bài này có GTLN thôi bạn 

\(\Delta'=\left(m+1\right)^2-\left(m^2+4m+3\right)=-2m-2\)

Để pt luôn có 2 nghiệm 

\(-2m-2\ge0\Leftrightarrow m+1\le0\Leftrightarrow m\le-1\)

Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=-2\left(m+1\right)\\x_1x_2=m^2+4m+3\end{matrix}\right.\)

\(A=\left|x_1x_2-2\left(x_1+x_2\right)\right|\)

\(=\left|m^2+4m+3+4\left(m+1\right)\right|=\left|m^2+8m+7\right|\)

\(=\left|m^2+8m+16-9\right|=\left|\left(m+4\right)^2-9\right|\)

Ta có : \(m\le-1\Rightarrow m+4\le3\Leftrightarrow\left(m+4\right)^2\le9\Leftrightarrow\left(m+4\right)^2-9\le0\Rightarrow\left|\left(m+4\right)^2-9\right|\le\left|0\right|=0\)

Vậy với m = -1 thì A đạt GTNN là 0 

Lời giải:

$x^2+2(m+1)x+m^2+4m+3=0$

Để pt có 2 nghiệm $x_1,x_2$ thì:

$\Delta'=(m+1)^2-(m^2+4m+3)\geq 0$

$\Leftrightarrow -2m-2\geq 0\Leftrightarrow m\leq -1$

Áp dụng định lý Viet:

$x_1+x_2=-2(m+1)$

$x_1x_2=m^2+4m+3$
Khi đó:

$A=|x_1x_2-2x_1-2x_2|$

$=|x_1x_2-2(x_1+x_2)|=|m^2+4m+3+4(m+1)|=|m^2+8m+7|$

$=|(m+1)(m+7)|\geq 0$ với mọi $m\leq -1$

Vậy GTNN của $A$ là $0$ khi $m=-1$

\(\Delta'=\left(m-1\right)^2+m+3=m^2-m+4=\left(m-\dfrac{1}{2}\right)^2+\dfrac{7}{2}>0;\forall m\)

\(\Rightarrow\) Phương trình luôn có 2 nghiệm với mọi m

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=-m-3\end{matrix}\right.\)

a.

\(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)

\(=4\left(m-1\right)^2+2\left(m+3\right)=4m^2-6m+10\)

\(=4\left(m-\dfrac{3}{4}\right)^2+\dfrac{31}{4}\ge\dfrac{3}{4}\)

Dấu = xảy ra khi \(m=\dfrac{3}{4}\)

b.

\(x_1^2+x_2^2=8m^3-8m^2\)

\(\Leftrightarrow4m^2-6m+10=8m^3-8m^2\)

\(\Leftrightarrow8m^3-12m^2+6m-1=9\)

\(\Leftrightarrow\left(2m-1\right)^3=9\)

\(\Leftrightarrow2m-1=\sqrt[3]{9}\)

\(\Rightarrow m=\dfrac{1+\sqrt[3]{9}}{2}\)

a: Δ=(2m-2)^2-4(-m-3)

=4m^2-8m+4+4m+12

=4m^2-4m+16

=4m^2-4m+1+15=(2m-1)^2+15>0

=>Phương trình luôn có 2 nghiệm pb

A=x1^2+x2^2

=(x1+x2)^2-2x1x2

=(2m-2)^2-2(-m-3)

=4m^2-8m+4+2m+6

=4m^2-6m+10

=4(m^2-3/2m+5/2)

=4(m^2-2*m*3/4+9/16+31/16)

=4(m-3/4)^2+31/4>=31/4

Dấu = xảy ra khi m=3/4

b: x1^2+x2^=8m^3-8m^2

=>4m^2-6m+10=8m^3-8m^2

=>8m^3-8m^2-4m^2+6m-10=0

=>8m^3-12m^2+6m-10=0

=>\(m\simeq1,54\)

1:

a: =x^2-7x+49/4-5/4

=(x-7/2)^2-5/4>=-5/4

Dấu = xảy ra khi x=7/2

b: =x^2+x+1/4-13/4

=(x+1/2)^2-13/4>=-13/4

Dấu = xảy ra khi x=-1/2

e: =x^2-x+1/4+3/4=(x-1/2)^2+3/4>=3/4

Dấu = xảy ra khi x=1/2

f: x^2-4x+7

=x^2-4x+4+3

=(x-2)^2+3>=3

Dấu = xảy ra khi x=2

2:

a: A=2x^2+4x+9

=2x^2+4x+2+7

=2(x^2+2x+1)+7

=2(x+1)^2+7>=7

Dấu = xảy ra khi x=-1

b: x^2+2x+4

=x^2+2x+1+3

=(x+1)^2+3>=3

Dấu = xảy ra khi x=-1

1) Áp dụng bđt Cauchy cho 3 số dương ta có

 \(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+x^3\ge4\sqrt[4]{\dfrac{1}{x}.\dfrac{1}{x}.\dfrac{1}{x}.x^3}=4\) (1)

\(\dfrac{3}{y^2}+y^2\ge2\sqrt{\dfrac{3}{y^2}.y^2}=2\sqrt{3}\) (2)

\(\dfrac{3}{z^3}+z=\dfrac{3}{z^3}+\dfrac{z}{3}+\dfrac{z}{3}+\dfrac{z}{3}\ge4\sqrt[4]{\dfrac{3}{z^3}.\dfrac{z}{3}.\dfrac{z}{3}.\dfrac{z}{3}}=4\sqrt{3}\) (3)

Cộng (1);(2);(3) theo vế ta được

\(\left(\dfrac{3}{x}+\dfrac{3}{y^2}+\dfrac{3}{z^3}\right)+\left(x^3+y^2+z\right)\ge4+2\sqrt{3}+4\sqrt{3}\)

\(\Leftrightarrow3\left(\dfrac{1}{x}+\dfrac{1}{y^2}+\dfrac{1}{z^3}\right)\ge3+4\sqrt{3}\)

\(\Leftrightarrow P\ge\dfrac{3+4\sqrt{3}}{3}\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{x}=x^3\\\dfrac{3}{y^2}=y^2\\\dfrac{3}{z^3}=\dfrac{z}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\sqrt[4]{3}\\z=\sqrt{3}\end{matrix}\right.\) (thỏa mãn giả thiết ban đầu)

2) Ta có \(4\sqrt{ab}=2.\sqrt{a}.2\sqrt{b}\le a+4b\)

Dấu"=" khi a = 4b

nên \(\dfrac{8}{7a+4b+4\sqrt{ab}}\ge\dfrac{8}{7a+4b+a+4b}=\dfrac{1}{a+b}\)

Khi đó \(P\ge\dfrac{1}{a+b}-\dfrac{1}{\sqrt{a+b}}+\sqrt{a+b}\)

Đặt \(\sqrt{a+b}=t>0\) ta được

\(P\ge\dfrac{1}{t^2}-\dfrac{1}{t}+t=\left(\dfrac{1}{t^2}-\dfrac{2}{t}+1\right)+\dfrac{1}{t}+t-1\)

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

Có \(\dfrac{1}{t}+t\ge2\sqrt{\dfrac{1}{t}.t}=2\) (BĐT Cauchy cho 2 số dương)

nên \(P=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\ge\left(\dfrac{1}{t}-1\right)^2+1\ge1\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{t}-1=0\\t=\dfrac{1}{t}\end{matrix}\right.\Leftrightarrow t=1\)(tm)

khi đó a + b = 1

mà a = 4b nên \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

Vậy MinP = 1 khi \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

\(A=\dfrac{b^2}{b-1}=\dfrac{b^2-1+1}{b-1}=b+1+\dfrac{1}{b-1}=b-1+\dfrac{1}{b-1}+2\)

Áp dụng BĐT cosi cho \(b>0\left(b>1\right)\)

\(A=b-1+\dfrac{1}{b-1}+2\ge2\sqrt{\left(b-1\right)\cdot\dfrac{1}{b-1}}+2=2+2=4\)

Dấu \("="\Leftrightarrow\left(b-1\right)^2=1\Leftrightarrow\left[{}\begin{matrix}b-1=1\\b-1=-1\left(ktm\right)\end{matrix}\right.\Leftrightarrow b=2\left(tm\right)\)

\(A=\dfrac{\left(x+1\right)^2+2+7}{\left(x+1\right)^2+2}=1+\dfrac{7}{\left(x+1\right)^2+2}< =1+\dfrac{7}{2}=\dfrac{9}{2}\)

Dấu = xảy ra khi x=-1

Áp dụng Bunyakovsky, ta có :

\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x.1+y.1\right)^2=1\)

=> \(\left(x^2+y^2\right)\ge\frac{1}{2}\)

=> \(Min_C=\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\)

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