Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, Thay m=2 vào pt ta có:
(1)\(\Leftrightarrow2x^2+\left(2.2-1\right)x+2-1=0\)
\(\Leftrightarrow2x^2+3x+1=0\\ \Leftrightarrow\left(2x^2+2x\right)+\left(x+1\right)=0\\ \Leftrightarrow2x\left(x+1\right)+\left(x+1\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=-1\end{matrix}\right.\)
b,\(\Delta=\left(2m-1\right)^2-4.2\left(m-1\right)=4m^2-4m+1-8\left(m-1\right)=4m^2-4m+1-8m+8=4m^2-12m+9\)
Để pt có 2 nghiệm thì \(\Delta\ge0\Leftrightarrow4m^2-12m+9\ge0\left(luôn.đúng\right)\)
Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{1-2m}{2}\\x_1x_2=\dfrac{m-1}{2}\end{matrix}\right.\)
\(4x^2_1+4x^2_2+2x_1x_2=1\\ \Leftrightarrow4\left(x^2_1+x^2_2\right)+2.\dfrac{m-1}{2}=1\\ \Leftrightarrow4\left(x_1+x_2\right)^2-8x_1x_2+m-1=1\\ \Leftrightarrow4.\left(\dfrac{1-2m}{2}\right)^2-8.\dfrac{m-1}{2}+m-2=0\)
\(4.\dfrac{\left(1-2m\right)^2}{4}-4\left(m-1\right)+m-2=0\\ \Leftrightarrow4\left(1-4m+4m^2\right)-4m+4+m-2=0\\ \Leftrightarrow4-16m+16m^2-3m+2=0\\ \Leftrightarrow16m^2-19m+6=0\)
Ta có:\(\Delta=\left(-19\right)^2-4.16.6=361-384=-23< 0\)
Suy ra pt vô nghiệm
c) Ta có: \(\text{Δ}=\left[-2\left(m+1\right)\right]^2-4\cdot1\cdot\left(2m+1\right)\)
\(=\left(-2m-2\right)^2-4\left(2m+1\right)\)
\(=4m^2+8m+4-8m-4\)
\(=4m^2\ge0\forall m\)
Do đó, phương trình luôn có nghiệm
Áp dụng hệ thức Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{2\left(m+1\right)}{1}=2m+2\\x_1\cdot x_2=2m+1\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1-2x_2=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x_2=2m-1\\x_1=2m+2+x_2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_2=\dfrac{2m-1}{3}\\x_1=2m+3+\dfrac{2m-1}{3}=\dfrac{8m+8}{3}\end{matrix}\right.\)
Ta có: \(x_1\cdot x_2=2m+1\)
\(\Leftrightarrow\dfrac{2m-1}{3}\cdot\dfrac{8m+8}{3}=2m+1\)
\(\Leftrightarrow\left(2m-1\right)\left(8m+8\right)=9\left(2m+1\right)\)
\(\Leftrightarrow16m^2+16m-8m-8-18m-9=0\)
\(\Leftrightarrow16m^2-10m-17=0\)
\(\text{Δ}=\left(-10\right)^2-4\cdot16\cdot\left(-17\right)=1188\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}m_1=\dfrac{10-6\sqrt{33}}{32}\\m_2=\dfrac{10+6\sqrt{33}}{32}\end{matrix}\right.\)
\(\Delta=\left(2m+1\right)^2-4\left(m^2+m-2\right)=9>0;\forall m\)
Phương trình luôn có 2 nghiệm pb với mọi m
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2m+1\\x_1x_2=m^2+m-2\end{matrix}\right.\)
\(x_1\left(x_1-2x_2\right)+x_2\left(x_2-2x_1\right)=9\)
\(\Leftrightarrow x_1^2+x_2^2-4x_1x_2=9\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-6x_1x_2=9\)
\(\Leftrightarrow\left(2m+1\right)^2-6\left(m^2+m-4\right)=9\)
\(\Leftrightarrow2m^2+2m-4=0\)
\(\Rightarrow\left[{}\begin{matrix}m=1\\m=-2\end{matrix}\right.\)
b: \(\text{Δ}=\left(2m+3\right)^2-4\left(4m+2\right)\)
\(=4m^2+12m+9-16m-8\)
\(=4m^2-4m+1=\left(2m-1\right)^2>=0\)
Do đó: Phương trình luôn có hai nghiệm
Theo đề, ta có:
\(\left\{{}\begin{matrix}2x_1-5x_2=6\\x_1+x_2=2m+3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x_1-5x_2=6\\2x_1+2x_2=4m+6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-7x_2=-4m\\2x_1=5x_2+6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_2=\dfrac{4}{7}m\\2x_1=\dfrac{20}{7}m+6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_2=\dfrac{4}{7}m\\x_1=\dfrac{10}{7}m+3\end{matrix}\right.\)
Theo đề, ta có: \(x_1x_2=4m+2\)
\(\Rightarrow4m+2=\dfrac{40}{49}m^2+\dfrac{12}{7}m\)
\(\Leftrightarrow m^2\cdot\dfrac{40}{49}-\dfrac{16}{7}m-2=0\)
\(\Leftrightarrow40m^2-112m-98=0\)
\(\Leftrightarrow40m^2-140m+28m-98=0\)
=>\(20m\left(2m-7\right)+14\left(2m-7\right)=0\)
=>(2m-7)(20m+14)=0
=>m=7/2 hoặc m=-7/10
Lời giải:
Để pt có 2 nghiệm $x_1,x_2$ thì:
$\Delta'=(m-1)^2+2m+1=m^2+2\geq 0$
$\Leftrightarrow m\in\mathbb{R}$
Áp dụng định lý Viet:
$x_1+x_2=2(m-1)$
$x_1x_2=-2m-1$
Khi đó:
$2x_1+3x_2+3x_1x_2=-11$
$\Leftrightarrow 2(x_1+x_2)+3x_1x_2+x_2=-11$
$\Leftrightarrow 4(m-1)+3(-2m-1)+x_2=-11$
$\Leftrightarrow x_2=2m-4$
$x_1=2(m-1)-x_2=2m-2-(2m-4)=2$
$-2m-1=x_1x_2=2(2m-4)$
$\Leftrightarrow -2m-1=4m-8$
$\Leftrightarrow 7=6m$
$\Leftrightarrow m=\frac{7}{6}$
Lời giải:
Để pt có 2 nghiệm $x_1,x_2$ thì:
$\Delta'=(m-1)^2+2m+1=m^2+2\geq 0$
$\Leftrightarrow m\in\mathbb{R}$
Áp dụng định lý Viet:
$x_1+x_2=2(m-1)$
$x_1x_2=-2m-1$
Khi đó:
$2x_1+3x_2+3x_1x_2=-11$
$\Leftrightarrow 2(x_1+x_2)+3x_1x_2+x_2=-11$
$\Leftrightarrow 4(m-1)+3(-2m-1)+x_2=-11$
$\Leftrightarrow x_2=2m-4$
$x_1=2(m-1)-x_2=2m-2-(2m-4)=2$
$-2m-1=x_1x_2=2(2m-4)$
$\Leftrightarrow -2m-1=4m-8$
$\Leftrightarrow 7=6m$
$\Leftrightarrow m=\frac{7}{6}$
`Delta=(2m-1)^2-8(m-1)`
`=4m^2-4m+1-8m+8`
`=4m^2-12m+9`
`=(2m-3)^2>=0(AA m)`
Vi-ét:
`x_1+x_2=(1-2m)/2,x_1.x_2=(m-1)/2`
`4x_1+4x_2^2+2x_1x_2=1`
`<=>(2x_1+2x_2)^2-6x_1x_2=1`
`<=>(2m-1)^2-3(m-1)-1=0`
`<=>4m^2-4m+1-3m+3-1=0`
`<=>4m^2-7m+3=0`
`<=>m=1\or\m=3/4`