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1)
\(\frac{7.8^3-5.2^{10}}{\left(-16\right)^2}\)
= \(\frac{7.2^8.2-5.2^8.2^2}{16^2}\)
= \(\frac{2^8.\left(2.7-5.2^2\right)}{2^8}\)
= \(\frac{2^8.\left(-6\right)}{2^8}\)
= \(-6\)
\(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+3^{10}.2^{20}}\)
\(B=\frac{2^{19}.3^9+3^9.5.2^{18}}{2^{19}.3^9+3^{10}.2^{20}}\)
\(B=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9\left(1+3.2\right)}\)
\(B=\frac{7}{2.7}\)
\(B=\frac{1}{2}\)
\(C=\frac{2^{13}.4^{11}-16^9}{\left(3.2^{17}\right)^2}\)
\(C=\frac{2^{13}.2^{22}-2^{36}}{3^2.2^{34}}\)
\(C=\frac{2^{35}-2^{36}}{3^2.2^{34}}\)
\(C=\frac{2^{35}\left(1-2\right)}{3^2.2^{34}}\)
\(C=\frac{-2}{9}\)
\(D=\frac{4^7.2^8}{3.2^{15}.16^2-5.2^2.\left(2^{10}\right)^2}\)
\(D=\frac{2^{14}.2^8}{3.2^{15}.2^8-5.2^2.2^{20}}\)
\(D=\frac{2^{14}.2^8}{3.2^{23}-5.2^{22}}\)
\(D=\frac{2^{22}}{2^{22}\left(3.2-5\right)}\)
\(D=1\)
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) =\(\left[\left(12+1\right)^2+\left(12+2\right)^2\right]:\left(13^2+14^2\right)\)
=1
b)=(1.2.3....8).(9-1-8)
=(1.2.3....8).0
=0
mik chỉ giải được zậy thôi.
t mik nha.
\(H=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(H=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(H=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(H=\frac{2^{19}.3^9+2^{18}.5.3^9}{2^{19}.3^9+2^{20}.3^{10}}\)
\(H=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9.\left(1+2.3\right)}\)
\(H=\frac{2^{18}.3^9.7}{2^{19}.3^9.\left(1+6\right)}\)
\(H=\frac{2^{18}.3^9.7}{2^{19}.3^9.7}=\frac{1}{2}\)
\(K=\frac{4^7.2^8}{3.2^{15}.16^2-5.2^2.\left(2^{10}\right)^2}\)
\(K=\frac{\left(2^2\right)^7.2^8}{3.2^{15}.\left(2^4\right)^2-5.2^2.2^{20}}\)
\(K=\frac{2^{14}.2^8}{3.2^{15}.2^8-5.2^{22}}\)
\(K=\frac{2^{22}}{3.2^{23}-5.2^{22}}\)
\(K=\frac{2^{22}}{2^{22}.\left(3.2-5\right)}=\frac{2^{22}}{2^{22}.1}=1\)
ta có: 4^7.2^8 / 3.2^15.16^2 - 5.2^2.(2^10)^2
= 2^14.2^8 / 3.2^15.2^8 - 5.2^2.2^20
= 2^22 / 6.2^22 - 5.2^22
= 2^22 / 2^22 ( 6 - 5 )
= 1
\(\frac{7.8^3-5.2^{10}}{\left(-16\right)^2}\)\(=\frac{7.512-5.1024}{\left(-256\right)}\)\(=\frac{3583-5120}{\left(-256\right)}\)\(=\frac{1537}{256}\)
Mình học hơi kém nếu sai mong mn đừng gạch đá .