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a: \(\Leftrightarrow x^3=\dfrac{539}{64}\)
hay \(x=\dfrac{7\sqrt{11}}{4}\)
c: \(\Leftrightarrow2^{2x-1}=2^9\cdot2^2=2^{11}\)
=>2x-1=11
hay x=6
d: \(\Leftrightarrow x^{17}-x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
hay \(x\in\left\{0;1;-1\right\}\)
Bài 3 :
A = 26 + 27 + 28 + 29 + 30 + 31 + 32 + 33
=> A = ( 33 + 26 ) . 8 : 2 = 236
Vậy A = 236
\(\text{#Hok tốt!}\)
a) 2 . 31 . 12 + 4 . 6 . 42 + 8 . 27 . 3
= 24 . 31 + 24 . 42 + 24 . 27
= 24 . ( 31 + 42 + 27 )
= 24 . 100
= 2400
\(F=\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^6}\)
\(\Rightarrow2F=1+\frac{1}{2}+....+\frac{1}{2^5}\)
\(\Rightarrow2F-F=F=\left(1+\frac{1}{2}+....+\frac{1}{2^5}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^6}\right)\)
\(\Rightarrow F=1-\frac{1}{2^6}\)
phamthiminhtrang
\(a,x=\frac{3}{6}-\frac{8}{16}\)
\(\Rightarrow x=0\)
\(b,\frac{12}{16}:x=\frac{32}{64}\)
\(x=\frac{12}{16}:\frac{32}{64}\)
\(x=\frac{12}{16}\cdot\frac{64}{32}\)
\(x=\frac{3}{8}\)
\(a,\)\(x\)\(=\frac{3}{6}-\frac{8}{16}=\frac{1}{2}-\frac{1}{2}=0\)
\(b,\)\(\frac{12}{16}\)\(:\)\(x\)\(=\frac{32}{64}\)
\(=>\) \(x\)\(=\)\(\frac{12}{16}:\frac{32}{64}\)
\(x\) \(=\)\(\frac{12}{16}.\frac{64}{32}\)
\(x\)\(=\)\(\frac{3}{4}.2\)
\(x\)\(=\)\(\frac{6}{4}=\frac{3}{2}\)
Đặt 1/4 + 1/8 + 1/16 + 1/32 + 1/64 =A
=> 2A = 2.(1/4 + 1/8 + 1/16 + 1/32 + 1/64)
=> 2A = 1/2 +1/4+1/8+1/16+1/32
=> A= 2A-A = 1/2 +1/4+1/8+1/16+1/32 - 1/4 - 1/8 - 1/16 - 1/32 - 1/64
=> A = 1/2 - 1/64 =31/64
Ta có: 1/4 + 1/8 + 1/16 + 1/32 + 1/64
= 1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64
= 1/2 - 1/64
= 31/64
Nhấn đúng cho mk nha!!!!!!!!!!!
\(\frac{6}{13}+\frac{2}{26}=\frac{6}{13}+\frac{1}{13}=\frac{7}{13}\)
\(\frac{64}{8}+\frac{32}{8}=\frac{96}{8}=12\)
Ủng hộ nha..!
Giải:
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
\(\Leftrightarrow A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}\)
\(\Leftrightarrow\dfrac{1}{2}A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}+\dfrac{1}{2^7}\)
Lấy vế trừ vế, ta được:
\(A-\dfrac{1}{2}A=\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{2^7}\)
\(\Leftrightarrow\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{2^7}\)
\(\Leftrightarrow A=\dfrac{\dfrac{1}{2}-\dfrac{1}{2^7}}{\dfrac{1}{2}}\)
\(\Leftrightarrow A=\dfrac{\dfrac{1}{2}\left(1-\dfrac{1}{2^6}\right)}{\dfrac{1}{2}}\)
\(\Leftrightarrow A=1-\dfrac{1}{2^6}\)
Vậy \(A=1-\dfrac{1}{2^6}\).
Chúc bạn học tốt!!!
Đặt:
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
\(A=\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}\)
\(2A=2\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}\right)\)
\(2A=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\)
\(2A-A=\left(1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\right)-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}\right)\)
\(A=1-\dfrac{1}{2^6}=1-\dfrac{1}{64}=\dfrac{63}{64}\)
\(A=\left(4^2\right)^{17}.\frac{64^{36}}{8^{35}.32^{34}}\)
\(A=16^{17}.\frac{64^{2.36}}{\left(2^3\right)^{35}.\left(2^5\right)^{34}}\)
\(A=...6.\frac{...6^{36}}{2^{105}.2^{170}}\)
Đáp án là B
Ta có:
4 = (-2).(-2)
-8 = 4.(-2)
16 = (-8).(-2)
Do đó, trong dãy số trên số hạng sau là tích của số hạng trước với – 2
Vậy hai số hạng tiếp theo là:
16.(-2) = -32
(-32).(-2) = 64