a. (102+ 112 + 122 ) : (132+142)
b. 1.2.3...9-1.2.3...8-1.2.3...7.82
Những câu hỏi liên quan
1.2.3...9 - 1.2.3...8 - 1.2.3...7.82
\(1.2.3.....9-1.2.3.....8-1.2.3.....7.8^2\)
\(=1.2.3.....8\left(9-1-8\right)\)
\(=1.2.3.....8\cdot0\)
\(=0\)
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1.2.3.....9−1.2.3.....8−1.2.3.....7.821.2.3.....9−1.2.3.....8−1.2.3.....7.82
=1.2.3.....8(9−1−8)=1.2.3.....8(9−1−8)
=1.2.3.....8⋅0=1.2.3.....8⋅0
=0
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1.2.3...9-1.2.3...8-1.2.3...7.82
A= 1.2.3...9- 1.2.3...8-1.2.3...8.8
=1.2.3...8(9-1-8)
=1.2.3...8.0=0
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\(A=1.2.3...9-1.2.3...8-1.2.3...8.8\)
\(=1.2.3...8.\left(9-1-8\right)\)
\(=1.2.3...8.0\)
\(=0\)
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b) 1.2.3...9-1.2.3...8-1.2.3...7.8^ 2
1.2.3...9 - 1.2.3...8 - 1.2.3....7.82
= 1.2.3.4.5.6.7.8.(9 - 1 - 8)
= 1.2.3.4.5.6.7.8.0
= 0
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A = 1.2.3....9 - 1.2.3...8 - 1.2.3.....8.8
b) 1.2.3...9-1.2.3...8-1.2.3... 0.7 .0.8 ^ 2
Giúp mình với
a/ 1.2.3...9-1.2.3...8-1.2.3...7.82
1.2.3...9-1.2.3...8-1.2.3...7.82
= (1.2.3.4.5.6.7).9 - (1.2.3.4.5.6.7).8 - (1.2.3.4.5.6.7).1
= (1.2.3.4.5.6.7).(9-8-1)
=(1.2.3.4.5.6.7).0
= 0
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tính hợp lý:
( 102 + 112 + 122 ) : ( 132 + 142 )
(102 + 112 + 122) : (132 + 142)
= (100 + 121 + 144) :( 169 + 196)
= 365: 365
= 1
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Thực hiện phép tính
a, 1.2.3. ... .9-1.2.3. ... .8-1.2.3. ... .8^2
1.2.3.4.....9-1.2.3.4....8-1.2.3...8.8=1.2.3.....8.(9-1)-1.2.3.4....8.8=1.2.3....8.8-1.2.3....8.8=0
1.2.3...9-1.2.3...8-1.2.3...8^2
Giải thích giùm mình nha
1.2.3...9 - 1.2.3...8 - 1.2.3...82
= 1.2.3...8.(9 - 1) - 1.2.3...82
= 1.2.3...8.8 - 1.2.3...8.8
= 0
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1.2.3...9 - 1.2.3...8 - 1.2.3...82
= 1.2.3...8.(9 - 1) - 1.2.3...82
= 1.2.3...8.8 - 1.2.3...8.8
= 0
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\(1.2.3.....9-1.2.3.....8-1.2.3.....8^2\)
\(=1.2.3.....8.\left(9-1\right)-1.2.3.....8.8\)
\(=1.2.3.....8.8-1.2.3.....8.8\)
\(=0\)
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