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(102 +112+122):(132 +142)
=(100+121+144):(169+196)
=365:365
=1
\(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)
\(=\left(100+121+144\right):\left(169+196\right)\)
\(=\left(221+144\right):365\)
\(=365:365\)
\(=1\)Vậy kết quả = 1
\(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)
\(=\left(100+121+144\right):\left(169+196\right)\)
\(=365:365\)
\(=1\)
\(a,\dfrac{121.75.130.169}{39.60.11.198}=\dfrac{11.11.25.3.13.10.169}{13.3.6.10.11.11.18}=\dfrac{25.169}{6.18}\)
a) =\(\left[\left(12+1\right)^2+\left(12+2\right)^2\right]:\left(13^2+14^2\right)\)
=1
b)=(1.2.3....8).(9-1-8)
=(1.2.3....8).0
=0
mik chỉ giải được zậy thôi.
t mik nha.
a) Đặt \(A=\left(10^2+11^2+12^2\right)\div\left(13^2+14^2\right)\)
- Ta có: \(A=\left(100+121+144\right)\div\left(169+196\right)\)
\(\Leftrightarrow A=365\div365=1\)
Vậy \(A=1\)
b) Đặt \(B=1.2.3.....9-1.2.3.....8-1.2.3.....8^2\)
- Ta có: \(B=1.2.3.....8.\left(9-1\right)-1.2.3.....8^2\)
\(\Leftrightarrow B=1.2.3.....8.8-1.2.3.....8.8=0\)
Vậy \(B=0\)
c) Đặt \(C=\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
- Ta có: \(C=\frac{3^2.4^2.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
\(\Leftrightarrow C=\frac{3^2.2^4.2^{32}}{11.2^{35}-2^{36}}\)
\(\Leftrightarrow C=\frac{3^2.2^{36}}{2^{35}.\left(11-2\right)}\)
\(\Leftrightarrow C=\frac{9.2^{36}}{2^{35}.9}\)
\(\Leftrightarrow C=2\)
Vậy \(C=2\)
d) Đặt \(D=1152-\left(374+1152\right)+\left(-65+374\right)\)
- Ta có: \(D=1152-374-1152-65+374\)
\(\Leftrightarrow D=\left(1152-1152\right)+\left(374-374\right)-65\)
\(\Leftrightarrow D=-65\)
Vậy \(D=-65\)
a,(10^2+11^2+12^2):(13^2+14^2)
=(100+121+144);(169+196)
=365:365
=1
b,1.2.3...9-1.2.3...8-1.2.3...8^2=0
a)=(100+121+144):(169+196)=365:365=1
b)=1.2.3...8.(9-1-8)=1.2.3...8.0=0
k mk nhaa
\(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)
\(=\left(100+121+144\right):\left(169+196\right)\)
\(=\frac{365}{365}=1\)
(102 + 112 + 122) : (132 + 142)
= (100 + 121 + 144) : (169 + 196)
= 365 : 365
= 1
Ủng hộ mk nha ^_-