Tìm x biết:
A = 22x-11 và A = 4+22+23+24+....+220
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a: \(280-\left(x-140\right):35=270\)
=>(x-140):35=280-270=10
=>x-140=350
=>x=350+140
=>x=490
b: \(\left(190-2x\right):35-32=16\)
=>\(\left(190-2x\right):35=32+16=48\)
=>\(190-2x=35\cdot48=1680\)
=>2x=190-1680=-1490
=>x=-745
c: \(720:\left\lbrack41-\left(2x-5\right)\right\rbrack=2^3\cdot5\)
=>\(720:\left\lbrack41-\left(2x-5\right)\right\rbrack=8\cdot5=40\)
=>41-(2x-5)=720:40=18
=>2x-5=41-18=23
=>2x=28
=>x=14
d: \(\left(x:23+45\right)\cdot37-22=2^4\cdot105\)
=>\(\left(\frac{x}{23}+45\right)\cdot37=16\cdot105+22=1702\)
=>\(\frac{x}{23}+45=46\)
=>\(\frac{x}{23}=1\)
=>x=23
e: \(\left(3x-4\right)\left(x-1\right)^3=0\)
=>\(\left[\begin{array}{l}3x-4=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac43\\ x=1\end{array}\right.\)
f: \(2^{2x-1}:4=8^3\)
=>\(2^{2x-1-2}=2^9\)
=>2x-3=9
=>2x=12
=>x=6
g: \(x^{17}=x\)
=>\(x^{17}-x=0\)
=>\(x\left(x^{16}-1\right)=0\)
=>\(\left[\begin{array}{l}x=0\\ x^{16}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^{16}=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\\ x=-1\end{array}\right.\)
h: \(\left(x-5\right)^4=\left(x-5\right)^6\)
=>\(\left(x-5\right)^6-\left(x-5\right)^4=0\)
=>\(\left(x-5\right)^4\cdot\left\lbrack\left(x-5\right)^2-1\right\rbrack=0\)
=>\(\left(x-5\right)^4\cdot\left(x-4\right)\left(x-6\right)=0\)
=>x∈{4;5;6}
i: \(\left(x+2\right)^5=2^{10}\)
=>\(\left(x+2\right)^5=\left(2^2\right)^5=4^5\)
=>x+2=4
=>x=2
k: 1+2+3+...+x=78
=>\(\frac{x\left(x+1\right)}{2}=78\)
=>x(x+1)=156
=>\(x^2+x-156=0\)
=>(x+13)(x-12)=0
=>x=-13(loại) hoặc x=12(nhận)
l: \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)
=>\(3x-16=2\cdot\frac{7^4}{7^3}=2\cdot7=14\)
=>3x=16+14=30
=>\(x=\frac{30}{3}=10\)
n: \(5^{x}:5^2=125\)
=>\(5^{x-2}=5^3\)
=>x-2=3
=>x=5
m: \(\left(x+1\right)^2=\left(x+1\right)^0\)
=>\(\left(x+1\right)^2=1\)
=>\(\left[\begin{array}{l}x+1=1\\ x+1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-2\end{array}\right.\)
o: Số số hạng của dãy số 2;4;..;52 là:
(52-2):2+1=50:2+1=25+1=26(số)
Tổng của dãy số 2;4;...;52 là:
\(\left(52+2\right)\cdot\frac{26}{2}=54\cdot13=702\)
(2+x)+(4+x)+...+(52+x)=780
=>26x+702=780
=>26x=78
=>x=3
p: \(70=2\cdot5\cdot7;80=2^4\cdot5\)
=>ƯCLN(70;80)=\(2\cdot5=10\)
70⋮x; 80⋮x
=>x∈ƯC(70;80)
=>x∈Ư(10)
mà x>8
nên x=10
q: \(12=2^2\cdot3;25=5^2;30=2\cdot3\cdot5\)
=>BCNN(12;25;30)=\(2^2\cdot3\cdot5^2=300\)
x⋮12; x⋮25; x⋮30
=>x∈BC(12;25;30)
=>x∈B(300)
mà 0<x<500
nên x=300
Đặt \(B=2^2+2^3+\cdots+2^{20}\)
=>\(2B=2^3+2^4+\cdots+2^{21}\)
=>2B-B=\(2^3+2^4+\cdots+2^{21}-2^2-2^3-\cdots-2^{20}\)
=>B=\(2^{21}-4\)
\(A=4+2^2+2^3+\cdots+2^{20}\)
\(=4+2^{21}-4\)
\(=2^{21}\)
A=2+22+23+...+220A=2+22+23+...+220
2A=22+23+24+...+2212A=22+23+24+...+221
2A−A=(22+23+24+...+221)−(2+22+23+...+220)2A−A=(22+23+24+...+221)−(2+22+23+...+220)
A=221−2=24.5+1−2=(24)5.2−2=165.2−2A=221−2=24.5+1−2=(24)5.2−2=165.2−2
A=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯.......6.2−2=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯........2−2=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯...........0A=.......6¯.2−2=........2¯−2=...........0¯
Vậy chữ số tận cùng cả A là 0
\(-\dfrac{15}{23}:\dfrac{22x}{7}=-\dfrac{14x}{11}:\left(13+\dfrac{4}{5}\right)\)
=>\(-\dfrac{15}{23}\cdot\dfrac{7}{22x}=\dfrac{-14x}{11}:\dfrac{69}{5}\)
=>\(-\dfrac{105}{23\cdot22x}=\dfrac{-70x}{11\cdot69}\)
=>\(\dfrac{-3}{2x}=\dfrac{-2x}{3}\)
=>\(4x^2=9\)
=>\(x^2=\dfrac{9}{4}\)
=>\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(\dfrac{-15}{23}:\dfrac{22x}{7}=\dfrac{-14x}{11}:\left(13+\dfrac{4}{5}\right)\)
\(\Leftrightarrow-\dfrac{15}{23}\cdot\dfrac{7}{22x}=\dfrac{-14x}{11}:\dfrac{69}{5}\)
=>\(\dfrac{-15\cdot7}{23\cdot22x}=\dfrac{-14x}{11}\cdot\dfrac{5}{69}\)
=>\(\dfrac{5\cdot3\cdot7}{23\cdot2\cdot11x}=\dfrac{2\cdot7x}{11}\cdot\dfrac{5}{3\cdot23}\)
=>\(\dfrac{3}{2x}=\dfrac{2x}{3}\)
=>\(4x^2=9\)
=>\(x^2=\dfrac{9}{4}\)
=>\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Đổi 4 thành 2 mũ 2
Thử xem cs đúng ko . Vì mik chữ thầy toán giả thầy toán hết r
Dễ:đổi 4=22
B=22+23+24+...+220
ta có:B=2B-B=(23+24+25+...+221)-(22+23+24+...+220)
= 221-22
Nói trước: đây là mình rút gọn chứ viết mà theo cơ số 2 thì khó quá
a) Đặt: \(A=1+2^2+2^3+...+2^{10}\)
\(\Rightarrow2A=2\left(1+2^2+2^3+...+2^9+2^{10}\right)\)
\(\Rightarrow2A=2+2^3+2^4+...+2^{10}+2^{11}\)
\(\Rightarrow2A-A=\left(2+2^3+2^4+...+2^{10}+2^{11}\right)-\left(1+2^2+2^3+...+2^{10}\right)\)
\(\Rightarrow A=\left(2^3-2^3\right)+\left(2^4-2^4\right)+...+\left(2-1\right)+\left(2^{11}-2^2\right)\)
\(\Rightarrow A=0+0+...+1+\left(2^{11}-2^2\right)\)
\(\Rightarrow A=1+2^{11}-2^2=1+2048-4=2045\)
Vậy: \(1+2^2+2^3+...+2^{10}=2045\)
b)
a] \(60-3\left(x-1\right)=2^3\cdot3\)
\(\Rightarrow60-3\left(x-1\right)=24\)
\(\Rightarrow3\left(x-1\right)=36\)
\(\Rightarrow x-1=12\)
\(\Rightarrow x=13\)
b] \(\left(3x-2\right)^3=2\cdot2^5\)
\(\Rightarrow\left(3x-2\right)^3=2^6\)
\(\Rightarrow\left(3x-2\right)^3=\left(2^2\right)^3\)
\(\Rightarrow3x-2=2^2\)
\(\Rightarrow3x=6\)
\(x=2\)
c] \(5^{x+1}-5^x=500\)
\(\Rightarrow5^x\left(5-1\right)=500\)
\(\Rightarrow5^x\cdot4=500\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
d] \(x^2=x^4\)
\(\Rightarrow x=x^2\)
\(\Rightarrow x-x^2=0\)
\(\Rightarrow x\left(1-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\1-x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Sửa đề: \(A=2+2^2+2^3+2^4+...+2^{19}+2^{20}\)
=>\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{19}+2^{20}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{19}\left(1+2\right)\)
\(=3\left(2+2^3+...+2^{19}\right)⋮3\)
A = 2 + 22 + 23 + 24 + ... + 219 + 220
A = (2 + 22) + (23 + 24) +... + (219 + 220)
A = 2.(1+2) + 23.(1 + 2) +... + 219.(l + 2)
A = 2.3 + 23.3 +...+ 219.3 Do đó A chia hết cho 3