Bài 1:

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

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

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

Dấu "=" xảy ra khi \(x=-\frac{1}{2}\)

Bài 2:

Bình phương 2 vế 

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

\(\Leftrightarrow3x^2-4x+3=4x^2-4x+1\)

\(\Leftrightarrow2-x^2\Leftrightarrow x^2=2\Leftrightarrow x=-\sqrt{2}\) (tm)

\(x=-\sqrt{a}\Rightarrow-\sqrt{2}=-\sqrt{a}\Rightarrow a=2\)

4x^2+4x-1

=4x^2+4x+1-2

=(2x+1)^2-2

=> (2x+1)^2\(\ge\)0 voi moi x

=> (2x+1)^2 \(\ge\)2

=> GTNN la 2

-2 mà bạn

a, (x-1)(x-3)+11

=x²-3x-x+3+11

=(x-2)²+10

Vì..................................

b,5-4x²+4x

=-(4x²-4x+4)+9

=-(2x-2)²+9

...........................................................

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 \(x=-1\)

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

ai k mình k lại [ chỉ 3 người đầu tiên mà trên 10 điểm hỏi đáp ]

k lại đi

1)

\(A=3-4x-x^2=-\left(x^2+4x-3\right)=-\left(x+2\right)^2+7\)

=> \(Max_A=7\Leftrightarrow x=-2\)

2)

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

=> \(Max_B=4\Leftrightarrow x=2\)

I zì:vv

a) Ta có: \(A=4x^2+4x+11=4x^2+4x+1=10=\left(2x+1\right)^2+10\ge10\forall x\)

Vậy Min_A=10 khi \(x=-\dfrac{1}{2}\)

b) Ta có: \(B=5-8x-x^2=-\left(x^2+8x-5\right)=-\left(x^2+8x+16-21\right)\)

\(=-\left(x+4\right)^2+21\le21\forall x\)

Vậy Max_B=21 khi x=-4

A=4x²-4x+3

<=> A=4x²-4x+1+2

<=> A=(2x-1)²+2

Vì (2x-1)²\(\ge0\)nên \(\left(2x-1\right)^2+2\ge2\)

Vậy Min_A=2 khi x=\(\frac{1}{2}\)

\(a,A=\left(x^2-x\right)\left(x^2-x-12\right)\\ A=\left(x^2-x\right)^2-12\left(x^2-x\right)\\ A=\left(x^2-x\right)^2-12\left(x^2-x\right)+36-36\\ A=\left(x^2-x+6\right)^2-36\ge-36\\ A_{min}=-36\Leftrightarrow x^2-x+6=0\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\ b,B=4x^4+4x^3+5x^2+4x+3\\ B=\left(4x^4+4x^3+x^2\right)+\left(x^2+4x+4\right)-1\\ B=x^2\left(2x+1\right)^2+\left(x+2\right)^2-1\ge-1\\ B_{min}=-1\Leftrightarrow\left\{{}\begin{matrix}x\left(2x+1\right)=0\\x+2=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

Vậy dấu \("="\) không xảy ra

\(A=\left(x^2-4x+3\right)\left(x^2-4x+50\right)\)

\(=\left[\left(x-2\right)^2-1\right]\left[\left(x-2\right)^2+46\right]\)

\(=\left(x-2\right)^4+45\left(x-2\right)^2-46\ge-46\)

\(A_{min}=-46\) khi \(x=2\)

=2x-1+2x-3=4x-4=x^2-x^2+4x-4=x^2-(x^2-4x+4)=x^2-(x-2)^2 vay gtnn la x^2

Tìm GTNN

\(\sqrt{4X^2-4X+1}+\sqrt{4X^2-12X+9}\)