Đáp án :

Là C

# Hok tốt !

Trả lời:

a, x² + 4y² + 4xy - 16

= ( x² + 4xy + 4y² ) - 16

= [ x² + 2.x.2y + ( 2y )² ] - 16

= ( x + 2y )² - 4²

= ( x + 2y - 4 ) ( x + 2y + 4 )

b, 5x² - 10xy + 5y²

= 5 ( x² - 2xy + y² )

= 5 ( x - y )²

đáp án A sai

đáp án C sai

Trả lời:

( x - 2 ) ( x² + 2x + 4 ) 

= x³ + 2x² + 4x - 2x² - 4x - 8 

= x³ - 8 ( đpcm )

\(\left(x-2\right)\left(x^2+2x+4\right)=x^3-8\)

Mà: \(\left(x-2\right)\left(x^2+2x+4\right)\)

\(\Leftrightarrow x^3+2x^2+4x-2x^2-4x-8\)

\(\Leftrightarrow x^3-8\) \(\left(đpcm\right)\)

\(A=x^2-2x+3\left(\frac{1}{2}x-5\right)\)

\(A=\frac{1}{2}x^3-5x^2-x^2+10x+\frac{3}{2}x-15\)

\(A=\frac{1}{2}x^3-6x^2+\frac{23}{2}x-15\)

a) 2(3x - 1)(2x + 5) - 6(2x - 1)(x + 2) = -6

<=> 2(6x² + 13x - 5) - 6(2x² + 3x - 2) = -6

<=> 12x² + 26x - 10 - 12x² - 18x + 12 = -6

<=> 8x = -8

<=> x = -1

Vậy S = {-1}

b)Đk: x \(\ge\)0

 \(3\left(2\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)-\left(2\sqrt{x}-3\right)\left(9\sqrt{x}-1\right)-3=-3\)

<=> \(3\left(6x-5\sqrt{x}+1\right)-18x+19\sqrt{x}-3=0\)

<=> \(18x-15\sqrt{x}+3-18x+19\sqrt{x}-3=0\)

<=> \(4\sqrt{x}=0\) <=> x = 0 (tm)

vậy S = {0)

Ta có:\(\frac{a}{a+b+c+d}< \frac{a}{a+b+c}< \frac{a+d}{a+b+c+d}\)

Tương tự ta có:\(\frac{b}{a+b+c+d}< \frac{b}{b+c+d}< \frac{b+a}{a+b+c+d}\)

                        \(\frac{c}{a+b+c+d}< \frac{c}{c+d+a}< \frac{c+b}{a+b+c+d}\)

                         \(\frac{d}{a+b+c+d}< \frac{d}{d+a+b}< \frac{d+c}{a+b+c+d}\)

Cộng vế với vế ta được: \(1< \frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}< 2\)  (đpcm0

bn ghi ngoặc đi,để ri khó nhìn