\(\dfrac{1}{2}a^2+1=2\) 

\(\Rightarrow\dfrac{1}{2}a^2=2-1\)

\(\Rightarrow\dfrac{1}{2}a^2=1\)

\(\Rightarrow a^2=2\)

\(\Rightarrow a^2=\left(\sqrt{2}\right)^2\)

\(\Rightarrow a\in\left\{\sqrt{2};-\sqrt{2}\right\}\)

\(\dfrac{1}{2a^2+1}=2\)

\(2a^2+1=1\)

\(2a^2=0\)

\(a^2=0\)

\(a=0\)

1.2.3.4.5.6.7.8.9 - 1.2.3.4.5.6.7.8 - 1.2.3.4.5.6.7 - 8²

= 1.2.3.4.5.6.7.(8.9 - 8 - 1) - 64

= 5040.63 - 64

= 317520 - 64

= 317456

\(1\times2\times3\times4\times5\times6\times7\times8\times9-1\times2\times3\times4\times5\times6\times7\times8-1\times2\times3\times4\times5\times6\times7-8^2\)

\(=1\times2\times3\times4\times5\times6\times7\times\left(8\times9-8-1\right)-64\)

\(=5040\times63-64\)

\(=317520-64\)

\(=317456\)

Lời giải:
a.

$\frac{1}{a}+\frac{-1}{b}=\frac{1}{a-b}$

$\Rightarrow \frac{b-a}{ab}=\frac{1}{a-b}$

$\Rightarrow (b-a)(a-b)=ab$

$\Rightarrow -(a-b)^2=ab$
Với $a,b\in\mathbb{N}^*$, $ab>0$ còn $-(a-b)^2\leq 0$

Do đó $-(a-b)^2\neq ab$

$\Rightarrow$ không tồn tại $a,b\in\mathbb{N}^*$ thỏa mãn điều kiện đề.

b.

$\frac{5}{2a}=\frac{1}{6}+\frac{b}{3}$

$\frac{15}{6a}=\frac{1+2b}{6}=\frac{a+2ab}{6a}$

$\Rightarrow a+2ab=15$

$\Rightarrow a(1+2b)=15$

Do $a,b$ nguyên nên $a, 1+2b$ nguyên. Mà tích $a(1+2b)=15$ nên xét các TH sau:

TH1: $a=1, 2b+1=15\Rightarrow a=1; b=7$

TH2: $a=-1, 2b+1=-15\Rightarrow a=-1; b=-8$

TH3: $a=15, 2b+1=1\Rightarrow a=15; b=0$

TH4: $a=-15; 2b+1=-1\Rightarrow a=-15; b=-1$

TH5: $a=3, 2b+1=5\Rightarrow a=3; b=2$

TH6: $a=-3, 2b+1=-5\Rightarrow a=-3; b=-3$

TH7: $a=5; 2b+1=3\Rightarrow a=5; b=1$

TH8: $a=-5; 2b+1=-3\Rightarrow a=-5; b=-2$

e cảm ơn

giúp mình bài 3

\(\dfrac{-5}{7}:\dfrac{15}{31}\)

\(=\dfrac{-5}{7}\cdot\dfrac{31}{15}\)

\(=\dfrac{-5\cdot31}{7\cdot15}\)

\(=\dfrac{-31}{7\cdot3}\)

\(=\dfrac{-31}{21}\)

-1-2-3-4-5-6-7-8-9-10=-55

\(\dfrac{-49}{81}\cdot\dfrac{27}{-77}\)

\(=\dfrac{-7^2}{3^2\cdot9}\cdot\dfrac{3\cdot9}{-7\cdot11}\)

\(=\dfrac{7}{3\cdot11}\)

\(=\dfrac{7}{33}\)

Bài 3:

\(A=\dfrac{1}{2}+\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+...+\left(\dfrac{3}{2}\right)^{2012}\)

Đặt: \(C=\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+...+\left(\dfrac{3}{2}\right)^{2012}\)

\(\dfrac{3}{2}C=\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+...+\left(\dfrac{3}{2}\right)^{2013}\)

\(\dfrac{3}{2}C-C=\left[\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+...+\left(\dfrac{3}{2}\right)^{2013}\right]-\left[\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+...+\left(\dfrac{3}{2}\right)^{2012}\right]\)

\(\dfrac{1}{2}C=\left(\dfrac{3}{2}\right)^{2013}-\dfrac{3}{2}\)

\(C=2\left[\left(\dfrac{3}{2}\right)^{2013}-\dfrac{3}{2}\right]\)

\(A=\dfrac{1}{2}+2\left[\left(\dfrac{3}{2}\right)^{2013}-\dfrac{3}{2}\right]\)  

\(\Rightarrow B-A=\left(\dfrac{3}{2}\right)^{2023}:2-\dfrac{1}{2}-2\left[\left(\dfrac{3}{2}\right)^{2013}-\dfrac{3}{2}\right]\)

\(=\dfrac{\left(\dfrac{3}{2}\right)^{2023}}{2}-\dfrac{1+4\left[\left(\dfrac{3}{2}\right)^{2013}-\dfrac{3}{2}\right]}{2}\)

\(=\dfrac{\left(\dfrac{3}{2}\right)^{2023}-4\left(\dfrac{3}{2}\right)^{2013}-5}{2}\)