Lời giải:

$=3159\times 36+11\times 4\times 2584+64\times 3159+2584\times 56$

$=(3159\times 36+3159\times 64)+(44\times 2584+2584\times 56)$

$=3159\times (36+64)+2584\times (44+56)$

$=3159\times 100+2584\times 100=315900+258400$

$=574300$

a) {132 - [116- (16 - 8)]:2}.5

= [132 - (116 - 8): 2] .5

= (132 - 108 : 2). 54

= (132 - 54).5

= 78.5 = 390

b) 36: {136 : 200 - (12+ 8. 20)]}

= 36: {336 : [200 - ( 12 + 160)]}

= 36 : [336 : ( 200 - 172)

= 36 : ( 336 : 28) = 3

c) 86 - [15. (64 - 39): 75+11] 

= 86 - (15.25 : 75 + 11) 

= 86 - ( 5 + 11) 

= 70

d)  55 - [ 49 - ( 2 3 . 17 - 2 3 . 14 ) ]

= 55 - 49 - 2 3 . 3

= 55- ( 49 - 24)

= 30

Bạn ơi ở chỗ trước số 200 có dấu này [mà

Phần b/

a) {132 – [116 – (16 – 8)]:2}.5

= [132 – (116 – 8): 2] .5

= (132 – 108 : 2). 5

= (132 –54).5

= 78.5 = 390

b) 36: {136 : 200 – (12+ 8. 20)]}

= 36: {336 : [200 – ( 12 + 160)]}

= 36 : [336 : ( 200 – 172)

= 36 : ( 336 : 28)

= 36 : 12 = 3

c) 86 – [15. (64 - 39): 75+11]

= 86 – (15.25 : 75 + 11)

= 86 – ( 5 + 11)

= 70.

d) 55 – [49 – (2³ . 17 – 2³.14)]

= 55 – (49 – 2³ .3)

= 55 – ( 49 – 24)

= 30

\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{63}{64}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}...\frac{7.9}{8.8}\)

\(=\frac{1.3.2.4.3.5.4.6...7.9}{2.2.3.3.4.4.5.5...8.8}\)

\(=\frac{1.9}{2.8}=\frac{9}{16}\)

Hê lô

a ) − 7 2 . b ) 58 9 . c ) 5 18 . d ) 9 14 .

\(\dfrac{3}{4}+\dfrac{3}{28}+\dfrac{3}{70}+\dfrac{3}{130}+\dfrac{3}{208}+\dfrac{3}{304}\\ =\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+\dfrac{3}{13.16}+\dfrac{3}{16.19}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}\\ =1-\dfrac{1}{19}=\dfrac{18}{19}\)

\(\dfrac{3}{4}+\dfrac{3}{28}+\dfrac{3}{70}+\dfrac{3}{130}+\dfrac{3}{208}+\dfrac{3}{304}\)

\(=\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{10\times13}+\dfrac{3}{13\times16}+\dfrac{3}{16\times19}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}\)

\(=1-\dfrac{1}{19}\)

=\(\dfrac{18}{19}\)

\(\sqrt{\dfrac{49}{100}}=\dfrac{7}{10}\\ \sqrt{\dfrac{144}{289}}=\dfrac{12}{17}\\ \dfrac{\sqrt{36}}{\sqrt{225}}=\dfrac{6}{15}=\dfrac{2}{5}\\ \dfrac{\sqrt{25}}{\sqrt{121}}=\dfrac{5}{11}\)

`\sqrt{49/100}=\sqrt{(7/10)^2}=7/10`

`\sqrt{144/289}=\sqrt{(12/17)^2}=12/17`

`\sqrt{36/225}=\sqrt{(6/15)^2}=6/15`

`\sqrt{25/121}=\sqrt{(5/11)^2}=5/11`

\(4x^2-28x+49=\left(2x\right)^2-2\cdot2x\cdot7+7^2=\left(2x-7\right)^2\)

Khi x=4 thì \(4x^2-28x+49=\left(2x-7\right)^2=\left(2\cdot4-7\right)^2=1\)