\(A=-x^2+3x-7\)

\(=-\left(x^2-3x+7\right)\)

\(=-\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{19}{4}\right)\)

\(=-\left(x-\dfrac{3}{2}\right)^2-\dfrac{19}{4}< 0\forall x\)

\(3x-7-x^2=-\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{19}{4}=-\left(x-\dfrac{3}{2}\right)^2-\dfrac{19}{4}\le-\dfrac{19}{4}< 0\)

Bài 1: 

a) Ta có: \(A=-x^2-4x-2\)

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

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

\(=-\left(x+2\right)^2+2\le2\forall x\)

Dấu '=' xảy ra khi x=-2

b) Ta có: \(B=-2x^2-3x+5\)

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

\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)

\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{4}\)

c) Ta có: \(C=\left(2-x\right)\left(x+4\right)\)

\(=2x+8-x^2-4x\)

\(=-x^2-2x+8\)

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

\(=-\left(x^2+2x+1-9\right)\)

\(=-\left(x+1\right)^2+9\le9\forall x\)

Dấu '=' xảy ra khi x=-1

Bài 2: 
a) Ta có: \(=25x^2-20x+7\)

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

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

b) Ta có: \(B=9x^2-6xy+2y^2+1\)

\(=9x^2-6xy+y^2+y^2+1\)

\(=\left(3x-y\right)^2+y^2+1>0\forall x,y\)

c) Ta có: \(E=x^2-2x+y^2-4y+6\)

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

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

`B = x^2- 2xy + y^2 + 2x - 10y + 17

`2B = 2x^2 - 4xy + 2y^2 + 4x - 20y + 34`

`= (x-y)^2 + (x+2)^2 + (y-5)^2 + 5 >= 5`.

Mik cảm ơn

Bai 1:

(x-5)(3x+3)-3x(x-3)+3x+7

=x(3x+3)-5(3x+3)-(3x²-9x)+3x+7

=3x²+3x-15x+15-3x²+9x+3x+7

=22

=>biểu thức này không phụ thuộc vào giá trị của biến.

Bài 2:

(x+2)(x+1)-(x-3)(x+5)=0

x²+x+2x+2-x²-5x+3x+15=0

x+17=0

x= -17

Thanks bn nhiều.

1 + 1=

Ai có nhu cầu tình dục cao thì liên hẹ vs e nha, e làm cho, 20k thôi, e cần tiền chữa bệnh cho mẹ

\(B=3xy-8y-15x+40\)

\(=y\left(3x-8\right)-5\left(3x-8\right)\)

\(=\left(3x-8\right)\left(y-5\right)\)

Thay x=1999 và y=5 vào B ta được :

\(B=\left(3.1999-8\right)\left(5-5\right)\)

\(=0\)

A xác định khi 5x-10 ≠0 <=> X ≠ 2b) A = x²-4x+4/5x-10= (x-2)²/5(x-2)= x-2/5c) x= -2018<=> A = -2018-2/5= -2020/5 = -404

Chúc bạn học tốt

a) ĐKXĐ: \(x\ne2\)

b) Ta có: \(A=\dfrac{x^2-4x+4}{5x-10}\)

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

\(=\dfrac{x-2}{5}\)

\(A=\dfrac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}=\dfrac{3x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{3x}{2\left(x-3\right)}\)

Thay x=1 vào A ta được\(A=\dfrac{3x}{2\left(x-3\right)}=\dfrac{3.1}{2\left(1-3\right)}=\dfrac{3}{2.\left(-2\right)}=\dfrac{-3}{4}\)

Thay x=4 vào A ta được\(A=\dfrac{3x}{2\left(x-3\right)}=\dfrac{3.4}{2\left(4-3\right)}=\dfrac{12}{2.1}=\dfrac{12}{2}=6\)

\(A=\dfrac{3x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}=\dfrac{3x}{2\left(x-3\right)}\\ x=1\Leftrightarrow A=\dfrac{3}{2\left(-2\right)}=-\dfrac{3}{4}\\ x=4\Leftrightarrow A=\dfrac{12}{2}=6\)