\(M=x^2-4x+5=x^2-4x+4+1=\left(x-2\right)^2+1.\)

Ta có: \(\left(x-2\right)^2\ge0\) \(\forall x\in R.\)

           \(1>0.\)

\(\Rightarrow\left(x-2\right)^2+1\ge1.\Rightarrow M\ge1.\)

Dấu \("="\) xảy ra. \(\Leftrightarrow\left(x-2\right)^2+1=1.\Leftrightarrow\left(x-2\right)^2=0.\Leftrightarrow x=2.\)

Vậy GTNN của M = 1 khi x = 2.

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

vì \(\left(x-2\right)^2\ge0\) nên \(\left(x-2\right)^2+1\ge1\)

=>\(M\ge1\) dấu''='' xảy ra  khi M = 1<=>x-2=0<=>x=2

kl:\(M_{min}=1\) khi và chỉ khi x =2

a, Vì \(\left(x-2\right)^2\ge0\) nên \(A=\left(x-2\right)^2+24\ge24\)

Dấu '=' xảy ra khi và chỉ khi: \(\left(x-2\right)^2=0\Leftrightarrow x=2\)

Vậy GTNN của A là 24 khi x=2.

b,Vì \(-x^2\le0\) nên \(B=-x^2+\dfrac{13}{5}\le\dfrac{13}{5}\)

Dấu '=' xảy ra khi và chỉ khi: \(-x^2=0\Leftrightarrow x=0\)

Vậy GTLN của B là \(\dfrac{13}{5}\) khi x=0

Ai trả lời nhanh và đúng mik give tick xanh nhé.

Tìm giá trị nhỏ nhất của biểu thức:

a) Ta có: 

\(M=2x^2+4x+7\)

\(M=2\cdot\left(x^2+2x+\dfrac{7}{2}\right)\)

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

\(M=2\cdot\left[\left(x+1\right)^2+2,5\right]\)

\(M=2\left(x+1\right)^2+5\)

Mà: \(2\left(x+1\right)^2\ge0\forall x\) nên:

\(M=2\left(x+1\right)^2+5\ge5\forall x\)

Dấu "=" xảy ra:

\(2\left(x+1\right)^2+5=5\Leftrightarrow2\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Vậy: \(M_{min}=5\) khi \(x=-1\)

b) Ta có:

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

\(N=x^2-2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}+\dfrac{3}{4}\)

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

Mà: \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\) nên \(N=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu '=" xảy ra: 

\(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\dfrac{3}{4}\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(N_{min}=\dfrac{3}{4}\) khi \(x=\dfrac{1}{2}\)

Tìm giá trị lớn nhất của biểu thức

a) Ta có: 

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

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

\(E=-\left[\left(2x\right)^2-2\cdot2x\cdot\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{15}{16}\right]\)

\(E=-\left[\left(2x-\dfrac{1}{4}\right)^2+\dfrac{15}{16}\right]\)

Mà: \(\left(2x+\dfrac{1}{4}\right)^2+\dfrac{15}{16}\ge\dfrac{15}{16}\forall x\) nên 

\(\Rightarrow E=-\left[\left(2x+\dfrac{1}{4}\right)^2+\dfrac{15}{16}\right]\le-\dfrac{15}{16}\forall x\)

Dấu "=" xảy ra:

\(-\left[\left(2x+\dfrac{1}{4}\right)^2+\dfrac{15}{16}\right]=-\dfrac{15}{16}\Leftrightarrow-\left(2x+\dfrac{1}{4}\right)^2-\dfrac{15}{16}=-\dfrac{15}{16}\)

\(\Leftrightarrow-\left(2x+\dfrac{1}{4}\right)^2=0\Leftrightarrow2x-\dfrac{1}{4}=0\Leftrightarrow x=\dfrac{1}{16}\)

Vậy: \(E_{max}=-\dfrac{15}{16}\) khi \(x=\dfrac{1}{16}\)

b) Ta có:

\(F=5x-3x^2+6\)

\(F=-3x^2+5x-6\)

\(F=-\left(3x^2-5x-6\right)\)

\(F=-3\left(x^2-\dfrac{5}{3}x-2\right)\)

\(F=-3\left[\left(x-\dfrac{5}{6}\right)^2-\dfrac{97}{36}\right]\)

\(F=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{36}\)

Mà: \(-3\left(x-\dfrac{5}{6}\right)^2\le0\forall x\) nên:

\(F=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{36}\le\dfrac{97}{36}\forall x\)

Dấu "=" xảy ra:

\(-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{36}=\dfrac{97}{36}\Leftrightarrow-3\left(x-\dfrac{5}{6}\right)^2=0\)

\(\Leftrightarrow x-\dfrac{5}{6}=0\Leftrightarrow x=\dfrac{5}{6}\)

Vậy: \(F_{max}=\dfrac{97}{36}\) khi \(x=\dfrac{5}{6}\)

\(M=2x^2+4x+7\)

\(=2\left(x^2+2x+\dfrac{7}{2}\right)\)

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

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

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

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

\(\Rightarrow2\left(x+1\right)^2+5\ge5\forall x\)

\(\Rightarrow M_{min}=5\Leftrightarrow2\left(x+1\right)^2=0\Leftrightarrow x=-1\)

Tương tự: \(N=x^2-x+1\)

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

\(\Rightarrow N_{min}=\dfrac{3}{4}\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)

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

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

\(=-4\left[x^2-2.x.\dfrac{1}{8}+\left(\dfrac{1}{8}\right)^2-\left(\dfrac{1}{8}\right)^2+\dfrac{1}{4}\right]\)

\(=-4\left[\left(x-\dfrac{1}{8}\right)^2+\dfrac{15}{64}\right]\)

\(=-4\left(x-\dfrac{1}{8}\right)^2-\dfrac{15}{16}\)

Vì \(-4\left(x-\dfrac{1}{8}\right)^2\le0\forall x\)

\(\Rightarrow-4\left(x-\dfrac{1}{8}\right)^2-\dfrac{15}{16}\le-\dfrac{15}{16}\forall x\)

\(\Rightarrow E_{max}=-\dfrac{15}{16}\Leftrightarrow-4\left(x-\dfrac{1}{8}\right)^2=0\Leftrightarrow x=\dfrac{1}{8}\)

Tương tự: \(F=5x-3x^2+6\)

\(=-3x^2+5x+6\)

\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{12}\le\dfrac{97}{12}\forall x\)

\(\Rightarrow F_{max}=\dfrac{97}{12}\Leftrightarrow-3\left(x-\dfrac{5}{6}\right)^2=0\Leftrightarrow x=\dfrac{5}{6}\)

Ta có C = x² - 4x + y² - y + 5 

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

= \(\left(x-2\right)^2+\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)

=> Min C = 3/4

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-2=0\\y-\frac{1}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=\frac{1}{2}\end{cases}}\)

Vậy Min C = 3/4 <=> x = 2 ; y = 1/2 

C = ( x² - 4x + 4 ) + ( y² - y + 1/4 ) + 3/4

= ( x - 2 )² + ( y - 1/2 )² + 3/4 ≥ 3/4 ∀ x.y 

Dấu "=" xảy ra <=> x = 2 ; y = 1/2 . Vậy MinC = 3/4

\(P=\left(x-2\right)^2+\left|y-x\right|+3\)

\(\left(x-2\right)^2>=0\forall x\)

\(\left|y-x\right|>=0\forall x,y\)

Do đó: \(\left(x-2\right)^2+\left|y-x\right|>=0\forall x,y\)

=>\(\left(x-2\right)^2+\left|y-x\right|+3>=3\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-2=0\\y-x=0\end{matrix}\right.\)

=>x=y=2

Ai giúp mik vs

Huhu ai giúp vs

                                                      Bài giải

\(M=\left(3x-\frac{1}{2}\right)^2-4\)

Do \(\left(3x-\frac{1}{2}\right)^2\ge0\text{ với mọi }x\text{ }\Rightarrow\text{ }\left(3x-\frac{1}{2}\right)^2-4\ge-4\)

Dấu " = " xảy ra khi \(3x-\frac{1}{2}=0\text{ }\Rightarrow\text{ }3x=\frac{1}{2}\text{ }\Rightarrow\text{ }x=\frac{1}{6}\)

Vậy GTNN của \(M=-4\text{ khi }x=\frac{1}{6}\)

a) \(A=\left|x+19\right|+\left|y-5\right|+1890\)

TA có: \(\hept{\begin{cases}\left|x+19\right|\ge0;\forall x,y\\\left|y-5\right|\ge0;\forall x,y\end{cases}\Rightarrow\left|x+19\right|+\left|y-5\right|\ge}0;\forall x,y\)

\(\Rightarrow\left|x+19\right|+\left|y-5\right|+1890\ge1890;\forall x,y\)

Dấu"="xảy ra \(\Leftrightarrow\hept{\begin{cases}\left|x+19\right|=0\\\left|y-5\right|=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-19\\y=5\end{cases}}\)

Vậy \(A_{min}=1890\Leftrightarrow\hept{\begin{cases}x=-19\\y=5\end{cases}}\)

b) \(B=-\left|x-7\right|-\left|y+13\right|+1945\)

Ta có: \(\hept{\begin{cases}-\left|x-7\right|\le0;\forall x,y\\-\left|y+13\right|\le0;\forall x,y\end{cases}}\)\(\Rightarrow-\left|x-7\right|-\left|y+13\right|\le0;\forall x,y\)

\(\Rightarrow-\left|x-7\right|-\left|y+13\right|+1945\le1945;\forall x,y\)

Dấu"="Xảy ra \(\Leftrightarrow\hept{\begin{cases}\left|x-7\right|=0\\\left|y+13\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=7\\y=-13\end{cases}}\)

Vậy MAX\(B=1945\Leftrightarrow\hept{\begin{cases}x=7\\y=-13\end{cases}}\)