Theo vi ét: \(\left\{{}\begin{matrix}x_1+x_2=6\\x_1x_2=8\end{matrix}\right.\)

Theo đề:

\(B=\dfrac{x_1\sqrt{x_1}-x_2\sqrt{x_2}}{x_1-x_2}=\dfrac{\left(\sqrt{x_1}-\sqrt{x_2}\right)\left(x_1+\sqrt{x_1x_2}+x_2\right)}{\left(\sqrt{x_1}-\sqrt{x_2}\right)\left(\sqrt{x_1}+\sqrt{x_2}\right)}\left(x_1,x_2\ge0\right)\)

\(=\dfrac{6+\sqrt{8}}{\sqrt{x_1}+\sqrt{x_2}}\)

Tính: \(\left(\sqrt{x_1}+\sqrt{x_2}\right)^2=x_1+x_2+2\sqrt{x_1x_2}=6+2\sqrt{8}=6+4\sqrt{2}=\left(\sqrt{4}+\sqrt{2}\right)^2\)

\(\Rightarrow\sqrt{x_1}+\sqrt{x_2}=\sqrt{4}+\sqrt{2}\) (thỏa mãn \(x_1,x_2\ge0\))

Khi đó: \(P=\dfrac{6+\sqrt{8}}{\sqrt{4}+\sqrt{2}}=4-\sqrt{2}\)

\(\Delta=\left(-5\right)^2-4\left(m-1\right)\)

   \(=25-4m+4\)

   \(=29-4m\)

Để pt có 2 nghiệm thì \(\Delta>0\)

                                    \(\Leftrightarrow m< \dfrac{29}{4}\)

Theo hệ thức Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=5\\x_1x_2=m-1\end{matrix}\right.\) (1)

\(2x_2=\sqrt{x_1}\) ; \(ĐK:x_1;x_2\ge0\)

\(\Leftrightarrow4x_2^2=\left|x_1\right|\)

\(\Leftrightarrow4x_2^2=x_1\) (2)

Thế \(x_1=4x^2_2\) vào \(\left(1\right)\), ta được:

\(\left\{{}\begin{matrix}4x_2^2+x_2-5=0\\4x_2^3-m+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x_2=-\dfrac{5}{4}\left(ktm\right)\\x_2=1\left(tm\right)\end{matrix}\right.\\4.1^3-m+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_2=1\\m=5\end{matrix}\right.\)

\(\left(2\right)\Rightarrow x_1=4\)

Vậy \(\left\{{}\begin{matrix}m=5\\x_1=4\\x_2=1\end{matrix}\right.\)

Lời giải:

Để pt có 2 nghiệm dương phân biệt thì:

\(\left\{\begin{matrix} \Delta=25-4(m-2)>0\\ S=5>0\\ P=m-2>0\end{matrix}\right.\Leftrightarrow 2< m< \frac{33}{4}\)

Khi đó:

\(2\left(\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}\right)=3\Leftrightarrow 4(\frac{1}{x_1}+\frac{1}{x_2}+\frac{2}{\sqrt{x_1x_2}})=9\)

\(\Leftrightarrow 4\left(\frac{5}{m-2}+\frac{2}{\sqrt{m-2}}\right)=9\)

\(\Leftrightarrow 4(5t^2+2t)=9\) với $t=\frac{1}{\sqrt{m-2}}$

$\Rightarrow t=\frac{1}{2}$

$\Leftrightarrow m=6$ (thỏa)

a: A=x1+x2=-5/2

b: \(=\dfrac{x_1+x_2}{x_1x_2}=\dfrac{-5}{2}:\left(-1\right)=\dfrac{5}{2}\)

c: \(=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\)

\(=\left(-\dfrac{5}{2}\right)^3-3\cdot\dfrac{-5}{2}\cdot\left(-1\right)\)

\(=-\dfrac{125}{8}-\dfrac{15}{2}=\dfrac{-185}{8}\)

e: \(E=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\)

\(=\sqrt{\left(-\dfrac{5}{2}\right)^2-4\cdot\left(-1\right)}=\sqrt{\dfrac{25}{4}+4}=\dfrac{\sqrt{41}}{2}\)

1. Theo hệ thức Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{4}{3}\\x_1.x_2=\dfrac{1}{3}\end{matrix}\right.\)

\(C=\dfrac{x_1}{x_2-1}+\dfrac{x_2}{x_1-1}=\dfrac{x_1\left(x_1-1\right)+x_2\left(x_2-1\right)}{\left(x_1-1\right)\left(x_2-1\right)}\)

   \(=\dfrac{x_1^2-x_1+x_2^2-x_2}{x_1x_2-x_1-x_2+1}=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2-\left(x_1+x_2\right)}{x_1x_2-\left(x_1+x_2\right)+1}\)

  \(=\dfrac{\left(-\dfrac{4}{3}\right)^2-2.\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)}{\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)+1}=\dfrac{\dfrac{22}{9}}{\dfrac{8}{3}}=\dfrac{11}{12}\)

\(1,3x^2+4x+1=0\)

Do pt có 2 nghiệm \(x_1,x_2\) nên theo đ/l Vi-ét ta có :

\(\left\{{}\begin{matrix}S=x_1+x_2=\dfrac{-b}{a}=-\dfrac{4}{3}\\P=x_1x_2=\dfrac{c}{a}=\dfrac{1}{3}\end{matrix}\right.\)

Ta có :

\(C=\dfrac{x_1}{x_2-1}+\dfrac{x_2}{x_1-1}\)

\(=\dfrac{x_1\left(x_1-1\right)+x_2\left(x_2-1\right)}{\left(x_2-1\right)\left(x_1-1\right)}\)

\(=\dfrac{x_1^2-x_1+x_2^2-x_2}{x_1x_2-x_2-x_1+1}\)

\(=\dfrac{\left(x_1^2+x_2^2\right)-\left(x_1+x_2\right)}{x_1x_2-\left(x_1+x_2\right)+1}\)

\(=\dfrac{S^2-2P-S}{P-S+1}\)

\(=\dfrac{\left(-\dfrac{4}{3}\right)^2-2.\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)}{\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)+1}\)

\(=\dfrac{11}{12}\)

Vậy \(C=\dfrac{11}{12}\)

\(3,3x^2-7x-1=0\)

Do pt có 2 nghiệm \(x_1,x_2\) nên theo đ/l Vi-ét ta có :

\(\left\{{}\begin{matrix}S=x_1+x_2=-\dfrac{b}{a}=\dfrac{7}{3}\\P=x_1x_2=\dfrac{c}{a}=-\dfrac{1}{3}\end{matrix}\right.\)

Ta có :

\(B=\dfrac{2x_2^2}{x_1+x_2}+2x_1\)

\(=\dfrac{2x_2^2+2x_1\left(x_1+x_2\right)}{x_1+x_2}\)

\(=\dfrac{2x_2^2+2x_1^2+2x_1x_2}{x_1+x_2}\)

\(=\dfrac{2\left(x_1^2+x_2^2\right)+2x_1x_2}{x_1+x_2}\)

\(=\dfrac{2\left(S^2-2P\right)+2P}{S}\)

\(=\dfrac{2\left(\dfrac{7}{3}^2-2\left(-\dfrac{1}{3}\right)\right)+2\left(-\dfrac{1}{3}\right)}{\dfrac{7}{3}}\)

\(=\dfrac{104}{21}\)

Vậy \(B=\dfrac{104}{21}\)

Câu 1

a) Xét phương trình : 2x² +5x - 8 = 0

Có \(\Delta=5^2-4.2.\left(-8\right)=89>0\)

=> Phương trình luôn có 2 nghiệm phân biệt x₁, x₂

b) Do phương trình luôn có 2 nghiệm x₁,x₂

=> Theo định lí viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{5}{2}\\x_1.x_2=-4\end{matrix}\right.\)

A = \(\dfrac{2}{x_1}+\dfrac{2}{x_2}=\dfrac{2.x_2}{x_1x_2}+\dfrac{2x_1}{x_1x_2}=\dfrac{2\left(x_1+x_2\right)}{x_1x_2}=\dfrac{2.\left(-\dfrac{5}{2}\right)}{-4}=\dfrac{-5}{-4}=\dfrac{5}{4}\)

Vậy A = \(\dfrac{5}{4}\)

Câu 2

Ta có \(P=\dfrac{a+4\sqrt{a}+4}{\sqrt{x}+2}+\dfrac{4-a}{2-\sqrt{a}}\left(a\ge0;a\ne4\right)\)

\(=\dfrac{\left(2+\sqrt{a}\right)^2}{2+\sqrt{a}}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{2-\sqrt{a}}\)

\(=\sqrt{a}+2+\left(2+\sqrt{a}\right)=2\sqrt{a}+4\)

Vậy P = \(2\sqrt{a}+4\left(a\ge0;a\ne4\right)\)

b) Ta có a² - 7a + 12 = 0

\(\Leftrightarrow a^2-4a-3a+12=0\)

\(\Leftrightarrow a\left(a-4\right)-3\left(a-4\right)=0\)

\(\Leftrightarrow\left(a-4\right)\left(a-3\right)=0\Leftrightarrow\left[{}\begin{matrix}a=4\left(loại\right)\\a=3\end{matrix}\right.\)

Với a = 3 thay vào P ta được P = \(2\sqrt{3}+4\)

\(\Rightarrow\sqrt{P}=\sqrt{2\sqrt{3}+4}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

Vậy \(\sqrt{P}=\sqrt{3}+1\) tại a² -7a + 12 =0

Câu 3

Xét hệ phương trình: \(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{3}{2}\\3x-2y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=3y\\3x-2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3y=0\\3x-2y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x-9y=0\\6x-4y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5y=-10\\2x-3y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=3\end{matrix}\right.\)

Vậy hpt có nghiệm  (x;y) = (2;3)

\(x^2-\left(m+1\right)x+m+4=0\left(1\right)\)

\(\Rightarrow\Delta>0\Leftrightarrow\left(m+1\right)^2-4\left(m+4\right)>0\Leftrightarrow\left[{}\begin{matrix}m< -3\\m>5\end{matrix}\right.\)\(\left(2\right)\)

\(ddkt-thỏa:\sqrt{x1}+\sqrt{x2}=2\sqrt{3}\)

\(x1=0\Rightarrow\left(1\right)\Leftrightarrow m=-4\Rightarrow\left(1\right)\Leftrightarrow x^2+3x=0\Leftrightarrow\left[{}\begin{matrix}x1=0\\x2=-3< 0\left(loại\right)\end{matrix}\right.\)

\(x1\ne0\) \(\Rightarrow0< x1< x2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x1+x2>0\\x1x2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m+1>0\\m+4>0\end{matrix}\right.\)\(\Rightarrow m>-1\)\(\left(3\right)\)

\(\left(2\right)\left(3\right)\Rightarrow m>5\)

\(\Rightarrow\sqrt{x1}+\sqrt{x2}=2\sqrt{3}\)

\(\Leftrightarrow x1+x2+2\sqrt{x1x2}=12\Leftrightarrow m+1+2\sqrt{m+4}=12\)

\(\Leftrightarrow m+4+2\sqrt{m+4}-15=0\)

\(đặt:\sqrt{m+4}=t>5\Rightarrow t^2+2t-15=0\Leftrightarrow\left[{}\begin{matrix}t=-5\left(ktm\right)\\t=3\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow m\in\phi\)

Để pt có 2 nghiệm pb 

\(\left(m+1\right)^2-4\left(m+4\right)=m^2+2m+1-4m-16\)

\(=m^2-2m-15>0\)

Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=m+1\\x_1x_2=m+4\end{matrix}\right.\)

Ta có : \(\left(\sqrt{x_1}+\sqrt{x_2}\right)^2=12\Leftrightarrow x_1+2\sqrt{x_1x_2}+x_2=12\)

Thay vào ta được \(m+1+2\sqrt{m+4}=12\Leftrightarrow2\sqrt{m+4}=11-m\)đk : m >= -4 

\(\Leftrightarrow4\left(m+4\right)=121-22m+m^2\Leftrightarrow m^2-26m+105=0\)

\(\Leftrightarrow m=21\left(ktm\right);m=5\left(ktm\right)\)

a. Với m=6 thì phương trình (1) có dạng 

x^2 - 5x +4= 0

<=> (x-1)(x-4)=0

<=> x=1 hoặc x=4

Vậy m=6 thì phương trình có nghiệm x=1 hoặc x=4

b. Xét \(\text{ Δ}=\left(-5\right)^2-4\cdot1\cdot\left(m-2\right)=33-4m\)

Để (1) có nghiệm phân biệt khi \(m< \dfrac{33}{4}\)

Theo Vi-et ta có: \(x_1x_2=m-2;x_1+x_2=5\)

Để 2 nghiệm phương trình (1) dương khi m>2

Ta có:

\(\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}=\dfrac{3}{2}\Leftrightarrow\dfrac{1}{x_1}+\dfrac{1}{x_2}+\dfrac{2}{\sqrt{x_1x_2}}=\dfrac{9}{4}\\ \Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}+\dfrac{2}{\sqrt{x_1x_2}}=\dfrac{9}{4}\\ \Leftrightarrow\dfrac{5}{m-2}+\dfrac{2}{\sqrt{m-2}}=\dfrac{9}{4}\Leftrightarrow20+8\sqrt{m-2}=9\left(m-2\right)\\ \Leftrightarrow\left(\sqrt{m-2}-2\right)\left(9\sqrt{m-2}+10\right)=0\Leftrightarrow\sqrt{m-2}=2\Leftrightarrow m-2=4\Leftrightarrow m=6\left(t.m\right)\)

a: Khi m = -4 thì:

\(x^2-5x+\left(-4\right)-2=0\)

\(\Leftrightarrow x^2-5x-6=0\)

\(\Delta=\left(-5\right)^2-5\cdot1\cdot\left(-6\right)=49\Rightarrow\sqrt{\Delta}=\sqrt{49}=7>0\)

Pt có 2 nghiệm phân biệt:

\(x_1=\dfrac{5+7}{2}=6;x_2=\dfrac{5-7}{2}=-1\)

b: \(\Delta=\left(-5\right)^2-4\left(m-2\right)=25-4m+8=33-4m\)

Theo viet:

\(x_1+x_2=-\dfrac{b}{a}=5\)

\(x_1x_2=\dfrac{c}{a}=m-2\)

Để pt có 2 nghiệm dương phân biệt:

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\x_1+x_2>0\\x_1x_2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}33-4m>0\\5>0\left(TM\right)\\m-2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< \dfrac{33}{4}\\x>2\end{matrix}\right.\Leftrightarrow m=2< m< \dfrac{33}{4}\)

Vậy \(2< m< \dfrac{33}{4}\) thì pt có 2 nghiệm dương phân biệt.

Theo đầu bài: \(\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}=\dfrac{3}{2}\)

\(\Leftrightarrow\sqrt{x_1}+\sqrt{x_2}=\dfrac{3}{2}\left(\sqrt{x_1x_2}\right)\)

\(\Leftrightarrow\left(\sqrt{x_1}+\sqrt{x_2}\right)^2=\dfrac{9}{4}x_1x_2\)

\(\Leftrightarrow x_1+2\sqrt{x_1x_2}+x_2=\dfrac{9}{4}x_1x_2\)

\(\Leftrightarrow x_1+x_2+2\sqrt{x_1x_2}=\dfrac{9}{4}x_1x_2\)

\(\Leftrightarrow5+2\sqrt{x_1x_2}=\dfrac{9}{4}\left(m-2\right)\)

\(\Leftrightarrow\dfrac{9}{4}\left(m-2\right)-2\sqrt{m-2}-5=0\)

Đặt \(\sqrt{m-2}=t\Rightarrow m-2=t^2\)

\(\Rightarrow\dfrac{9}{4}t^2-2t-5=0\)

\(\Leftrightarrow\dfrac{9}{4}t^2-2+\left(-5\right)=0\)

\(\Leftrightarrow\left(t-2\right)\left(9t+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t-2=0\\9t+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=2\left(TM\right)\\t=-\dfrac{10}{9}\left(\text{loại}\right)\end{matrix}\right.\)

Trả ẩn:

\(\sqrt{m-2}=2\)

\(\Rightarrow m-2=4\)

\(\Rightarrow m=6\)

Vậy m = 6 thì x₁ , x₂ thoả mãn hệ thức \(2\left(\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}\right)=\dfrac{3}{2}\).