a:

Sửa đề: \(M=\frac{a^2+2}{a^3-1}+\frac{a+1}{a^2+a+1}-\frac{1}{a-1}\)

ĐKXĐ: a<>1

Ta có: \(M=\frac{a^2+2}{a^3-1}+\frac{a+1}{a^2+a+1}-\frac{1}{a-1}\)

\(=\frac{a^2+2}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{a+1}{a^2+a+1}-\frac{1}{a-1}\)

\(=\frac{a^2+2+\left(a+1\right)\left(a-1\right)-a^2-a-1}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\frac{-a+1+a^2-1}{\left(a-1\right)\left(a^2+a+1\right)}=\frac{a^2-a}{\left(a-1\right)\left(a^2+a+1\right)}=\frac{a}{a^2+a+1}\)

b: \(a=\sqrt{7+4\sqrt3}+\sqrt{7-4\sqrt3}\)

\(=\sqrt{\left(2+\sqrt3\right)^2\left.\right.}+\sqrt{\left(2-\sqrt3\right)^2}\)

\(=2+\sqrt3+2-\sqrt3=4\)

Thay a=4 vào M, ta được:

\(M=\frac{4}{4^2+4+1}=\frac{4}{16+5}=\frac{4}{21}\)

c: \(M=\frac15\)

=>\(\frac{a}{a^2+a+1}=\frac15\)

=>\(a^2+a+1=5a\)

=>\(a^2-4a+1=0\)

=>\(a^2-4a+4-3=0\)

=>\(\left(a-2\right)^2=3\)

=>\(\left[\begin{array}{l}a-2=\sqrt3\\ a-2=-\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}a=\sqrt3+2\left(nhận\right)\\ a=-\sqrt3+2\left(nhận\right)\end{array}\right.\)

a: Thay a=9 và b=15 vào P, ta được:

\(P=\left(9+1\right)\cdot2+\left(15+1\right)\cdot3\)

\(=10\cdot2+16\cdot3=20+48=68\)

b: \(m=2\cdot a+3\cdot b+5=2\cdot9+3\cdot15+5=68\)

mà P=68

nên P=m

a: \(2\dfrac{3}{5}+1\dfrac{2}{5}\cdot\dfrac{31}{2}\)

\(=\dfrac{13}{5}+\dfrac{7}{5}\cdot\dfrac{31}{2}\)

\(=\dfrac{26}{10}+\dfrac{217}{10}=\dfrac{243}{10}\)

b: \(4\dfrac{3}{4}-3\dfrac{2}{3}:1\dfrac{1}{6}\)

\(=\dfrac{19}{4}-\dfrac{11}{3}:\dfrac{7}{6}\)

\(=\dfrac{19}{4}-\dfrac{11}{3}\cdot\dfrac{6}{7}\)

\(=\dfrac{19}{4}-\dfrac{22}{7}\)

\(=\dfrac{19\cdot7-22\cdot4}{28}=\dfrac{45}{28}\)

a/ = 33/4 x 23/5 = 759/20

b/ = 4/3 : 9/4 = 4/3 x 4/9 = 16/27

2:

a: \(=\dfrac{1}{3}\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)=-\dfrac{1}{3}\cdot2=-\dfrac{2}{3}\)

1:

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(=-4-\dfrac{1}{4}=-\dfrac{17}{4}\)

Bài 1:

\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(A=\left(7-6-5\right)-\left(\dfrac{3}{4}+\dfrac{5}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\)

\(A=-4-\dfrac{3+5-7}{4}+\dfrac{1+4-5}{3}\)

\(A=-4-\dfrac{1}{4}+\dfrac{0}{3}\)

\(A=-\dfrac{16}{4}-\dfrac{1}{4}+0\)

\(A=\dfrac{-16-1}{4}\)

\(A=-\dfrac{17}{4}\)

Bài 2:

\(\dfrac{1}{3}\cdot-\dfrac{4}{5}+\dfrac{1}{3}\cdot-\dfrac{6}{5}\)

\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{-4-6}{5}\)

\(=\dfrac{1}{3}\cdot\dfrac{-10}{5}\)

\(=\dfrac{1}{3}\cdot-2\)

\(=-\dfrac{2}{3}\)

M=\(\frac{1}{1\cdot2}\)- \(\frac{1}{2\cdot3}\)+ \(\frac{1}{2\cdot3}\) - \(\frac{1}{3\cdot4}\) +..........+ \(\frac{1}{10\cdot11}\) - \(\frac{1}{11\cdot12}\)

= \(\frac{1}{2}\) - \(\frac{1}{11\cdot12}\) 

=\(\frac{65}{132}\) 

A = 2⁵.(-5)² - 8² - 7

= 32.25 - 64 - 7

= 729

= 27²

B = 2³.(-4)² + (-3)².3² - 40

= 8.16 + 9.9 - 40

= 169

= 13²

C = (1/4 - 1/2 - 1)³ . (2 - 2/5)³

= (-5/4)³ . (8/5)³

= (-5/4 . 8/5)³

= (-2)³

D = (-1/4)² : (1/2 - 1/3)

= 1/16 : 1/6

= 3/8

E = 4 . (1/4)² + 25 . [(3/4)³ : (5/4)³] : (3/2)³

= 1/4 + 25 . (3/4 . 5/4)³ : (3/2)³

= 1/4 + 25 . (15/16)³ : 27/8

= 1/4 + 25 . 3375/4096 : 27/8

= 1/4 + 84375/4096 : 27/8

= 1/4 + 3125/512

= 3253/512

F = 2³ + 3.(1/2)⁰ - 1 + [(-2)² : 1/2] - 8

= 8 + 3.1 - 1 + (4 : 1/2) - 8

= 8 + 3 - 1 + 8 - 8

= 10

a)

\(=\dfrac{13}{5}+\dfrac{7}{5}\cdot\dfrac{7}{2}\)

\(=\dfrac{13}{5}+\dfrac{49}{10}\\ =\dfrac{26}{10}+\dfrac{49}{10}\\ =\dfrac{15}{2}\)

b)

\(=\dfrac{52}{4}-\dfrac{11}{3}:\dfrac{7}{6}\)

\(=\dfrac{52}{4}-\dfrac{22}{7}\\ =\dfrac{69}{7}\)

a) $2\dfrac35 + 1\dfrac25 . 3\dfrac12$

$= \dfrac{13}5 + \dfrac75.\dfrac72$

$= \dfrac{26}{10} + \dfrac{49}{10}$

$=\dfrac{15}2$.

b) $4\dfrac34 - 3\dfrac23 : 1\dfrac16$

$= \dfrac{19}4 - \dfrac{11}3 : \dfrac76$

$= \dfrac{19}4 - \dfrac{11}3 . \dfrac67$

$= \dfrac{19}4 - \dfrac{22}7$

$= \dfrac{45}{28}$.