Ptrình :  \(x^2-7x+10=0\)

Ta có : \(\Delta=\left(-7\right)^2-4.1.10=9>0\)

=> Phương trình có 2 nghiệm phân biệt \(x1\) và \(x2\)

\(x1=\dfrac{-\left(-7\right)+\sqrt{\Delta}}{2.1}=\dfrac{7+\sqrt{9}}{2}=5\)

\(x2=\dfrac{-\left(-7\right)-\sqrt{\Delta}}{2.1}=\dfrac{7-\sqrt{9}}{2}=2\)

Vậy :

A = \(x_1^2+x_2^2+3x_1x_2=5^2+2^2+3.5.2=59\)  

B = .................

.... (có x1 và x2 rồi thik thay vào lak tính đc, cái này bn tự tính nha)

Câu 1 :

\(P=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

Câu 2 :

Ta có :

\(\Delta=m^2+16>0\)

\(=>\) phương trình có 2 nghiệm phân biệt .

Theo định lý vi-ét ta có :

\(\left\{{}\begin{matrix}x_1+x_2=m\\x_1.x_2=-4\end{matrix}\right.\)

Thay vào ta được :

\(\dfrac{2m+7}{m^2+8}\ge-\dfrac{1}{8}\)

\(\Leftrightarrow16m+56\ge-m^2-8\)

\(\Leftrightarrow m^2+16m+64\ge0\)

\(\Leftrightarrow\left(m+8\right)^2\ge0\) ( đúng )

Lời giải:

Theo định lý Viet:

$x_1+x_2=19$

$x_1x_2=9$

Khi đó:
\(x_1\sqrt{x_1}+x_2\sqrt{x_2}=(\sqrt{x_1})^3+(\sqrt{x_2})^3=(\sqrt{x_1}+\sqrt{x_2})(x_1-\sqrt{x_1x_2}+x_2)\)

\(=(\sqrt{x_1}+\sqrt{x_2})(19-\sqrt{9})=16(\sqrt{x_1}+\sqrt{x_2})\)

\(=16\sqrt{x_1+x_2+2\sqrt{x_1x_2}}=16\sqrt{19+2\sqrt{9}}=80\)

\(x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=19^2-2.9=343\)

$\Rightarrow P=\frac{80}{343}$

\(đk:\left\{{}\begin{matrix}\Delta\ge0\\0< x1\le x2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5^2-4\left(-m^2+m+6\right)\ge0\\\left\{{}\begin{matrix}x1+x2>0\\x1x2>0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4m^2-4m+1=\left(2m-1\right)^2\ge0\left(đúng\right)\\\left\{{}\begin{matrix}5>0đúng\\-m^2+m+6>0\Leftrightarrow-2< m< 3\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-2< m< 3\)

\(\Rightarrow\dfrac{1}{\sqrt{x1}}+\dfrac{1}{\sqrt{x2}}=\dfrac{3}{2}\Leftrightarrow\dfrac{\sqrt{x1}+\sqrt{x2}}{\sqrt{x1x2}}=\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{x1+x2+2\sqrt{x1x2}}{x1x2}=\dfrac{9}{4}\Leftrightarrow\dfrac{5+2\sqrt{-m^2+m+6}}{-m^2+m+6}=\dfrac{9}{4}\)

\(đặt::\sqrt{-m^2+m+6}=t\ge0\Rightarrow\dfrac{5+2t}{t^2}=\dfrac{9}{4}\)

\(\Rightarrow9t^2-8t-20=0\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-\dfrac{10}{9}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{-m^2+m+6}=2\Leftrightarrow\left[{}\begin{matrix}m=2\left(tm\right)\\m=-1\left(tm\right)\end{matrix}\right.\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=\sqrt{5}\\x_1x_2=1\end{matrix}\right.\)

\(A=\left(x_1+x_2\right)^2-5x_1x_2=\left(\sqrt{5}\right)^2-5.1=0\)

\(B=\frac{1}{\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)}=\frac{1}{\left(\sqrt{5}\right)^3-3.1.\sqrt{5}}=\frac{1}{2\sqrt{5}}\)

\(C=\frac{x_1+x_2}{x_1x_2}=\sqrt{5}\)

\(D=\frac{x_1^2+x_2^2}{\left(x_1x_2\right)^2}=\frac{\left(x_1+x_2\right)^2-2x_1x_2}{\left(x_1x_2\right)^2}=\frac{5-2}{1^2}=3\)

\(E=\sqrt{x_1x_2}\left(\sqrt{x_1}+\sqrt{x_2}\right)\Rightarrow E^2=x_1x_2\left(x_1+x_2+2\sqrt{x_1x_2}\right)\)

\(\Rightarrow E^2=1\left(\sqrt{5}+2.1\right)\Rightarrow E=\sqrt{2+\sqrt{5}}\)

\(F=\frac{3\left(x_1+x_2\right)+5x_1x_2}{x_1x_2\left(x_1^2+x_2^2\right)}=\frac{3\left(x_1+x_2\right)-5x_1x_2}{x_1x_2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]}=\frac{3\sqrt{5}-5}{3}\)

Để (1) có 2 nghiệm dương \(\Rightarrow\left\{{}\begin{matrix}\Delta'=\left(m+3\right)^2-m-1\ge0\\x_1+x_2=2\left(m+3\right)>0\\x_1x_2=m+1>0\end{matrix}\right.\) \(\Rightarrow m>-1\)

\(P=\left|\dfrac{\sqrt{x_1}-\sqrt{x_2}}{\sqrt{x_1x_2}}\right|>0\Rightarrow P^2=\dfrac{\left(\sqrt{x_1}-\sqrt{x_2}\right)^2}{x_1x_2}\)

\(P^2=\dfrac{x_1+x_2-2\sqrt{x_1x_2}}{x_1x_2}=\dfrac{2\left(m+3\right)-2\sqrt{m+1}}{m+1}=\dfrac{4}{m+1}-\dfrac{2}{\sqrt{m+1}}+2\)

\(P^2=\left(\dfrac{2}{\sqrt{m+1}}-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\Rightarrow P\ge\dfrac{\sqrt{7}}{2}\)

Dấu "=" xảy ra khi \(\sqrt{m+1}=4\Rightarrow m=15\)

(căn x1+căn x2)^2=x1+x2+2*căn x1x2

=12+2*căn 4=16

=>căn x1+căn x2=4

\(T=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{4}=\dfrac{12^2-2\cdot4}{4}=34\)

Lời giải:

Để pt có 2 nghiệm $x_1,x_2$ thì:

$\Delta'=1+(3+m)=4+m\geq 0\Leftrightarrow m\geq -4$ (chứ không phải với mọi m như đề bạn nhé)!

Áp dụng định lý Viet: \(\left\{\begin{matrix} x_1+x_2=-2\\ x_1x_2=-(m+3)\end{matrix}\right.\)

$x_1, x_2\neq 0\Leftrightarrow -(m+3)\neq 0\Leftrightarrow m\neq -3$

$\frac{x_1}{x_2}-\frac{x_2}{x_1}=\frac{-8}{3}$

$\Leftrightarrow \frac{x_1^2-x_2^2}{x_1x_2}=\frac{-8}{3}$

$\Leftrightarrow \frac{-2(x_1-x_2)}{-(m+3)}=\frac{-8}{3}$
$\Leftrightarrow x_1-x_2=\frac{4}{3}(m+3)$

$\Rightarrow (x_1-x_2)^2=\frac{16}{9}(m+3)^2$

$\Leftrightarrow (x_1+x_2)^2-4x_1x_2=\frac{16}{9}(m+3)^2$
$\Leftrightarrow 4+4(m+3)=\frac{16}{9}(m+3)^2$

$\Leftrightarrow m+3=3$ hoặc $m+3=\frac{-3}{4}$

$\Leftrightarrow m=0$ hoặc $m=\frac{-15}{4}$ (đều thỏa mãn)