Biểu thức nào em?

cả hai ạ

a: \(-x^2+2x-4\)

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

\(=-\left(x^2-2x+1+3\right)\)

\(=-\left\lbrack\left(x-1\right)^2+3\right\rbrack=-\left(x-1\right)^2-3\le-3\forall x\)

=>\(\frac{1}{-x^2+2x-4}\ge-\frac13\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

b: \(-4x^2+12x-13\)

\(=-\left(4x^2-12x+13\right)\)

\(=-\left(4x^2-12x+9+4\right)\)

\(=-\left\lbrack\left(2x-3\right)^2+4\right\rbrack=-\left(2x-3\right)^2-4\le-4\forall x\)

=>\(\frac{12}{-4x^2+12x-13}\ge\frac{12}{-4}=-3\forall x\)

Dấu '=' xảy ra khi 2x-3=0

=>2x=3

=>\(x=\frac32\)

c: Đặt \(A=\frac{x^2-4x-4}{x^2-4x+5}\)

\(=\frac{x^2-4x+5-9}{x^2-4x+5}\)

\(=1-\frac{9}{x^2-4x+5}\)

Ta có: \(x^2-4x+5\)

\(=x^2-4x+4+1\)

\(=\left(x-2\right)^2+1\ge1\forall x\)

=>\(\frac{9}{\left(x-2\right)^2+1}\le\frac91=9\forall x\)

=>\(-\frac{9}{\left(x-2\right)^2+1}\ge-9\forall x\)

=>\(A=-\frac{9}{\left(x-2\right)^2+1}+1\ge-9+1=-8\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

e: Đặt \(B=\frac{x^2-2011}{4\left(x^2+1\right)}\)

\(=\frac14\cdot\frac{4x^2-8044}{4x^2+4}=\frac14\cdot\frac{x^2-2011}{x^2+1}=\frac14\left(\frac{x^2+1-2012}{x^2+1}\right)=\frac14\left(1-\frac{2012}{x^2+1}\right)\)

Ta có: \(x^2+1\ge1\forall x\)

=>\(\frac{2012}{x^2+1}\le2012\forall x\)

=>\(-\frac{2012}{x^2+1}\ge-2012\forall x\)

=>\(1-\frac{2012}{x^2+1}\ge-2012+1=-2011\forall x\)

=>\(\frac14\left(1-\frac{2012}{x^2+1}\right)\ge-\frac{2011}{4}\forall x\)

Dấu '=' xảy ra khi x=0

a,

\(x^2+4x+6=x^2+4x+4+2=\left(x+2\right)^2+2\ge2\)

\(=>A=\dfrac{x^2+4x+6}{3}\ge\dfrac{2}{3}\)

Vậy giá trị nhỏ nhất của biểu thức là 2/3 , dấu ''='' xảy ra khi và chỉ khi x = -2 .

b, \(Ta,c\text{ó}:\left|1-2x\right|\ge0\)

\(=>4+\left|1-2x\right|\ge4\)

\(=>\dfrac{4+\left|1+2x\right|}{5}\ge\dfrac{4}{5}\)

Vậy giá trị nhỏ nhất của biểu thức là 4/5 , dấu bằng xảy ra khi và chỉ khi 1 - 2x = 0 => x = 1/2

c,

\(\dfrac{5}{4x^2+4x+2y+y^2+3}\)

\(=\dfrac{5}{\left(2x+1\right)^2+\left(y+1\right)^2+1}\ge\dfrac{5}{1}=5\)

Vậy giá trị nhỏ nhất của biểu thức là 5 , dấu '='' xảy ra khi và chỉ khi

\(\left\{{}\begin{matrix}2x+1=0\\y+1=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-1\end{matrix}\right.\)

\(A=-\dfrac{4}{x^2-4x+10}\\ =-\dfrac{4}{\left(x^2-2.x.2+4+6\right)}\\ =-\dfrac{4}{\left(x-2\right)^2+6}\)

\(\left(x-2\right)^2\ge0\\ \Rightarrow\left(x-2\right)^2+6\ge6\\ \Rightarrow\dfrac{4}{\left(x-2\right)^2+6}\le\dfrac{2}{3}\\ \Rightarrow A=-\dfrac{4}{\left(x-2\right)^2+6}\ge-\dfrac{2}{3}\)

Min A=-2/3 khi x=2

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

Vì \(\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2+1\ge1\)

\(\Rightarrow C\le2\)

Dấu ''='' xảy ra \(\Leftrightarrow x=-2\)

Vậy Min C = 2 kjhi x = -2

Bài 1: 

a: \(B=\left(x+2\right)^2+\left(y-\dfrac{1}{5}\right)^2-10\ge-10\)

Dấu '=' xảy ra khi x=-2 và y=1/5

b: \(C=\left(x+3\right)^4+1\ge1\)

Dấu '=' xảy ra khi x=-3

c: \(D=x^2-4x+4+11=\left(x-2\right)^2+11\ge11\)

Dấu '=' xảy ra khi x=2

Bài 1:

\(B=\dfrac{4\left(x+3\right)^2}{\left(3x+5\right)^2-4x^2}-\dfrac{\left(x^2-25\right)}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{\left(4x+15\right)^2-x^2}\)

\(=\dfrac{4\left(x+3\right)^2}{\left(3x+5-2x\right)\left(3x+5+2x\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{\left(4x+15-x\right)\left(4x+15+x\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{5\left(x-5\right)\left(x+1\right)}-\dfrac{3\left(x+3\right)\left(x+1\right)}{15\left(x+5\right)\left(x+3\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{x+5}{5\left(x+1\right)}-\dfrac{x+1}{5\left(x+5\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+5\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+1\right)^2}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{4\left(x^2+6x+9\right)-\left(x^2+10x+25\right)-\left(x^2+2x+1\right)}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{4x^2+24x+36-x^2-10x-25-x^2-2x-1}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{2x^2+12x+10}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{2\left(x^2+6x+5\right)}{5\left(x+5\right)\left(x+1\right)}\)

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

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

đc bn , nhg mà đề bài câu a b2 sao tự nhiên lại có " n "

bn xem lại đề đi

a.

\(A=\dfrac{2013}{x^2}-\dfrac{2}{x}+1=2013\left(\dfrac{1}{x}-\dfrac{1}{2013}\right)^2+\dfrac{2012}{2013}\ge\dfrac{2012}{2013}\)

Dấu "=" xảy ra khi \(x=2013\)

b.

\(B=\dfrac{4x^2+2-4x^2+4x-1}{4x^2+2}=1-\dfrac{\left(2x-1\right)^2}{4x^2+2}\le1\)

\(B_{max}=1\) khi \(x=\dfrac{1}{2}\)

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

\(B_{max}=-\dfrac{1}{2}\) khi \(x=-1\)

em cảm ơn ạ