https://friend20.com/vn/d20/quiz/69707687

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bài 5

Nhân phá tung ra ta có :

\(3x^2+2x+x^2+2x+1-4x^2+25=-12\Leftrightarrow4x=-38\Leftrightarrow x=-\frac{19}{2}\)

bài 6.

ta có : \(2n^2+5n+2-3=\left(2n+1\right)\left(n+2\right)-3\text{ chia hết cho }2n+1\)

Khi 3 chia hết cho 2n+1 hay \(2n+1\in\left\{\pm1,\pm3\right\}\Rightarrow n\in\left\{-2,-1,0,1\right\}\)

Bài 5. 

\(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=-12\)

\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=-12\)

\(\Leftrightarrow4x=-38\)

\(\Leftrightarrow x=-\frac{38}{5}\).

Bài 6. 

\(\left(2n^2+5n-1\right)⋮\left(2n-1\right)\Leftrightarrow\left(4n^2+10n-2\right)⋮\left(2n-1\right)\)(1)

Có \(4n^2+10n-2=4n^2-4n+1+14n-7+4=\left(2n-1\right)^2+7\left(2n-1\right)+4\)

Do đó (1) tương đương với \(4⋮\left(2n-1\right)\Leftrightarrow2n-1\in\left\{-1,1\right\}\)(vì \(2n-1\)là số lẻ)

\(\Leftrightarrow n\in\left\{0,1\right\}\).

 \(4x^2+4x+1-4x^2-8x-16=9\)

\(-4x-15=9\)

\(-4x=24\)

x = -6

học tốt

• k phải -6 nha

[2x + 1]²-4[x+2]²= 9

4x²+4x+1-4x²-16x-16=9

-12x-15=9

-12x=24

x=-2

\(\left(x+3\right)\left(5x-1\right)=5\left(x+1\right)\left(x-2\right)\)

\(\Leftrightarrow5x^2+14x-3=5\left(x^2-x-2\right)\)

\(\Leftrightarrow5x^2+14x-3=5x^2-5x-10\)

\(\Leftrightarrow19x+7=0\Leftrightarrow x=-\frac{7}{19}\)

a.\(x^2-y^2+x+y=\left(x+y\right)\left(x-y+1\right)\)

b.\(3\left(x^2-2xy+y^2\right)-12=3\left[\left(x-y\right)^2-4\right]=3\left(x-y+2\right)\left(x-y-2\right)\)

c.\(3\left(x+y\right)-\left(x^2+2xy+y^2\right)=\left(x+y\right)\left(3-x-y\right)\)

d.\(\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)

\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)

Số lớn hơn 100 là:

101;102;103;104;....

HT ~ K cho tớ nhaaa

#Army

100< 101,102,103,104,105,....

3x(x + 1) - (x - 2)(3x + 1) = 12

<=> 3x² + 3x - (3x² - 5x - 2) = 12 

<=> 3x² + 3x - 3x² + 5x + 2 = 12 

<=> 8x = 10 

<=> x = 1,25 

Vậy x = 1,25

3x(x+1)-(x-2)(3x+1)=12

x=5/4

Trả lời:

sửa: \(x^3-4x^2-8x+8\)

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

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

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

\(=\left(x+2\right)\left(x^2-6x+4\right)\)