\(3x^2-2\left(x^2+4x\right)+3x+2=0\)

\(\Leftrightarrow3x^2-2x^2-8x+3x+2=0\)

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

`3x^2-2x=x^2+3`

`<=>3x^2-x^2-2x-3=0`

`<=>2x^2-2x-3=0`

Ptr có: `\Delta'=(-1)^2-2.(-3)=7 > 0`

  `=>`Ptr có `2` nghiệm pb

`=>{(x_1=[-b'+\sqrt{\Delta'}]/a=[1+\sqrt{7}]/2),(x_1=[-b'-\sqrt{\Delta'}]/a=[1-\sqrt{7}]/2):}`

a.

⇔ \(5x^2-3x+\left(-7\right)-1=0\)

⇔ \(5x^2-3x-8=0\)

Δ=\(b^2-4ac\) \(=\left(-3\right)^2-4.5.\left(-8\right)=169\)>0

Vì Δ>0 nên pt có 2 nghiệm phân biệt:

\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+\sqrt{169}}{2.5}=\dfrac{8}{5}\)

\(x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{3-\sqrt{169}}{2.5}=-1\)

a.67 b có ko chắc

Theo Vi-ét, ta có:

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

\(A=-2\left(x_1-2\right)\left(x_2-2\right)\)

\(=\left(-2x_1+4\right)\left(x_2-2\right)\)

\(=-2x_1x_2+4x_1+4x_2-8\)

\(=-2x_1x_2+4\left(x_1+x_2\right)-8\)

\(=-2.\left(-\dfrac{1}{4}\right)+4.\left(-\dfrac{3}{4}\right)-8\)

\(=\dfrac{1}{2}-3-8\)

\(=\dfrac{1}{2}-11\)

\(=-\dfrac{21}{2}\)

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 \(B=\dfrac{104}{21}\)

\(\Delta'=\left(-2\right)^2-3.\left(-8\right)=4+24=28>0.\)

\(\Rightarrow\) Pt có 2 nghiệm phân biệt \(x_1;x_2.\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{2+2\sqrt{7}}{3}.\\x_2=\dfrac{2-2\sqrt{7}}{3}.\end{matrix}\right.\)

a) dễ rồi bạn chỉ việc bế x = 1/2 vào tìm m bình thường

b) mx - 2 + m = 3x

<=> ( m - 3 )x + m - 2 = 0

Để pt có nghiệm duy nhất thì m - 3 ≠ 0 <=> m ≠ 3

Khi đó nghiệm duy nhất là x = -m+2/m-3

\(\Leftrightarrow\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)=0\)

\(\Leftrightarrow\left(-6x-18\right)\left(8x^2-18\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x-3\right)\left(2x+3\right)=0\)

hay \(x\in\left\{-3;\dfrac{3}{2};-\dfrac{3}{2}\right\}\)

<=> 4x²-3x-18=4x²+3x

<=>6x=-18

<=>x=-3