\(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\)

\(=\left(5x-3y\right)^2-16z^2=25x^2-30xy+9y^2-16z^2=\)

\(=25x^2-30xy+9y^2-16\left(x^2-y^2\right)=\)

\(=9x^2-30xy+25y^2=\left(3x-5y\right)^2\)

+) \(\left(2x-1\right)^2-\left(2x+3\right)\left(2x-1\right)=0\)

\(\left(2x-1\right)\left(2x-1-2x-3\right)=0\)

\(-4\left(2x-1\right)=0\)

\(x=\dfrac{1}{2}\)

+) \(\left(x+5\right)\left(x-2\right)-\left(x-3\right)\left(x+3\right)\)

\(=x^2-2x+5x-10-\left(x^2-9\right)\)

\(=3x-1\)

a, <=> (2x-1)(2x-1-2x-3)=0

<=> (2x-1).-4=0

<=> -8x + 4= 0

<=> -8x=-4 <=> x= 1/2

Mình chỉ có đề thi học sinh giỏi các cấp của toán thôi 

ừ

The - the

more simply

Which

traditional

communal

craft

herding

reading

the

When

harder

a, 3

b, 0

c, 28

d, 2019,75

a,Có

A = \(x^2-4x+7\)

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

\(A=\left(x-2\right)^2+3\) \(\ge\)3 

Vì \(\left(x-2\right)^2\ge0\forall x\)

Dấu''='' xảy ra khi x=2

Vậy Amin = 3 khi x=2

b, Có

\(B=x^2+10x\)

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

\(B=\left(x+5\right)^2-25\ge-25\) 

Vì \(\left(x+5\right)^2\ge0\forall x\)

Dấu''='' xảy ra khi x=-5

Vậy Bmin = -25 khi x=-5

c,Có

\(C=2x^2-4x+15\)

\(C=2\left(x^2-2x+1\right)+13\)

\(C=2\left(x-1\right)^2+13\ge13\)

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

Dấu ''='' xảy ra khi x = 1

Vậy Cmin = 13 khi x=1

d,Có

\(M=x^2-3x+2022\)

\(M=\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{8079}{4}\)

\(M=\left(x-\dfrac{3}{2}\right)^2+\dfrac{8079}{4}\) \(\ge\) \(\dfrac{8079}{4}\)

Vì \(\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\)

Dấu ''='' xảy ra khi x= 3/2

Vậy Mmin = \(\dfrac{8079}{4}\) khi x= 3/2