ta có : pt \(x^2+bx+c=0\) có \(a=1;b=b;c=c\)

ta có : \(x=\dfrac{-b+\sqrt{b^2-4ac}}{2}=\dfrac{-b+\sqrt{\Delta}}{2a}\) đúng như trong công thức

\(\Rightarrow\left(đpcm\right)\)

Ta có : \(f\left(x\right)=ax^2+bx+c=a\left(x^2+\frac{bx}{a}\right)+c=a\left(x^2+2.x.\frac{b}{2a}+\frac{b^2}{4a^2}\right)-\frac{b^2}{4a}+c\)

\(=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a}\ge-\frac{b^2-4ac}{4a}\)(vì a>0)

Dấu đẳng thức xảy ra \(\Leftrightarrow x=-\frac{b}{2a}\)

Do đó : Min f(x) = \(\frac{4ac-b^2}{4a}\Leftrightarrow x=-\frac{b}{2a}\)

b) \(f\left(x\right)=-ax^2+bx+c=-a\left(x^2-bx\right)+c=-a\left(x^2-2.x.\frac{b}{2a}+\frac{b^2}{4a^2}\right)-\frac{b^2}{4a}+c=-a\left(x-\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}\le\frac{4ac-b^2}{4a}\)(vì a<0)

Dấu đẳng thức xảy ra \(\Leftrightarrow x=\frac{b}{2a}\)

Vậy Max f(x) = \(\frac{4ac-b^2}{4a}\Leftrightarrow x=\frac{b}{2a}\)

ĐKXĐ : \(0\le x\ne1\)

a) \(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{\left(1-x\right)^2}{2}\)

\(=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{\left(1-x\right)^2}{2}\)

\(=\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)

\(=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)

\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)

b) \(P=\sqrt{x}\left(1-\sqrt{x}\right)\)

Để P > 0 thì \(\hept{\begin{cases}\sqrt{x}>0\\1-\sqrt{x}>0\end{cases}\Rightarrow}0< x< 1\)

c) \(P=-x+\sqrt{x}=-\left(x-2\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

Vậy max P = 1/4 khi x = 1/4

`a) 7x^2 - 2x + 3 = 0`

`(a = 7; b = -2; c = 3)`

`Δ = b^2 - 4ac = (-2)^2 - 4.7.3 = -80 < 0`

`=>` phương trình vô nghiệm

`b) 6x^2 + x + 5 = 0`

`(a = 6;b = 1;c = 5)`

`Δ = b^2 - 4ac = 1^2 - 4.6.5 = -119 < 0`

`=>` phương trình vô nghiệm

`c) 6x^2 + x - 5 = 0`

`(a = 6;b=1;c=-5)`

`Δ = b^2 - 4ac = 1^2 - 4.6.(-5) = 121 > 0`

`=>` phương trình có 2 nghiệm phân biệt

`x_1 = (-b + sqrt{Δ})/(2a) = (-1+ sqrt{121})/(2.6) = (-1+11)/12 = 10/12 = 5/6`

`x_2 = (-b - sqrt{Δ})/(2a) = (-1- sqrt{121})/(2.6) = (-1-11)/12 = -12/12 = -1`

Vậy phương trình có 1 nghiệm `x_1 = 5/6; x_2 = -1`

ủa, mấy bài đó tương tự như ct mà:

\(7x^2-2x+3=0\) \(\left\{{}\begin{matrix}a=7\\b=-2\\c=3\end{matrix}\right.\)

\(\Delta=b^2-4ac=\left(-2\right)^2-4.7.3=-80\)

Vì \(\Delta< 0\) \(\Rightarrow\) pt vô nghiệm

a)

`7x^2 -2x+3=0`

có \(\Delta=b^2-4ac=\left(-2\right)^2-4\cdot7\cdot3=-80< 0\)

=> phương trình vô nghiệm

b)

`6x^2 +x+5=0`

có \(\Delta=b^2-4ac=1^2-4\cdot6\cdot5=-119< 0\)

=> phương trình vô nghiệm

c)

`6x^2 +x-5=0`

có \(\Delta=b^2-4ac=1^2-4\cdot6\cdot\left(-5\right)=121>0\)

\(=>x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-1+\sqrt{121}}{2\cdot6}=\dfrac{5}{6}\)

\(=>x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-1-\sqrt{121}}{2\cdot6}=-1\)

Answer:

a. \(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\left(\frac{1-x}{\sqrt{2}}\right)^2\)   ĐK: \(x\ge0;x\ne1\)

\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(1-x\right)^2}{2}\)

\(=\frac{-2\sqrt{x}}{\sqrt{x}+1}.\frac{x-1}{2}\)

\(=\frac{\sqrt{x}\left(1-x\right)}{\sqrt{x}+1}\)

\(=\frac{\sqrt{x}\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)

\(=\sqrt{x}\left(1-\sqrt{x}\right)\)

b. Vì \(0< x< 1\Rightarrow\hept{\begin{cases}\sqrt{x}\ge0\\1-\sqrt{x}>0\end{cases}}\Rightarrow\sqrt{x}\left(1-\sqrt{x}\right)>0\)

Do vậy \(\sqrt{x}\left(1-\sqrt{x}\right)>0\)

c. \(P=\sqrt{x}\left(1-\sqrt{x}\right)\)

\(=-\left(\sqrt{x}\right)^2+\sqrt{x}\)

\(=-\left(x-2\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}\)

\(=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)

Dấu "=" xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\Rightarrow x=\frac{1}{4}\)