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a/\(A=\left(3+5\right)^2=8^2=64\)
\(B=3^2+5^2=9+25=34\)
\(\Rightarrow A>B\)
b/ \(C=\left(3+5\right)^3=8^3=512\)
\(D=3^3+5^3=27+125=152\)
\(\Rightarrow C>D\)
a/ A= (3+5)2 = 82 = 64
B = 32 + 52 = 9 + 25 = 34
vì 64>34 => A > B
b/ C = (3+5)3 = 83 = 512
D = 33 + 53 = 27 + 125 = 152
Vì 512>152 => C > D
a/ A = 82
A = 64
B = 9 + 25, B = 34
b/ C = 83, CC = 512
D = 27 + 125
D = 152
Tìm x :
a) 35 + 3. |x| = 50
b)72 - [ 41 - (2x - 5) - 23.5
c) 70 - 5 (x - 3) = 45
giup mk voi mk dang vội
Tìm x :
a) 35 + 3. |x| = 50
3.|x|=50-35
3.|x|=15
|x|=15:3
|x|=5
x=5 hoặc x=-5
b)72 - [41 - (2x - 5) ]- 23.5
41-(2x-5)=72-40
41-(2x-5)=32
2x-5=41-32
2x-5=9
2x=9+5
2x=14
x=14:2
x=7
c) 70 - 5 (x - 3) = 45
5(x-3)=70-45
5(x-3)=25
x-3=25:5
x-3=5
x-5+3
x=8
A= 1+2+22+23+.......+298+299
A= (1+2)+(22+23)+.......+(298+299 )
A=3+22.(1+2)+...+298.(1+2)
A= 3+22.3+...+298.3
A=3.(22+...+298)
Vid 3 chia hết cho 3 nên A chia hết cho 3
Đơn giản như đang giỡn
HT
i) \(2345-1000\div\left[19-2\left(21-18\right)^2\right]\)
\(=\)\(2345-1000\div\left[19-2.3^2\right]\)
\(=\)\(2345-1000\div\left[19-2.9\right]\)
\(=\)\(2345-1000\div\left[19-18\right]\)
\(=\)\(2345-1000\div1\)
\(=\)\(2345-1000\)
\(=\)\(1345\)
j) \(128-\left[68+8\left(37-35\right)^2\right]\div4\)
\(=\)\(128-\left[68+8.2^2\right]\div4\)
\(=\)\(128-\left[68+8.4\right]\div4\)
\(=\)\(128-\left[68+32\right]\div4\)
\(=\)\(128-100\div4\)
\(=\)\(128-25\)
\(=\)\(3\)
k) \(568-\left\{5\left[143-\left(4-1\right)^2\right]+10\right\}\div10\)
\(=\)\(568-\left\{5\left[143-3^2\right]+10\right\}\div10\)
\(=\)\(568-\left\{5\left[143-9\right]+10\right\}\div10\)
\(=\)\(568-\left\{5.134+10\right\}\div10\)
\(=\)\(568-\left\{670+10\right\}\div10\)
\(=\)\(568-680\div10\)
\(=\)\(568-68\)
\(=\)\(500\)
a) \(107-\left\{38+\left[7.3^2-24\div6+\left(9-7\right)^3\right]\right\}\div15\)
\(=\)\(107-\left\{38+\left[7.3^2-24\div6+2^3\right]\right\}\div15\)
\(=\)\(107-\left\{38+\left[7.9-4+8\right]\right\}\div15\)
\(=\)\(107-\left\{38+\left[63-4+8\right]\right\}\div15\)
\(=\)\(107-\left\{38+67\right\}\div15\)
\(=\)\(107-105\div15\)
\(=\)\(107-7\)
\(=\)\(7\)
b) \(307-\left[\left(180-160\right)\div2^2+9\right]\div2\)
\(=\)\(307-\left[20\div4+9\right]\div2\)
\(=\)\(307-\left[5+9\right]\div2\)
\(=\)\(307-14\div2\)
\(=\)\(307-7\)
\(=\)\(300\)
c) \(205-\left[1200-\left(4^2-2.3\right)^3\right]\div40\)
\(=\)\(205-\left[1200-\left(16-6\right)^3\right]\div40\)
\(=\)\(205-\left[1200-10^3\right]\div40\)
\(=\)\(205-\left[1200-1000\right]\div40\)
\(=\)\(205-200\div40\)
\(=\)\(205-5\)
\(=\)\(200\)
a)\(72.19 - 36^2:18= 72.19- 36.36:18\)
\(=72.19 - 72=72.(19 - 1)\)
\(=72.18=1296\)
b)
\(\begin{array}{l}750{\rm{ }}:{\rm{ }}\left\{ {130{\rm{ }} - {\rm{ }}[{{\left( {5.{\rm{ }}14{\rm{ }} - {\rm{ }}65} \right)}^3} + {\rm{ }}3} \right\}\\ = 750{\rm{ }}:{\rm{ }}\left\{ {130{\rm{ }} - {\rm{ }}\left[ {{{\left( {{\rm{70 }} - {\rm{ }}65} \right)}^3} + {\rm{ }}3} \right]} \right\}\\ = 750{\rm{ }}:{\rm{ }}\left\{ {130{\rm{ }} - {\rm{ }}\left[ {{5^3} + {\rm{ }}3} \right]} \right\}\\ = 750{\rm{ }}:{\rm{ }}\left\{ {130{\rm{ }} - {\rm{ }}\left[ {125 + {\rm{ }}3} \right]} \right\}\\ = 750{\rm{ }}:{\rm{ }}\left\{ {130{\rm{ }} - {\rm{ 128}}} \right\}\\ = 750:2 = 375.\end{array}\)