tại lộn Ko thấy phân biệt

Gọi 2 số đó là a và b (a\(\ge0,b\ge0\) )

 câu a

Áp dụng  BĐT Bu-nhia -xkop-ki ,ta có

a+b\(\le\sqrt{\left(a^2+b^2\right)\left(1^2+1^2\right)}\)

\(\Leftrightarrow82\le\sqrt{\left(a^2+b^2\right)2}\) \(\Rightarrow\) \(6724\le\left(a^2+b^2\right)2\Leftrightarrow\left(a^2+b^2\right)\ge3362\)

Vậy Min a²+b²=3362\(\Leftrightarrow a=b=41\)

(m-1)x²-2mx+m-2=0(m\(\ne1\) )

\(\Delta\)^'=\(m^2-\left(m-2\right)\left(m-1\right)\)

   =\(m^2-m^2+m+2m-2\)

 =3m-2

Để pt có nghiệm 2 ngiệm trái dấu thì \(\Delta\) ^' =3m-2>0\(\Leftrightarrow m>\dfrac{2}{3}\)

Áp dụng hệ thức Viet, ta có 

\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{2m}{m-1}\\x_1.x_2=\dfrac{m-2}{m-1}\end{matrix}\right.\)

Để PT có 2 nghiệm trái dấu thì x₁x₂<0\(\Leftrightarrow\dfrac{m-2}{m-1}< 0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}m-2< 0\\m-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}m-2>0\\m-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}m< 2\\m>1\end{matrix}\right.\\\left\{{}\begin{matrix}m>2\\m< 1\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow1< m< 2\)

Vậy 1<m<2 thì pt có 2 nghiệm trái dấu 

câu b

.Với m=1\(\Rightarrow-2x-1=0\Leftrightarrow x=\dfrac{-1}{2}\left(l\right)\)

.Với \(m\ne1\)

\(\Rightarrow\Delta\)^'=3m-2\(\ge0\Leftrightarrow m\ge\dfrac{2}{3}\)

A=\(\dfrac{1}{x^2+x}+\dfrac{1}{x^2-3x+2}=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{x^2-x-2x+2}\)

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

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

rút gọn 

B=9-/5x\

Ta có :

/5x\\(\ge0\) \(\Rightarrow\) -/5x\\(\le0\)

\(\Rightarrow\)9-/5x\\(\le9\)

 Vậy Max B =9\(\Leftrightarrow x=0\)

\(\dfrac{x}{12457}=9\Leftrightarrow x=9.12457=112113\)

bài giải 

PT\(\Leftrightarrow\dfrac{\left(2^n\right)^3}{2^n.4}=4\)

\(\Leftrightarrow\dfrac{2^{2n}}{4}=4\)

\(\Leftrightarrow\dfrac{4^n}{4}=4\)

\(\Rightarrow n=2\)

x(x+1)-(x-1)(x+2)=8

\(\Leftrightarrow\)\(x^2+x-x^2-2x+x+2=8\)

\(\Leftrightarrow0x=6\left(ptvn\right)\)

\(\Rightarrow S=\varnothing\)

A=\(\dfrac{\sqrt{325}-\sqrt{117}+2\sqrt{208}}{\sqrt{13}}=\dfrac{\sqrt{13.25}-\sqrt{13.9}+2\sqrt{13.16}}{\sqrt{13}}\)

   =\(\dfrac{5\sqrt{13}-3\sqrt{13}+8\sqrt{13}}{\sqrt{13}}=\dfrac{10\sqrt{13}}{\sqrt{13}}=10\)