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

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) \(A=1-8x-x^2=-\left(x^2+8x+16\right)+17=-\left(x-4\right)^2+17\le17\)

\(ĐTXR\Leftrightarrow x=4\)

b) \(B=5-2x+x^2=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\)

\(ĐTXR\Leftrightarrow x=1\)

c) \(C=x^2+4y^2-6x+8y-2021=\left(x^2-6y+9\right)+\left(4y^2+8y+4\right)-2034=\left(x-3\right)^2+\left(2y+2\right)^2-2034\ge-2034\)

\(ĐTXR\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)

a: Ta có: \(A=-x^2-8x+1\)

\(=-\left(x^2+8x-1\right)\)

\(=-\left(x^2+8x+16-17\right)\)

\(=-\left(x+4\right)^2+17\le17\forall x\)

Dấu '=' xảy ra khi x=-4

b: Ta có: \(x^2-2x+5\)

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

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

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

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 \(x=y=-\dfrac{1}{2}\)

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

1.

a) [124 - (20 - 4x)] : 30 + 7 = 11

=> [124 - (20 - 4x)] : 30 = 11 - 7

=> [124 - (20 - 4x)] : 30 = 4

=> 124 - (20 - 4x) = 4 x 30

=> 124 - (20 - 4x) = 120

=> 20 - 4x = 124 - 120

=> 20 - 4x = 4

=> 4x = 20 - 4

=> 4x = 16

=> x = 16 : 4

=> x = 4

Vậy x = 4

b) |2x - 5| = 1

TH1: 2x - 5 = 1

=> 2x = 1 + 5 

=> 2x = 6

=> x = 6 : 2

=> x = 3

TH2: 2x - 5 = -1

=> 2x = -1 + 5

=> 2x = 4

=> x = 4 : 2

=> x = 2

Vậy x = 3 hoặc x = 2

     2x-5=1

     2x    =1+5

     2x    = 6

      x     = 6 : 2

      x     = 3

b)  (2x-6)(x+4)=0

c)  (x-3)(x+4)<0

d)  (x+2)(X-5)>0

bạn đăg tách ra cho m.n cùng giúp nhé

Bài 2 : 

a, \(A=\left|2x-4\right|+2\ge2\)

Dấu ''='' xảy ra khi x = 2 

Vậy GTNN A là 2 khi x = 2 

b, \(B=\left|x+2\right|-3\ge-3\)

Dấu ''='' xảy ra khi x = -2 

Vậy GTNN B là -3 khi x = -2 

\(a.A=\left(x-2\right)^2+\left(y+1\right)^2+1\ge1\forall x;y\) . " = " \(\Leftrightarrow x=2;y=-1\) 

b.\(B=7-\left(x+3\right)^2\le7\forall x\)  " = " \(\Leftrightarrow x=-3\)

c.\(C=\left|2x-3\right|-13\ge-13\forall x\)  " = " \(\Leftrightarrow x=\dfrac{3}{2}\)

d.\(D=11-\left|2x-13\right|\le11\forall x\)  " = " \(\Leftrightarrow x=\dfrac{13}{2}\)