thay số ta có

0 : ( 119 x 119 + 2005 ) + (119 ; 119 - 0X2005)

0 + 1-0= 1

****

Giải:

Thay a = 119 ; b = 0 vào biểu thức M 

Khi đó M = 0 : ( 119 x 119 + 2005 ) + ( 119 : 119 - 0 x 2005 )

M = 0 + ( 119 : 119 - 0 x 2005 )

M = 0 + ( 1 - 0 )

M = 0 + 1

M = 1

Vậy M = 1

Học tốt!!!

Tại a = 119 và b = 0 thì

\(M=0\div\left(119\cdot119+2005\right)+\left(119\div119-0\cdot2005\right)\)

\(M=0+1-0=1\)

Vậy M = 1

M = 0 : (119 x 119 + 2005 ) + ( 119 : 119 - 0 x 2005) 

= 0 + 1 

= 1

Thay a = 119 ; b = 0 vào biểu thức M ta có :

 \(M=0:\left(119.119+2005\right)+\left(119:a-b.2005\right)\)

 \(M=0+1\)

 \(M=1\)

\(M=b:\left(119.a+2005\right)+\left(119:a-b.2005\right)\)

\(M=0:\left(119.119+2005\right)+\left(119:119-0.2005\right)\)

\(M=0+\left(1-0\right)=1\)

Vì a = 119, b = 0. Thay vào M ta có:

\(M=b:\left(119\cdot a+2005\right)+\left(119:a-b\cdot2005\right)\)

\(=119:\left(119\cdot0+2005\right)+\left(119:119-0:2005\right)\)

\(=119:\left(0+2005\right)+\left(1-0\right)\)

\(=119:2005+1\)

a) thay a = 119 ;b = 0 vào bt : ( 119 x a +2005 ) +119 :a- b x 2005 ta có : 

( 119 x 199+2005) +119 :119 -0 x2005

<=> (14161+2005 ) +119 : 119 -0 x 2005

<=>16166+119 : 199 -0 x 2005

<=> 16166+1 - 0 x 2005 

<=> 16166 +1 - 0 

<=> 16167

b) 100 : {2 x [ 52-(25 - 8 )]}

<=> 100 : { 2 x [ 52 - 17] }

<=> 100 : { 2 x 35 }

<=> 100 : 70 

<=> \(\frac{10}{7}\)

c) 12 : { 390 : [ 500 - ( 125 +25 : 7 ) ] }

<=> 12 : { 390 : [ 500 - ( \(\frac{900}{7}\) ) ] }

<=> 12 : { 390 : [ \(\frac{2600}{7}\) ] }

<=> 12 : { \(\frac{21}{20}\) }

<=> \(\frac{80}{7}\)

a: Đặt 117=a; 119=b

\(A=3\frac{1}{117}\cdot4\frac{1}{119}-1\frac{116}{117}\cdot5\frac{118}{119}-\frac{5}{119}\)

\(=3\frac{1}{a}\cdot4\frac{1}{b}-\left(1+\frac{a-1}{a}\right)\cdot\left(5+\frac{b-1}{b}\right)-\frac{5}{b}\)

\(=\frac{3a+1}{a}\cdot\frac{4b+1}{b}-\frac{2a-1}{a}\cdot\frac{6b-1}{b}-\frac{5}{b}\)

\(=\frac{\left(3a+1\right)\left(4b+1\right)-\left(2a-1\right)\left(6b-1\right)-5a}{ab}\)

\(=\frac{12ab+3a+4b+1-\left(12ab-2a-6b+1\right)-5a}{ab}\)

\(=\frac{12ab+3a+4b+1-12ab+2a+6b-1-5a}{ab}=\frac{10b}{ab}=\frac{10}{a}\)

\(=\frac{10}{117}\)

b: Đặt 105=a; 651=b

\(B=2\frac{1}{315}\cdot\frac{1}{651}-\frac{1}{105}\cdot3\frac{650}{651}-\frac{4}{315\cdot651}+\frac{4}{105}\)

\(=\left(2+\frac{1}{3a}\right)\cdot\frac{1}{b}-\frac{1}{a}\cdot\left(3+\frac{b-1}{b}\right)-\frac{4}{3a\cdot b}+\frac{4}{a}\)

\(=\frac{6a+1}{3a}\cdot\frac{1}{b}-\frac{1}{a}\cdot\frac{3b+b-1}{b}-\frac{4}{3ab}+\frac{4}{a}\)

\(=\frac{6a+1}{3ab}-\frac{4b-1}{ab}-\frac{4}{3ab}+\frac{4}{a}=\frac{6a+1-3\left(4b-1\right)-4+12b}{3ab}\)

\(=\frac{6a-3+12b-12b+3}{3ab}=\frac{6a}{3ab}=\frac{2}{b}=\frac{2}{651}\)