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Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{5}{3}\\x_1x_2=-2\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}y_1+y_2=2x_1-x_2+2x_2-x_1\\y_1y_2=\left(2x_1-x_2\right)\left(2x_2-x_1\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y_1+y_2=x_1+x_2\\y_1y_2=-2x_1^2-2x_2^2+5x_1x_2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y_1+y_2=-\dfrac{5}{3}\\y_1y_2=-2\left(x_1+x_2\right)^2+9x_1x_2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y_1+y_2=-\dfrac{5}{3}\\y_1y_2=-2.\left(-\dfrac{5}{3}\right)^2+9.\left(-2\right)=-\dfrac{212}{9}\end{matrix}\right.\)
\(\Rightarrow y_1;y_2\) là nghiệm của:
\(y^2+\dfrac{5}{3}y-\dfrac{212}{9}=0\Leftrightarrow9y^2+10y-212=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3x^2+5x-6=0\\ \Delta=5^2-4.3.\left(-6\right)=97\\ \Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-5+\sqrt{97}}{2}\\x_2=\dfrac{-5-\sqrt{97}}{2}\end{matrix}\right.\)
\(\left(x_1-2x_2\right).\left(2x_1-x_2\right)=2x^2_1-4x_1x_2+2x_2^2\)
\(=2.\left(\dfrac{-5+\sqrt{97}}{2}\right)^2-4.\left(\dfrac{-5+\sqrt{97}}{2}\right).\left(\dfrac{-5-\sqrt{97}}{2}\right)+2.\left(\dfrac{-5-\sqrt{97}}{2}\right)^2\\ =\left(\dfrac{-5+\sqrt{97}}{2}\right)^2-2.\left(\dfrac{-5+\sqrt{97}}{2}\right).\left(\dfrac{-5-\sqrt{97}}{2}\right)+\dfrac{\left(-5-\sqrt{97}\right)^2}{2^2}\\ =\left(\dfrac{-5+\sqrt{97}}{2}-\dfrac{-5-\sqrt{97}}{2}\right)^2\\ =\left(\dfrac{-5+\sqrt{97}+5+\sqrt{97}}{2}\right)^2\\ =\left(\dfrac{2\sqrt{97}}{2}\right)^2\\ =\left(\sqrt{97}\right)^2=97\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Theo định lý Viet:
$x_1+x_2=2$
$x_1x_2=-6$
Khi đó:
$A=2x_1-x_1x_2+2x_2=2(x_1+x_2)-x_1x_2$
$=2.2-(-6)=4+6=10$
![](https://rs.olm.vn/images/avt/0.png?1311)
bạn đăng tách ra cho mn giúp nhé
a, Để pt có 2 nghiệm pb
\(\Delta'=1-m\ge0\Leftrightarrow m\le1\)
Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=-2\left(1\right)\\x_1x_2=m\left(2\right)\end{matrix}\right.\)
\(x_1-3x_2=0\)(3)
Từ (1) ; (3) ta có hệ \(\left\{{}\begin{matrix}x_1+x_2=-2\\x_1-3x_2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x_1=-2\\x_2=-2-x_1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=-\dfrac{1}{2}\\x_2=-\dfrac{3}{2}\end{matrix}\right.\)
Thay vào (2) ta được \(m=\left(-\dfrac{1}{2}\right)\left(-\dfrac{3}{2}\right)=\dfrac{3}{4}\)
\(b,\Delta=\left(m+5\right)^2-4\left(-m+6\right)\ge0\Leftrightarrow\left[{}\begin{matrix}m\le-7-4\sqrt{3}\\m\ge-7+4\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x1+x2=m+5\\2x1+3x2=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x1+2x2=2m+10\\2x1+3x2=13\end{matrix}\right.\)\(\)
\(\Rightarrow x2=13-2m-10=3-2m\Rightarrow x1=m+5-x2=m+5-3+2m=3m+2\)
\(x1x2=6-m\Rightarrow\left(3-2m\right)\left(3m+2\right)=6-m\Leftrightarrow\left[{}\begin{matrix}m=0\left(tm\right)\\m=1\left(tm\right)\end{matrix}\right.\)
\(c,\Delta'=\left(m+1\right)^2-\left(m^2-2m+29\right)\ge0\Leftrightarrow m\ge7\)
\(\Rightarrow\left\{{}\begin{matrix}x1+x2=2m+2\\x1=2x2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x2=\dfrac{2m+2}{3}\\x1=\dfrac{2\left(2m+2\right)}{3}\end{matrix}\right.\)
\(\Rightarrow x1.x2=\dfrac{\left(2m+2\right).2\left(2m+2\right)}{9}=m^2-2m+29\Leftrightarrow\left[{}\begin{matrix}m=11\left(tm\right)\\m=23\left(tm\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Δ=(2m-1)^2-4*(-1)(m-m^2)
=4m^2-4m+1+4m-4m^2=1>0
=>(1) luôn có hai nghiệm phân biệt
b: m=x1-2x1x2+x2-2x1x2
=x1+x2-4x1x2
=2m-1+4(m-m^2)
=>m-2m+1-4m+4m^2=0
=>4m^2-5m+1=0
=>m=1 hoặc m=1/4
c: x1+x2-2x1x2
=2m-1+2m-2m^2=-2m^2+4m-1
=-2m^2+4m-2+1
=-2(m-1)^2+1<=1
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Với m= 2, ta có phương trình: x 2 + 2 x − 3 = 0
Ta có: a + b + c = 1 + 2 − 3 = 0
Theo định lý Viet, phương trình có 2 nghiệm:
x 1 = 1 ; x 2 = − 3 ⇒ S = 1 ; − 3 .
b) Chứng minh rằng phương trình luôn có nghiệm ∀ m .
Ta có: Δ ' = m − 1 2 − 1 + 2 m = m 2 ≥ 0 ; ∀ m
Vậy phương trình luôn có nghiệm ∀ m .
c) Theo định lý Viet, ta có: x 1 + x 2 = − 2 m + 2 x 1 . x 2 = 1 − 2 m
Ta có:
x 1 2 . x 2 + x 1 . x 2 2 = 2 x 1 . x 2 + 3 ⇔ x 1 . x 2 x 1 + x 2 − 2 = 6 ⇒ 1 − 2 m − 2 m + 2 − 2 = 6 ⇔ 2 m 2 − m − 3 = 0
Ta có: a − b + c = 2 + 1 − 3 = 0 ⇒ m 1 = − 1 ; m 2 = 3 2
Vậy m= -1 hoặc m= 3/2
![](https://rs.olm.vn/images/avt/0.png?1311)
Ptr có: `\Delta' = b'^2-ac=(-1)^2-(-4)=5 > 0`
`=>` Ptr có `2` nghiệm pb
`=>` Áp dụng Vi-ét: `{(x_1+x_2=[-b]/a=2),(x_1.x_2=c/a=-4):}`
Có: `T=x_1(x_1-2x_2)+x_2(x_2-2x_1)`
`=>T=x_1 ^2 - 2x_1.x_2+x_2 ^2 - 2x_1.x_2`
`=>T=(x_1+x_2)^2-6x_1.x_2`
`=>T=2^2-6(-4)=28`
a: \(a_1+a_2=2x_1-x_2+2x_2-x_1=x_1+x_2=7\)
\(a_1a_2=\left(2x_1-x_2\right)\left(2x_2-x_1\right)\)
\(=4x_1x_2-2x_1^2-2x_2^2+x_1x_2\)
\(=5x_1x_2-2\left(x_1^2+x_2^2\right)\)
\(=5x_1x_2-2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\)
\(=5\cdot3-2\left[7^2-2\cdot3\right]\)
\(=15-2\left[49-6\right]\)
\(=15-2\cdot43=15-86=-71\)
Do đó: Pt cần tìm là \(a^2-7a-71=0\)
b: \(A^2=\left[\left(2x_1-x_2\right)^2+\left(2x_2-x_1\right)^2+2\left(2x_1-x_2\right)\left(2x_2-x_1\right)\right]\)
\(=\left[4x_1^2-4x_1x_2+x_2^2+4x_2^2-4x_2x_1+x_1^2+2\cdot\left(-71\right)\right]\)
\(=\left[5\left(x_1^2+x_2^2\right)-8x_1x_2+2\cdot\left(-71\right)\right]\)
\(=\left[5\cdot43-8\cdot3-142\right]\)
\(=49\)
=>A=7 hoặc A=-7