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\(100-\left(116-\left(13-5^2\right)\right)=100-\left(116-\left(-12\right)\right)\)
\(=100-\left(116+12\right)=100-128=-28\)
100-[116-(13-52)]
=100-[116-(13-25)]
=100-(116-13+25)
=100-(103+25)
=100-128
=-28
a) A= 100 - 116 : 4
=100-29
=71
b) B= 100 - [ 116 - (13 - 5)2]
=100-[116-82]
=100-(116-64)
=100-52
=48
c) C= |- 65| + (- 42) + |65|
= 65+(-42)+65
= 88
Câu 1
=> ( 2x-15)^3 - (2x -15)^5=0
(2x-15)^3 * ( -(2x-15)^2+1) =0
xét 2 TH : + (2x-15)^3=0
+ 1-(2x-15)^2 =0
=> ĐPCM
\(100-\)\(\left[116-\left(13-5^2\right)\right]\)
=\(100-\left[116-\left(13-25\right)\right]\)
=\(100-\left[116-\left(-12\right)\right]\)
=\(100-\left(116+12\right)\)
=\(100-128\)
=\(-28\)
\(611-\left[2.\left(x^2-13\right)+116\right]=449.\)
\(\Rightarrow2.\left(x^2-13\right)+116=611-449\)
\(\Rightarrow2.\left(x^2-13\right)+116=162\)
\(\Rightarrow2.\left(x^2-13\right)=162-116\)
\(\Rightarrow2.\left(x^2-13\right)=46\)
\(\Rightarrow x^2-13=46:2\)
\(\Rightarrow x^2-13=23\)
\(\Rightarrow x^2=23+13\)
\(\Rightarrow x^2=36\)
\(\Rightarrow x=\orbr{\begin{cases}6\\-6\end{cases}}\)
\(\left[116-\left(13-5^2\right)\right]=\left[116-\left(13-25\right)\right]=116-13+25=128\)
\(\frac{7^{25}}{7^{21}\times46+7^{21}\times3}=\frac{7^{25}}{7^{21}\times\left(46+3\right)}=\frac{7^4}{50}\)
\(57+212+43+288=\left(57+43\right)+\left(212+288\right)=100+500=600\)
\(28\times69+4\times18\times7+2\times14\times13=28\times69+18\times28+28\times13=28\times\left(69+18+13\right)=28\times100=2800\)
A=5+52+53+....+5100
->5A=53+54+55+....+5100+5101
4A=5101-5
->4A+5101-5+5=5101
->5n=5101
->n=101
#Hok_tốt
\(A=5+5^2+5^3+...+5^{100}\)
\(=\left(5+5^2+5^3+5^4\right)+\left(5^5+5^6+5^7+5^8\right)+...+\left(5^{97}+5^{98}+5^{99}+5^{100}\right)\)
\(=\left(5+5^2+5^3+5^4\right)+5^4.\left(5+5^2+5^3+5^4\right)+...+5^{96}.\left(5+5^2+5^3+5^4\right)\)
\(=780+5^4.780+...+5^{96}.780\)
\(=780.\left(1+5^4+...+5^{96}\right)\)
\(=13.60.\left(1+5^4+...+5^{96}\right)⋮13\)
\(\Rightarrow A⋮13\left(\text{ĐPCM}\right)\)
1) 3B - B = (32 + 33 + 34 + ... + 3101) - (3 + 32 + 33 + ... + 3100)
2B = 3101 - 3 => 2B + 3 = 3101 => n = 101
2) 52.C - C = (53 + 55 + 57 + 59 + ... + 5103) - (5 + 53 + 55 + 57 + ... + 5101)
24C = 5103 - 5
C =\(\frac{5^{103}-5}{24}\).Tương tự,\(D=\frac{13^{101}-13}{168}\Rightarrow C+D=\frac{5^{103}-5}{24}+\frac{13^{101}-13}{168}=\frac{7.\left(5^{103}-5\right)+\left(13^{101}-13\right)}{168}=\frac{7.5^{103}+13^{101}-48}{168}\)
a) (105 + 155 - 55) : 55
= 105 : 55 + 155 : 55 - 55 : 55
= 25 + 35 - 1
= 32 + 243 - 1
= 274
\(100-\left[116-\left(13-5\right)^2\right]\)
\(=100-\left(116-8^2\right)\)
\(=100-\left(116-64\right)\)
\(=100-52\)
\(=48\)
=100-(116-(9.9)
=100-(116-81)
=100-35=65