tính giúp mình phép 56+64+36+49 tính giá trị biểu thức please.
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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
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
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{63}{64}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}...\frac{7.9}{8.8}\)
\(=\frac{1.3.2.4.3.5.4.6...7.9}{2.2.3.3.4.4.5.5...8.8}\)
\(=\frac{1.9}{2.8}=\frac{9}{16}\)
\(\dfrac{3}{4}+\dfrac{3}{28}+\dfrac{3}{70}+\dfrac{3}{130}+\dfrac{3}{208}+\dfrac{3}{304}\\ =\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+\dfrac{3}{13.16}+\dfrac{3}{16.19}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}\\ =1-\dfrac{1}{19}=\dfrac{18}{19}\)
\(\dfrac{3}{4}+\dfrac{3}{28}+\dfrac{3}{70}+\dfrac{3}{130}+\dfrac{3}{208}+\dfrac{3}{304}\)
\(=\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{10\times13}+\dfrac{3}{13\times16}+\dfrac{3}{16\times19}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}\)
\(=1-\dfrac{1}{19}\)
=\(\dfrac{18}{19}\)
\(\sqrt{\dfrac{49}{100}}=\dfrac{7}{10}\\ \sqrt{\dfrac{144}{289}}=\dfrac{12}{17}\\ \dfrac{\sqrt{36}}{\sqrt{225}}=\dfrac{6}{15}=\dfrac{2}{5}\\ \dfrac{\sqrt{25}}{\sqrt{121}}=\dfrac{5}{11}\)
\(4x^2-28x+49=\left(2x\right)^2-2\cdot2x\cdot7+7^2=\left(2x-7\right)^2\)
Khi x=4 thì \(4x^2-28x+49=\left(2x-7\right)^2=\left(2\cdot4-7\right)^2=1\)
\(56+64+36+49=105+100=205\)
56 + 64 + 36 + 49 = 120 + 36 + 49 = 156 + 49 = 205