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\(C=2^0+2^1+2^2+2^3+2^0+2^1+2^2+2^3\)
\(=\left(2^0.2\right)+\left(2^1.2\right)+\left(2^2.2\right)+\left(2^3.2\right)\)
\(=2+4+8+16\)
\(=\left(2+8\right)+\left(4+16\right)\)
\(=10+20\)
\(=30\)
\(D=\left(2^0+2^1+2^2+2^3\right).2^0.2^1.2^2.2^3\)
\(=\left(1+2+4+8\right).1.2.4.8\)
\(=\left(8+2+4+1\right).1.2.4.8\)
\(=\left(10+4+1\right).1.2.4.8\)
\(=15.1.2.4.8\)
\(=\left(15.2\right).1.4.8\)
\(=30.1.4.8\)
\(=120.8\)
\(=960\)
(22+21+22+23).20.21.22.23
=(4+2+4+8).1.2.4.8
=18.1.2.4.8
=1152
1 3/8+1/8:(0,75-1/2)-25%.1/2
=11/8+1/8:(3/4-1/2)-1/4.1/2
=12/8:1/4-1/8
=6/1-1/8
=47/8
12 1/3-5/6:(24-23 5/7)
=37/3-5/6:(24-166/7)
=37/3-5/6:2/7
=37/3-35/2
=31/6
(-1/2)2-(-2)2-50
=1/4-4-1
=-19/4
\(\left(2^2+2^1+2^2+2^3\right)×2^0×2^1×2^2×2^3\)
\(=\left(4+2+4+8\right)×1×2×4×8\)
\(=18×1×2×4×8\)
\(=1152\)
\(1\frac{3}{8}+\frac{1}{8}:\left(0,75-\frac{1}{2}\right)-25\%×\frac{1}{2}\)
\(=\frac{11}{8}+\frac{1}{8}:\left(\frac{3}{4}-\frac{1}{2}\right)-\frac{1}{4}×\frac{1}{2}\)
\(=\frac{11}{8}+\frac{1}{8}:\frac{1}{4}-\frac{1}{8}\)
\(=\frac{11}{8}+\frac{1}{2}-\frac{1}{8}\)
\(=\frac{7}{4}\)
\(12\frac{1}{3}-\frac{5}{6}:\left(24-23\frac{5}{7}\right)\)
\(=\frac{37}{3}-\frac{5}{6}:\left(24-\frac{166}{7}\right)\)
\(=\frac{37}{3}-\frac{5}{6}:\frac{2}{7}\)
\(=\frac{37}{3}-\frac{35}{12}\)
\(=\frac{113}{12}\)
\(\left(\frac{-1}{2}\right)^2-\left(-2\right)^2-5^0\)
\(=\frac{1}{4}-4-1\)
\(=\frac{-19}{4}\)
Ta có F=(2^0+2^1+2^2+2^3).2^0.2^1.2^2.2^3
\(\Rightarrow F=\left(1+2+4+8\right).1.2.4.8\)
\(F=15.64\)
\(F=960\)
z lm lại, nhưng nó cứ sao ý!
\(\left(2^2+2^1+2^2+2^3\right).2^0.2^1.2^2.2^3\)
\(=18.64\)
\(=1152\)
Sửa đề:
đặt A = \(\left(2^0+2^1+2^2+2^3\right).2^0.2^1.2^2.2^3\)
\(A=\left(1+2+2^2+2^3\right).2^6\)
\(A=2^6+2^7+2^8+2^9\)
\(\Rightarrow2A=2^7+2^8+2^9+2^{10}\)
\(\Rightarrow2A-A=2^{10}-2^6\)
\(A=2^{10}-2^6\)
\(A=960\)
xem đề mk ghi như z cs đúng ko?
0.54166666666