không biet luon

Min A = -1 <-> x=2/3

Min B =2 <-> x=0 ; y=1

Max C = 5 <-> x=1/2

Max D = 1/3 <-> x=2

a)\(\left|3x-2\right|\ge0\Rightarrow2\left|3x-2\right|\ge0\Rightarrow A=2\left|3x-2\right|-1\ge-1\)

=>A_min=-1 <=>|3x-2|=0 <=>3x-2=0 <=>3x=2<=>x=2/3

b)\(x^2\ge0;\left|2y-2\right|\ge0\Leftrightarrow3\left|3y-2\right|\ge0\)

=>\(x^2+3\left|2y-2\right|\ge0\Rightarrow B=x^2+3\left|2y-2\right|-1\ge-1\)

=>B_min=-1 <=>x²=0 và |2y-2|=0 <=> x=0 và y=1

a) Ta có : \(1-4x-2x^2=-\left(2x^2+4x-1\right)=-[2(x^2+2x+1)-3]=-[2(x+1)^2-3]\)

Lại có \(2\left(x+1\right)^2\ge0=>-[2(x+1)^2-3]\le-3\)

Dấu"=" xảy ra khi và chỉ khi \(x+1=0=>x=-1\)

Vậy giá trị lớn nhất của biểu thức đã cho bằng -3 khi x=-1

b)\(x^2-4x+y^2+2y-5=\left(x-2\right)^2+\left(y+1\right)^2-10\)

Lại có : \(\left(x-2\right)^2\ge0;\left(y+1\right)^2\ge0=>\left(x-2\right)^2+\left(y+1\right)^2-10\ge-10\)

Dấu "=" xảy ra khi và chỉ khi \(x-2=y+1=0=>x=2;y=-1\)

\(\text{a) }1-4x-2x^2\)

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

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

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

\(\text{Vì }-2\left(x+1\right)^2\le0\)

\(\text{nên }-2\left(x+1\right)^2+3\le3\)

\(\text{Do đó: }GTLN=3\), dấu bằng  xảy ra khi \(x=-1\)

\(\text{b) }x^2-4x+y^2+2y-5\)

\(=\left(x^2-4x+4\right)+\left(y^2+2y+1\right)-10\)

\(=\left(x-2\right)^2+\left(y+1\right)^2-10\)

\(\text{Vì }\left(x-2\right)^2\ge0;\left(y+1\right)^2\ge0\)

\(\text{nên }\left(x-2\right)^2+\left(y+1\right)^2\ge0\)

\(\text{hay }\left(x-2\right)^2+\left(y+1\right)^2-10\ge-10\)

\(\text{Do đó: }GTNN=-10\), dấu bằng xảy ra tai \(x=2\)và  \(y=-1\)

\(A=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\\ A_{min}=4\Leftrightarrow x=1\\ B=2\left(x^2-3x\right)=2\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{9}{2}\\ B=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\\ B_{min}=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\\ C=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\\ C_{max}=7\Leftrightarrow x=2\)

a,\(A=x^2-2x+5=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\)

Dấu "=" \(\Leftrightarrow x=-1\)

b,\(B=2\left(x^2-3x\right)=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

Dấu "=" \(\Leftrightarrow x=\dfrac{3}{2}\)

c,\(=C=-\left(x^2-4x-3\right)=-\left[\left(x^2-4x+4\right)-7\right]=-\left(x-2\right)^2+7\le7\)

Dấu "=" \(\Leftrightarrow x=2\)

a) Ta có: \(\left(x-2\right)^2\ge0\forall x\)

nên Dấu '=' xảy ra khi x-2=0

hay x=2

Vậy: Gtnn của biểu thức \(\left(x-2\right)^2\) là 0 khi x=2

Đáp số A=-25

A = |x - 1| - 25

Ta có :

|x - 1| \(\ge\)0

|x - 1| - 25 \(\ge\)-25

=> Min_A = -25

<=> |x - 1| - 25 = -25

<=> |x - 1| = 0

<=> x = 1 

Vì \(\left|x-1\le\right|0\)

      Suy ra:\(\left|x-1\right|-25\le-25\)

Dấu = xảy ra khi x-1=0;x=1

              Vậy Min A=-25 khi x=1