100-[116-(13-5)2]
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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
\(100-\left(116-\left(13-5^2\right)\right)=100-\left(116-\left(-12\right)\right)\)
\(=100-\left(116+12\right)=100-128=-28\)
100-[116-(13-52)]
=100-[116-(13-25)]
=100-(116-13+25)
=100-(103+25)
=100-128
=-28
\(13\cdot16+13\cdot\left(-116\right)+17\cdot\left(-100\right)\)
\(=13\left(16+-116\right)+17\cdot\left(-100\right)\)
\(=13\cdot\left(-100\right)+17\cdot\left(-100\right)\)
\(=\left(-100\right)\cdot\left(13+17\right)=30\cdot\left(-100\right)=-3000\)
13.16+13.(-116)+17.(-100)
=13.(-116+16)+17.(-100)
=13.(-100)+17.(-100)
=(-100).(13+17)
=-100.30
=(-3000)
\(100-\)\(\left[116-\left(13-5^2\right)\right]\)
=\(100-\left[116-\left(13-25\right)\right]\)
=\(100-\left[116-\left(-12\right)\right]\)
=\(100-\left(116+12\right)\)
=\(100-128\)
=\(-28\)
\(611-\left[2.\left(x^2-13\right)+116\right]=449.\)
\(\Rightarrow2.\left(x^2-13\right)+116=611-449\)
\(\Rightarrow2.\left(x^2-13\right)+116=162\)
\(\Rightarrow2.\left(x^2-13\right)=162-116\)
\(\Rightarrow2.\left(x^2-13\right)=46\)
\(\Rightarrow x^2-13=46:2\)
\(\Rightarrow x^2-13=23\)
\(\Rightarrow x^2=23+13\)
\(\Rightarrow x^2=36\)
\(\Rightarrow x=\orbr{\begin{cases}6\\-6\end{cases}}\)
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
Ta co so so hang cua tong 2+4+6+8+...100 la
(100-2):2+1=50(so hang)
Tong 2+4+6+8+...100 la
(100+2)x50:2=2550
Vi Gia tri cua bieu thuc 58x8-116x4=0
nen tich 2550x0=0
Ta co so so hang cua tong 2+4+6+8+...100 la
(100-2):2+1=50(so hang)
Tong 2+4+6+8+...100 la
(100+2)x50:2=2550
Vi Gia tri cua bieu thuc 58x8-116x4=0
nen tich 2550x0=0
\(100-\left[116-\left(13-5\right)^2\right]\)
\(=100-\left(116-8^2\right)\)
\(=100-\left(116-64\right)\)
\(=100-52\)
\(=48\)
=100-(116-(9.9)
=100-(116-81)
=100-35=65