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\(x^2-\left(m+1\right)x+m+4=0\left(1\right)\)
\(\Rightarrow\Delta>0\Leftrightarrow\left(m+1\right)^2-4\left(m+4\right)>0\Leftrightarrow\left[{}\begin{matrix}m< -3\\m>5\end{matrix}\right.\)\(\left(2\right)\)
\(ddkt-thỏa:\sqrt{x1}+\sqrt{x2}=2\sqrt{3}\)
\(x1=0\Rightarrow\left(1\right)\Leftrightarrow m=-4\Rightarrow\left(1\right)\Leftrightarrow x^2+3x=0\Leftrightarrow\left[{}\begin{matrix}x1=0\\x2=-3< 0\left(loại\right)\end{matrix}\right.\)
\(x1\ne0\) \(\Rightarrow0< x1< x2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x1+x2>0\\x1x2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m+1>0\\m+4>0\end{matrix}\right.\)\(\Rightarrow m>-1\)\(\left(3\right)\)
\(\left(2\right)\left(3\right)\Rightarrow m>5\)
\(\Rightarrow\sqrt{x1}+\sqrt{x2}=2\sqrt{3}\)
\(\Leftrightarrow x1+x2+2\sqrt{x1x2}=12\Leftrightarrow m+1+2\sqrt{m+4}=12\)
\(\Leftrightarrow m+4+2\sqrt{m+4}-15=0\)
\(đặt:\sqrt{m+4}=t>5\Rightarrow t^2+2t-15=0\Leftrightarrow\left[{}\begin{matrix}t=-5\left(ktm\right)\\t=3\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow m\in\phi\)
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
\(\Leftrightarrow4\left(m+4\right)=121-22m+m^2\Leftrightarrow m^2-26m+105=0\)
\(\Leftrightarrow m=21\left(ktm\right);m=5\left(ktm\right)\)
\(a,ĐK:\left\{{}\begin{matrix}x\ge5\\x\le3\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Vậy pt vô nghiệm
\(b,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow0x=2\Leftrightarrow x\in\varnothing\)
\(c,ĐK:x\ge-\dfrac{3}{2}\\ PT\Leftrightarrow x^2+4x+5-2\sqrt{2x+3}=0\\ \Leftrightarrow\left(2x+3-2\sqrt{2x+3}+1\right)+\left(x^2+2x+1\right)=0\\ \Leftrightarrow\left(\sqrt{2x+3}-1\right)^2+\left(x+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x+3=1\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\left(tm\right)\\ d,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
a: =>\(x\cdot\left(\sqrt{3}-1\right)=16\)
=>\(x=\dfrac{16}{\sqrt{3}-1}=8\left(\sqrt{3}+1\right)\)
b: =>(x-căn 15)^2=0
=>x-căn 15=0
=>x=căn 15
Lời giải:
Ta có: $\Delta=(m-3)^2+16>0$ với mọi $m$ nên pt luôn có 2 nghiệm phân biệt $x_1,x_2$ với mọi $m$.
Theo định lý Viet:
$x_1+x_2=m-3$
$x_1x_2=-4$
Có:
$\sqrt{x_1^2+2020}-x_1=\sqrt{x_2^2+2020}+x_2$
$\Leftrightarrow \sqrt{x_1^2+2020}-\sqrt{x_2^2+2020}=x_1+x_2$
$\Leftrightarrow \frac{x_1^2-x_2^2}{\sqrt{x_1^2+2020}+\sqrt{x_2^2+2020}}=x_1+x_2$
$\Leftrightarrow (x_1+x_2)\left[\frac{x_1-x_2}{\sqrt{x_1^2+2020}+\sqrt{x_2^2+2020}}-1\right]=0$
$\Leftrightarrow x_1+x_2=0$ hoặc $x_1-x_2=\sqrt{x_1^2+2020}+\sqrt{x_2^2+2020}$
Với $x_1+x_2=0$
$\Leftrightarrow m-3=0\Leftrightarrow m=3$ (tm)
Với $x_1-x_2=\sqrt{x_1^2+2020}+\sqrt{x_2^2+2020}$
$\Rightarrow (x_1-x_2)^2=(\sqrt{x_1^2+2020}+\sqrt{x_2^2+2020})^2$
$\Leftrightarrow -2x_1x_2=4040+2\sqrt{(x_1^2+2020)(x_2^2+2020)}$
$\Leftrightarrow 8=4040+2\sqrt{(x_1^2+2020)(x_2^2+2020)}$
$\Leftrightarrow \sqrt{(x_1^2+2020)(x_2^2+2020)}=-2016<0$ (vô lý - loại)
Vậy $m=3$
Ta có: \(\Delta=4m^2+4m-11\)
Để phương trình có 2 nghiệm phân biệt \(\Leftrightarrow4m^2+4m-11>0\)
Theo Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=2m+3\\x_1x_2=2m+5\end{matrix}\right.\)
Để phương trình có 2 nghiệm dương phân biệt
\(\Leftrightarrow\left\{{}\begin{matrix}4m^2+4m-11>0\\2m+3>0\\2m+5>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m< \dfrac{-1-2\sqrt{3}}{2}\\m>\dfrac{-1+2\sqrt{3}}{2}\end{matrix}\right.\\m>-\dfrac{3}{2}\\m>-\dfrac{5}{2}\end{matrix}\right.\) \(\Leftrightarrow m>\dfrac{-1+2\sqrt{3}}{2}\)
Mặt khác: \(\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}=\dfrac{4}{3}\)
\(\Rightarrow\dfrac{x_1+x_2+2\sqrt{x_1x_2}}{x_1x_2}=\dfrac{16}{9}\) \(\Rightarrow\dfrac{2m+3+2\sqrt{2m+5}}{2m+5}=\dfrac{16}{9}\)
\(\Rightarrow18m+27+18\sqrt{2m+5}=32m+80\)
\(\Leftrightarrow14m-53=18\sqrt{2m+5}\)
\(\Rightarrow\) ...
2:
\(A=\dfrac{x_2-1+x_1-1}{x_1x_2-\left(x_1+x_2\right)+1}\)
\(=\dfrac{3-2}{-7-3+1}=\dfrac{1}{-9}=\dfrac{-1}{9}\)
B=(x1+x2)^2-2x1x2
=3^2-2*(-7)
=9+14=23
C=căn (x1+x2)^2-4x1x2
=căn 3^2-4*(-7)=căn 9+28=căn 27
D=(x1^2+x2^2)^2-2(x1x2)^2
=23^2-2*(-7)^2
=23^2-2*49=431
D=9x1x2+3(x1^2+x2^2)+x1x2
=10x1x2+3*23
=69+10*(-7)=-1
Ta có: \(\Delta=\left[-\left(m+3\right)\right]^2-4\left(4m-4\right)=m^2+6m+9-16m+16=\left(m-5\right)^2\ge0\)
=> pt luôn có 2 nghiệm x1, x2
=> \(x_1=\frac{-b-\sqrt{\Delta}}{2a}=\frac{m+3-m+5}{2}=4\)
\(x_2=\frac{-b+\sqrt{\Delta}}{2a}=\frac{m+3+m-5}{2}=m-1\)
Theo bài ra, ta có: \(\sqrt{x_1}+\sqrt{x_2}+x_1x_2=20\)
ĐK: \(x_1\ge0\); \(x_2\ge0\) <=> 4 \(\ge\) 0 và m - 1 \(\ge\)0 <=> m \(\ge\)1
<=> \(\sqrt{4}+\sqrt{m-1}+4\left(m-1\right)=20\)
<=> \(\sqrt{m-1}=22-4m\left(m\le\frac{11}{2}\right)\)
<=> \(m-1=16m^2-176m+484\)
<=> \(16m^2-177m+485=0\)
<=> \(16m^2-80m-97m+485=0\)
<=> \(\left(m-5\right)\left(16m-97\right)=0\)
<=> \(\orbr{\begin{cases}m=5\left(tm\right)\\m=\frac{97}{16}\left(ktm\right)\end{cases}}\)
Vậy ...