\(x^2 +5x+6=0\)

\(<=>x^2+3x+2x+6=0\)

\(<=>x(x+3)+2(x+3)=0\)

\(<=>(x+3)(x+2)=0\)

\(<=>\) $\begin{cases} x=-3\\x=-2 \end{cases}$

Vậy `S=`{`-3;-2`}

Lời giải:

$x^2+5x+6=0$

$\Leftrightarrow (x^2+2x)+(3x+6)=0$

$\Leftrightarrow x(x+2)+3(x+2)=0$

$\Leftrightarrow (x+2)(x+3)=0$

$\Leftrightarrow x+2=0$ hoặc $x+3=0$

$\Leftrightarrow x=-2$ hoặc $x=-3$

Bài 27.

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

b) \(8-12x+6x^2-x^3=\left(2-x\right)^3\)

Bài 28.

a) \(x^3+12x^2+48x+64=\left(x+4\right)^3\)

Tại \(x=6\) ta có \(\left(x+4\right)^3=\left(6+4\right)^3=10^3=1000\)

b) \(x^3-6x^2+12x-8=\left(x-2\right)^3\)

Tại \(x=22\) ta có \(\left(x-2\right)^3=\left(22-2\right)^3=20^3=8000\)

Ta có :

2n²+5n-1 ⋮ 2n-1

=> 2n²-n+6n-3+2 ⋮ 2n-1

=> n(2n-1)+3(2n-1)+2 ⋮ 2n-1

=> (n+3)(2n-1) + 2 ⋮ 2n-1

Mà (n+3)(2n-1) ⋮ 2n-1

=> 2 ⋮ 2n-1 mà 2n-1 là số lẻ

=> 2n-1 ∈ {-1;1}

=> 2n∈ {0;2}

=> n ∈ {0;1}

Lời giải:

$x^2+2y^2+2xy-2x-2y+40=0$

$\Leftrightarrow (x^2+2xy+y^2)+y^2-2x-2y+40=0$

$\Leftrightarrow (x+y)^2-2(x+y)+1+y^2+40=0$

$\Leftrightarrow (x+y-1)^2+y^2=-40<0$ (vô lý) 

Do đó không tồn tại $x,y$ thỏa mãn đề. Do đó không tính được $P$

\(\dfrac{20009^2}{20008^2+20010^2-2}=\dfrac{20009^2}{\left(20008^2-1^2\right)+\left(20010^2-1^2\right)}=\dfrac{20009^2}{\left(20008+1\right)\left(20008-1\right)+\left(20010+1\right)\left(20010-1\right)}\)

\(=\dfrac{20009^2}{20009.20007+20011.20009}=\dfrac{20009^2}{20009\left(20007+20011\right)}\)

\(=\dfrac{20009}{20007+20011}=\dfrac{20009}{40018}=\dfrac{20009}{2.20009}=\dfrac{1}{2}\)

1/2

Lời giải:
a. 

$(5a+5)^2+10(a-3)(1+a)+a^2-6a+9$

$=(5a+5)^2+2(a-3)(5a+5)+(a-3)^2$

$=(5a+5+a-3)^2$

$=(6a+2)^2$

b.

$B=(a-b)^3-3(b-a)^2c+3(a-b)c^2-c^3$

$=(a-b)^3-3(a-b)^2c+3(a-b)c^2-c^3$

$=(a-b-c)^3$

1) (3x² + \(\dfrac{y}{4}\) )³ = (3x²)³ + 3.(3x²)². \(\dfrac{y}{4}\) + 3.3x² . \(\dfrac{y^2}{16}\) + \(\dfrac{y^3}{64}\)

2) (4 - 26)³ = 4³ - 3.4².26 + 3.4.26² - 26³

3) (2a + 3b)² = (2a)² + 2.2a.3b + (3b)²

4) (\(\dfrac{b}{2}\) - 5).(\(\dfrac{b}{2}\) + 5) = \(\left(\dfrac{b}{2}\right)^2\) - 5²

a.

\(\left(x^2+x\right)^2-6x^2-6x+8=0\)

\(\Leftrightarrow\left(x^2+x\right)^2-6\left(x^2+x\right)+8=0\)

\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)-4\left(x^2+x\right)+8=0\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-4\left(x^2+x-2\right)=0\)

\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2+x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\dfrac{-1\pm\sqrt{17}}{2}\end{matrix}\right.\)

b.

\(\left(x^2+10x+16\right)\left(x^2+10x+24\right)-20=0\)

\(\Leftrightarrow\left(x^2+10x+16\right)\left(x^2+10x+16+8\right)-20=0\)

\(\Leftrightarrow\left(x^2+10x+16\right)^2+8\left(x^2+10x+16\right)-20=0\)

\(\Leftrightarrow\left(x^2+10x+16\right)^2-2\left(x^2+10x+16\right)+10\left(x^2+10x+16\right)-20=0\)

\(\Leftrightarrow\left(x^2+10x+16\right)\left(x^2+10x+14\right)+10\left(x^2+10x+14\right)=0\)

\(\Leftrightarrow\left(x^2+10x+14\right)\left(x^2+10x+26\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+10x+14=0\\x^2+10x+26=0\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow x=-5\pm\sqrt{11}\)

\(\Leftrightarrow x^4+2x^3+2x^2-2x^2-4x-4=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+2x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\\\left(x+1\right)^2+1=0\left(\text{vô nghiệm}\right)\end{matrix}\right.\) 

\(\Leftrightarrow x=\pm\sqrt{2}\)