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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Gọi CTHH của hợp chất M là $N_xO_y$
Ta có : 

$\dfrac{m_N}{m_O} = \dfrac{14x}{16y} = \dfrac{7}{16}$

và $PTK = 14x + 16y = 46$

Suy ra : x = 1 ; y = 2

Vậy CTHH của M là $NO_2$

\(CaSiO_3:20+14+8.3=58\\ H_3PO_4:3+15+8.4=50\\ Al_2\left(SO_4\right)_3:13.2+\left(16+8.4\right).3=170\\ KMnO_4:19+25+8.4=76\\ Na_2S:11.2+16=38\\ BaCl_2:56+17.2=90\)