Bài 1:

a: \(M=x^2-10x+3\)

\(=x^2-10x+25-22\)

\(=\left(x^2-10x+25\right)-22\)

\(=\left(x-5\right)^2-22>=-22\forall x\)

Dấu '=' xảy ra khi x-5=0

=>x=5

b: \(N=x^2-x+2\)

\(=x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi x-1/2=0

=>x=1/2

c: \(P=3x^2-12x\)

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

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

\(=3\left(x-2\right)^2-12>=-12\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

TA CÓ

\(\left(X-1\right)^2\ge0\)VỚI MỌI X \(\in\)R

\(\left(Y+3\right)^2\ge0\)VỚI MỌI Y\(\in\)R

=>\(\left(X-1\right)^2+\left(Y+3\right)^2+2002\ge2002\) VỚI MỌI X,Y \(\in\)R

=>BIỂU THỨC M ĐẠT GIA TRỊ NHỎ NHẤT LÀ 2002 KHI \(\left(X-1\right)^2=0\) VÀ \(\left(Y+3\right)^2=0\)=>X=1;Y=-3

Ta có: \(A=\left(x-3\right)^2+\left(11-x\right)^2\)

\(=x^2-6x+9+x^2-22x+121\)

\(=2x^2-28x+130\)

\(=2\left(x^2-14x+49+16\right)\)

\(=2\left(x-7\right)^2+32\ge32\forall x\)

Dấu '=' xảy ra khi x=7

\(A=\sqrt{x-2}+\sqrt{4-x}\ge\sqrt{x-2+4-x}=\sqrt{2}\)

\(A_{min}=\sqrt{2}\) khi \(\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

\(y=4x^2+\dfrac{9}{x^2}-3\ge2\sqrt{\dfrac{36x^2}{x^2}}-3=9\)

\(y_{min}=9\) khi \(x^2=\dfrac{3}{2}\)

\(P=\dfrac{x-1}{4}+\dfrac{1}{x-1}+\dfrac{1}{4}\ge2\sqrt{\dfrac{x-1}{4\left(x-1\right)}}+\dfrac{1}{4}=\dfrac{5}{4}\)

\(P_{min}=\dfrac{5}{4}\) khi \(x=\dfrac{3}{2}\)

Bài 3: 

a) Ta có: \(A=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)(đpcm)

d) Ta có: \(D=x^2-2x+2\)

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

\(=\left(x-1\right)^2+1>0\forall x\)(đpcm)

Bài 1: 

a) Ta có: \(A=x^2-2x+5\)

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

\(=\left(x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi x=1

b) Ta có: \(B=x^2-x+1\)

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

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

a) Ta có: \(A=x^2-2x+5\)

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

\(=\left(x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi x=1

b) Ta có: \(B=x^2-x+1\)

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

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

c) Ta có: \(C=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

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

\(=\left(x^2+5x\right)^2-36\ge-36\forall x\)

Dấu '=' xảy ra khi x(x+5)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

d) Ta có: \(x^2+5y^2-2xy+4y+3\)

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

\(=\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\forall x,y\)

Dấu '=' xảy ra khi \(x=y=-\dfrac{1}{2}\)

3. Tìm giá trị nhỏ nhất của các biểu thứca. A = 4x2  4x 11b. B = (x - 1) (x 2) (x 3) (x 6)c. C = x2 - 2x y2 - 4y 7Ai nha... - Hoc24

a. Ta có: ( x-2)² \(\ge\) 0 , \(\forall\) x

=> ( x-2)² +2023 \(\ge\) 2023

Vậy ...

Dấu bằng xảy ra khi x-2 = 0

b. (x-3)²+(y-2)²-2018

Ta có: \((x-3)^2 \ge0,\forall x\)

           \((y-2) ^2 \ge0,\forall y\) 

=> ( x-3)² + ( y-2)² \(\ge\) 0

=>  ( x-3)² + ( y-2)²-2018 \(\ge\) -2018, \(\forall\) x,y 

Vậy ...

Dấu bằng xảy ra khi x-3=0

                                 y-2=0

c. ( x+1)² +100

Ta có : ( x+1)² \(\ge0,\forall x\) 

=> ( x+1)²+100 \(\ge\) 100

Vậy ...

Dấu bằng xảy ra khi x+1=0

1 )Vì \(\left(x+2\right)^2\ge0;\left(y-3\right)^2\ge0\)

\(\Rightarrow\left(x+2\right)^2+\left(y-3\right)^2\ge0\)

\(\Rightarrow\left(x+2\right)^2+\left(y-3\right)^2+1\ge1\)

Dấu "=: xảy ra <=> \(\orbr{\begin{cases}\left(x+2\right)^2=0\\\left(y-3\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\y=3\end{cases}}}\)

Vậy ........

2 ) \(\frac{1}{\left(x-2\right)^2+2}\ge\frac{1}{2}\)

Dấu "=" xảy ra <=> x = 2

Vậy ..........