(\left\{\begin{matrix}x=\frac{5}{2}\\\frac{1}{6-x}=\frac{1}{x+1}\end{matrix}\right.\)<=> x=2,5

vậy Amax=.....

NX \(A=\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}\)

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

\(=\frac{a^2\left(a^2+2a+1+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}\)=\(\frac{a^4+2a^3+2a^2+\left(a+1\right)^2}{a^2\left(a+1\right)^2}\)

\(=\frac{a^4+2a^2\left(a+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}=\frac{\left(a^2+a+1\right)^2}{a^2\left(a+1\right)^2}\)=\(\left[\frac{a^2+a+1}{a\left(a+1\right)}\right]^2\)suy ra A=\(\frac{a^2+a+1}{a\left(a+1\right)}\)

                                                                                                =\(\frac{a\left(a+1\right)+1}{a\left(a+1\right)}=1+\frac{1}{a\left(a+1\right)}=1+\frac{1}{a}-\frac{1}{a+1}\)

ap dung vao bai ta co =\(\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2012}-\frac{1}{2013}\right)\)

=\(2011+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)= \(2011+\frac{1}{2}-\frac{1}{2013}=2011,499503\)

xin lỗi bn mik mới học lớp 6 thôi

a, ĐK: \(x\ge0;x\ne1\)

\(P=\left(1+\dfrac{2}{\sqrt{x}+1}+\dfrac{3}{\sqrt{x}-1}\right).\left(1-\dfrac{6}{\sqrt{x}+5}\right)\)

\(=\left[\dfrac{x-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right].\dfrac{\sqrt{x}+5-6}{\sqrt{x}+5}\)

\(=\dfrac{x+5\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}-1}{\sqrt{x}+5}\)

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

b, \(P=\dfrac{\sqrt{x}}{\sqrt{x}+1}=1-\dfrac{1}{\sqrt{x}+1}\in Z\)

\(\Leftrightarrow\sqrt{x}+1\in\left\{-1;1\right\}\)

\(\Leftrightarrow\sqrt{x}=0\)

\(\Leftrightarrow x=0\left(tm\right)\)

Vậy ta có điều phải chứng minh.

a: Ta có: \(P=\left(1+\dfrac{2}{\sqrt{x}+1}+\dfrac{3}{\sqrt{x}-1}\right)\cdot\left(1-\dfrac{6}{\sqrt{x}+5}\right)\)

\(=\dfrac{x-1+2\sqrt{x}-2+3\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+5-6}{\sqrt{x}+5}\)

\(=\dfrac{x+5\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+5}\)

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

1, Với \(x\ge0,x\ne1\) ta có :

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

   \(=\dfrac{\sqrt{x}+1+\sqrt{x}}{x-1}:\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}\)

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

   \(=\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)

2, Ta có \(P=\dfrac{7}{4}\)

          \(\Rightarrow\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}=\dfrac{7}{4}\)

         \(\Leftrightarrow4\left(2\sqrt{x}+1\right)=7\left(\sqrt{x}+1\right)\)

         \(\Leftrightarrow8\sqrt{x}+4=7\sqrt{x}=7\)

          \(\Leftrightarrow\sqrt{x}=3\)

          \(\Leftrightarrow x=9\left(tm\right)\)

1) Ta có: \(P=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-1\right)\)

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

\(=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}-\sqrt{x}+1}\)

\(=\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)

2) Để \(P=\dfrac{7}{4}\) thì \(\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}=\dfrac{7}{4}\)

\(\Leftrightarrow4\cdot\left(2\sqrt{x}+1\right)=7\left(\sqrt{x}+1\right)\)

\(\Leftrightarrow8\sqrt{x}+4=7\sqrt{x}+7\)

\(\Leftrightarrow8\sqrt{x}-7\sqrt{x}=7-4\)

\(\Leftrightarrow\sqrt{x}=3\)

hay x=9(nhận)

Vậy: Để \(P=\dfrac{7}{4}\) thì x=9