2: \(-4x^2+5x-2\)

\(=-4\left(x^2-\dfrac{5}{4}x+\dfrac{1}{2}\right)\)

\(=-4\left(x^2-2\cdot x\cdot\dfrac{5}{8}+\dfrac{25}{64}+\dfrac{7}{64}\right)\)

\(=-4\left(x-\dfrac{5}{8}\right)^2-\dfrac{7}{16}< =-\dfrac{7}{16}< 0\forall x\)

Sửa đề:\(f\left(x\right)=\dfrac{-x^2+4\left(m+1\right)x+1-4m^2}{-4x^2+5x-2}\)

Để f(x)>0 với mọi x thì \(\dfrac{-x^2+4\left(m+1\right)x+1-4m^2}{-4x^2+5x-2}>0\forall x\)

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

\(\text{Δ}=\left[\left(4m+4\right)\right]^2-4\cdot\left(-1\right)\left(1-4m^2\right)\)

\(=16m^2+32m+16+4\left(1-4m^2\right)\)

\(=32m+20\)

Để BĐT(1) luôn đúng với mọi x thì \(\left\{{}\begin{matrix}\text{Δ}< 0\\a< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}32m+20< 0\\-1< 0\left(đúng\right)\end{matrix}\right.\)

=>32m+20<0

=>32m<-20

=>\(m< -\dfrac{5}{8}\)

(x + 1)(x + 2) = (2 - x)(x + 2)

<=> x² + 2x + x + 2 = 4 - x²

<=> x² + 3x + 2 = 4 - x²

<=> x² + 3x + 2 - 4 + x² = 0

<=> 2x² + 3x - 2 = 0

<=> 2x² + 4x - x - 2 = 0

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

<=> (x + 2)(2x - 1) = 0

<=> x + 2 = 0 hoặc 2x - 1 = 0

<=> x = -2 hoặc x = 1/2

còn bài 2 nữa bạn

ư ư :<

a) 2(m + 3/5) - (m + 13/5) = 5

<=> 2m + 6/5 - m - 13/5 = 5

<=> m - 7/5 = 5

<=> m = 5 + 7/5

<=> m = 32/5

b) \(\frac{2\left(3m+1\right)+1}{5}-\frac{2\left(3m-1\right)}{5}+\frac{3m+1}{10}=5\)

<=> 15(2m + 1) - 8(3m - 1) + 3(3m + 2) = 100

<=> 30m + 15 - 24m + 1 + 9m + 6 = 100

<=> 12m + 27 = 100

<=> 12m = 100 - 27

<=> 12m = 73

<=> m = 73/12

refer

Hai đồ thị \(y=\left(3m+2\right)x+5\) và \(y=-x-1\) cắt nhau

\(\Rightarrow3m+2\ne-1\Rightarrow m\ne-1\)

Khi đó ta có giao điểm 2 đồ thị là \(A=\left(x;y\right)=\left(x;-x-1\right)\)

\(P=y^2+2x-2019=\left(-x-1\right)^2+2x-2019=x^2+4x-2018\\ =\left(x+2\right)^2-2022\ge-2022\)

Dấu = xảy ra \(\Leftrightarrow x+2=0\Leftrightarrow x=-2\Leftrightarrow y=1\)

\(\Rightarrow1=\left(3m+2\right)\left(-2\right)+5\Rightarrow-6m=0\Rightarrow m=0\left(TM\right)\)

1. Xét x = - 2, thay vào pt ta dc: -1.0 = 4.0 (Hợp lí)

Vậy x = -2 là 1 nghiệm của pt

Xét x \(\ne\)- 2, ta có: x + 1 = 2 - x

<=> 2x = 1 <=> x = 1/2

Vậy S = {1/2; -2}

2. a. \(2\left(m+\frac{3}{5}\right)-\left(m+\frac{13}{5}\right)=5\)

<=> \(2m+\frac{6}{5}-m-\frac{13}{5}=5\)

<=> m = \(\frac{32}{5}\)

b. \(2\left(3m+1\right)+\frac{1}{4}-\frac{2\left(3m-1\right)}{5}+3m+\frac{1}{5}=5\)

<=> \(6m+2+\frac{1}{4}-\frac{6m-2}{5}+3m+\frac{1}{5}=5\)

<=> \(6m-\frac{6m-2}{5}+3m=5-2-\frac{1}{4}-\frac{1}{5}\)

<=> \(9m-\frac{6m-2}{5}=\frac{51}{20}\)

<=> \(\frac{45m-6m+2}{5}=\frac{51}{20}\)

<=> \(20\left(39m+2\right)=51.5\)

<=> 780m + 40 = 255

<=> 780m = 215

<=> m = \(\frac{43}{156}\)

thanks

\(\Delta=\left(3m+2\right)^2-12m=9m^2+4>0\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-3m-2\\x_1x_2=3m\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1+1+x_2+1=-3m\\x_1x_2+x_1+x_2+1=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1+1+x_2+1=-3m\\\left(x_1+1\right)\left(x_2+1\right)=-1\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x_1+1=a\\x_2+1=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=-3m\\ab=-1\end{matrix}\right.\)

\(Q=a^4+b^4\ge2a^2b^2=2\)

Dấu "=" xảy ra khi \(a^2=b^2\Rightarrow\left[{}\begin{matrix}a=b\left(loại\right)\\a=-b\end{matrix}\right.\)

\(\Rightarrow-3m=0\Rightarrow m=0\)

a/ \(\left\{{}\begin{matrix}3m+1>0\\\Delta=\left(3m+1\right)^2-4\left(3m+1\right)\left(m+4\right)\le0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m>-\frac{1}{3}\\\left(3m+1\right)\left(-m-15\right)\le0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m>-\frac{1}{3}\\\left[{}\begin{matrix}m\ge-\frac{1}{3}\\m\le-15\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow m>-\frac{1}{3}\)

b/\(\left\{{}\begin{matrix}m+1>0\\\Delta'=\left(m-1\right)^2-\left(m+1\right)\left(3m-3\right)\le0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m>-1\\\left(m-1\right)\left(-2m-4\right)\le0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m>-1\\\left[{}\begin{matrix}m\ge1\\m\le-2\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow m\ge1\)