\(a,=x^2+2x+1+2019=\left(x+1\right)^2+2019\ge2019\) dấu"=" xảy ra<=>x=-1

b,\(=m^2+2.2m+4-5=\left(m+2\right)^2-5\ge-5\) dấu"=" xảy ra<=>m=-2

c, \(=x-2\sqrt{x}+10=x-2\sqrt{x}+1+9=\left(\sqrt{x}-1\right)^2+9\ge9\)

dấu"=" xảy ra<=>x=1

b, \(4x-8\sqrt{x}+2020=4x-2.2.2\sqrt{x}+4+2016=\left(2\sqrt{x}-2\right)^2+2016\ge2016\)

dấu"=" xảy ra<=>x=1

a) Ta có: \(x^2+2x+2020\)

\(=x^2+2x+1+2019\)

\(=\left(x+1\right)^2+2019\ge2019\forall x\)

Dấu '=' xảy ra khi x=-1

b) Ta có: \(m^2+4m-1\)

\(=m^2+4m+4-5\)

\(=\left(m+2\right)^2-5\ge-5\forall m\)

Dấu '=' xảy ra khi m=-2

c) Ta có: \(m^2+m\)

\(=m^2+2\cdot m\cdot\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}\)

\(=\left(m+\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\forall m\)

Dấu '=' xảy ra khi \(m=-\dfrac{1}{2}\)

a: \(\dfrac{1}{m-2}\cdot\sqrt{m^2-4m+4}\)

\(=\dfrac{1}{m-2}\cdot\sqrt{\left(m-2\right)^2}\)

\(=\dfrac{1}{m-2}\cdot\left|m-2\right|\)

\(=\dfrac{1}{m-2}\cdot\left(m-2\right)\left(m>2\right)\)

=1

b: \(2\sqrt{x}=14\)

=>\(\sqrt{x}=7\)

=>x=49

\(x+2\sqrt{x}+1=4\)

=>\(\left(\sqrt{x}+1\right)^2=4\)

=>\(\left[{}\begin{matrix}\sqrt{x}+1=2\\\sqrt{x}+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=-3\left(loại\right)\end{matrix}\right.\)

=>x=1(nhận)

binh phuong 2 ve giai ra

`A=sqrt{x-2}+sqrt{6-x}(2<=x<=6)`
Áp dụng BĐT `sqrtA+sqrtB>=sqrt{A+B}`
`=>A>=sqrt{x-2+6-x}=2`
Dấu "=" `<=>x=2` hoặc `x=6`
Áp dụng BĐT bunhia
`=>A<=sqrt{2(x-2+6-x)}=2sqrt2`
Dấu "=" `<=>x=4`
`C=sqrt{1+x}+sqrt{8-x}(-1<=x<=8)`
Áp dụng BĐT `sqrtA+sqrtB>=sqrt{A+B}`
`=>A>=sqrt{1+x+8-x}=3`
Dấu "=" `<=>x=-1` hoặc `x=8`
Áp dụng BĐT bunhia
`=>A<=sqrt{2(1+x+8-x)}=3sqrt2`
Dấu "=" `<=>x=7/2`

`D=2sqrt{x+5}+sqrt{1-2x}(-5<=x<=1/2)`
`=sqrt{4x+20}+sqrt{1-2x}`
Áp dụng BĐT `sqrtA+sqrtB>=sqrt{A+B}`
`=>D>=sqrt{4x+20+1-2x}=sqrt{2x+21}`
Mà `x>=-5`
`=>D>=sqrt{-10+21}=sqrt{11}`
Dấu "=" `<=>x=-5`

tự làm đi

Theo định lý Vi-ét:

\(x_1\cdot x_2=\dfrac{c}{a}=\dfrac{m^2+4m+3}{2}\)

\(x_1+x_2=-\dfrac{b}{a}=1-m\)

Ta có: \(A=\left|x_1\cdot x_2-2x_1-2x_2\right|=\left|\dfrac{m^2+4m+3}{2}-\left(1-m\right)\right|\)

\(=\left|\dfrac{m^2+6m+1}{2}\right|\)

Mặt khác \(\left|m^2+6m+1\right|=\left|\left(m+3\right)^2-8\right|\)

=> Min _{|m^2 +6m+1|} =8 khi x=-3

Mà A đạt gtnn khi |m^2 +6m+1| đạt gtnn

Vậy Max_A = 8/2 = 4 khi x=-3

--thay x=1 thì A=4 ; x=-3 cũng A=4;; và x=0 => A= 0,5 (gtnn)---

(Giải trật lất??!! Thay số 1 hồi tớ có linh cảm A không có gtln nhưng trên đề........)

----anh chị đi qua đi lại xin chỉ cho em biết em sai ở đâu ah----

Thank you for your reading and your instructing!!

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

\(\dfrac{3}{4-2x}=\dfrac{-3}{2\left(x-2\right)}=\dfrac{-3\left(x^2+2x+4\right)}{2\left(x-2\right)\left(x^2+2x+4\right)}\)

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

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

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

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

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

d) \(\dfrac{1}{3x+3y}=\dfrac{1}{3\left(x+y\right)}=\dfrac{\left(x-y\right)^2}{3\left(x+y\right)\left(x-y\right)^2}\)

\(\dfrac{2x}{x^2-y^2}=\dfrac{2x}{\left(x+y\right)\left(x-y\right)}=\dfrac{6x\left(x-y\right)}{3\left(x+y\right)\left(x-y\right)^2}\)

\(\dfrac{x^2-xy+y^2}{x^2-2xy+y^2}=\dfrac{x^2-xy+y^2}{\left(x-y\right)^2}=\dfrac{3\left(x^2-xy+y^2\right)\left(x+y\right)}{3\left(x+y\right)\left(x-y\right)^2}=\dfrac{3\left(x^3+y^3\right)}{3\left(x+y\right)\left(x-y\right)^2}\)