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Đây là dạng toán tính nhanh phân số
B = 3/2 + 3/8 + 3/32 + 3/128 + 3/512.
Giải chi tiết giúp mình nha!
1+2+4+8+16+32+64+128+256+512+1024+2048
=1+(2+8)+(4+16)+(32+128)+(64+256)+(512+2048)+1024
=1+10+20+160+320+2560+1024
=4095
1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 + 2048 = 4095
k nha công chúa nụ cười =_= ^_^
\(P+\frac{1}{512}=\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{4}{512}=\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{4}{128}=\)
\(=\frac{3}{2}+\frac{3}{8}+\frac{4}{32}=\frac{3}{2}+\frac{4}{8}=\frac{4}{2}=2\)
\(\Rightarrow P=2-\frac{1}{512}=\frac{1023}{512}\)
\(P=\frac{3}{2}+\frac{3}{8}+...+\frac{3}{512}\)
\(=3.\left(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\right)\)
\(4P=3\left(\frac{1}{2^3}+\frac{1}{2^5}+...+\frac{1}{2^{11}}\right)\)
\(4P-P=3\left(\frac{1}{2}-\frac{1}{2^{11}}\right)\)
\(3P=3\left(\frac{1}{2}-\frac{1}{2^{11}}\right)\)
\(P=\frac{1}{2}-\frac{1}{2^{11}}=\frac{2^{10}-1}{2^{11}}\)
P= 3/2 + 3/8 + 3/32 + 3/128 + 3/512
= 768/512 + 192/512 + 48/512 + 12/512 + 3/512
=1023/512
1+(-2)+3+(-4)+.....+(-19)+20=[1+(-2)]+[3+(-4)]+....+[(-19)+20]= (-1)+(-1)+....+(1) (có 10 thừa số -1)
=(-1).10= -10
Đặt A=3/2+3/8+...+3/512
bn tách
3/2=3/2^1
3/8=3/2^3
....
3/512=3/2^9
Rồi nhân nó lên trừ đc bao nhiêu - đi A ban đầu là đc
Chúc bạn học tốt
=768/512+192/512+48/512+12/512+3/512
=768+192+48+12+3/512
=1023/512
Đặt thừa số chung x.
Ta có:
\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+9\right)\Leftrightarrow x+\left(1+2+3+...+9\right)=135\)
\(\Leftrightarrow x+\left(1+2+3+4+5+6+7+8+9\right)\Leftrightarrow x+45=135\)
\(\Rightarrow x=135-45=90\)
Đs:
x * 10 + (1 + 2 + 3 ... +9) = 135
x *10 + 45 =135
x * 10 = 135 - 45
x * 10 = 90
x =90 : 10
x = 9
Vậy x = 9
thử lại:(tự làm)
\(a.\dfrac{7}{8}+\dfrac{1}{2}\text{=}\dfrac{11}{8}\)
\(b.\dfrac{5}{6}+\left(-2\right)\text{=}\dfrac{-7}{6}\)
\(c.\dfrac{2}{5}+\dfrac{-3}{8}\text{=}\dfrac{1}{40}\)
\(d.\dfrac{5}{7}-\dfrac{3}{8}\text{=}\dfrac{19}{56}\)
\(e.\dfrac{3}{4}-\dfrac{1}{2}+\dfrac{7}{6}\text{=}\dfrac{17}{12}\)
\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)
\(=\left(\frac{12}{8}+\frac{3}{8}\right)+\left(\frac{12}{128}+\frac{3}{128}\right)+\frac{3}{512}\)
\(=\frac{15}{8}+\frac{15}{128}+\frac{3}{512}\)
\(=\frac{240}{128}+\frac{15}{128}+\frac{3}{512}\)
\(=\frac{255}{128}+\frac{3}{512}\)
\(=\frac{1020}{512}+\frac{3}{512}\)
\(=\frac{1023}{512}\)
\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}=\frac{3}{1.2}+\frac{3}{2.4}+\frac{3}{4.8}+\frac{3}{8.16}+\frac{3}{16.32}\)
\(=\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{4}+\frac{3}{8}-\frac{3}{8}+\frac{3}{16}-\frac{3}{16}+\frac{3}{32}\)
\(=3+\frac{3}{32}=\frac{3.32}{32}+\frac{3}{32}=\frac{96+3}{32}=\frac{99}{32}\)