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Câu 1.
C = 5 + 42 + 43 + ... + 42020
a) Xét A = 42 + 43 + ... + 42020
=> 4A = 43 + 44 + ... + 42021
=> 4A - A = 3A
= 43 + 44 + ... + 42021 - ( 42 + 43 + ... + 42020 )
= 43 + 44 + ... + 42021 - 42 - 43 - ... - 42020
= 42021 - 42
=> A = \(\frac{4^{2021}-4^2}{3}\)
Thế vào C ta được : \(C=5+\frac{4^{2021}-4^2}{3}=\frac{15}{3}+\frac{4^{2021}-4^2}{3}=\frac{4^{2021}+15-16}{3}=\frac{4^{2021}-1}{3}\)
b) D = 42021 => \(\frac{D}{3}=\frac{4^{2021}}{3}\)
Vì 42021 - 1 < 42021 => \(\frac{4^{2021}-1}{3}< \frac{4^{2021}}{3}\)
=> C < D/3
c) Dùng kết quả ý a) ta được :
3C + 1 = 42x-6
<=> \(3\cdot\frac{4^{2021}-1}{3}+1=4^{2x-6}\)
<=> 42021 - 1 + 1 = 42x-6
<=> 42021 = 42x-6
<=> 2021 = 2x - 6
<=> 2x = 2027
<=> x = 2027/2
Câu 2.
( x - 1 )( 4 + 22 + 23 + ... + 220 ) = 222 - 221
Xét A = 22 + 23 + ... + 220
=> 2A = 23 + 24 + ... + 221
=> A = 2A - A
= 23 + 24 + ... + 221 - ( 22 + 23 + ... + 220 )
= 23 + 24 + ... + 221 - 22 - 23 - ... - 220
= 221 - 4
Thế vô đề bài ta được
( x - 1 )( 4 + 221 - 4 ) = 222 - 221
<=> ( x - 1 ).221 = 221( 2 - 1 )
<=> x - 1 = 1
<=> x = 2
Bài 1 :
\(2^x.8=512\)
\(2^x=512:8\)
\(2^x=64\)
\(2^x=2^6\)
\(\Rightarrow x=6\)
\(b,\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
\(c,x^{20}=x\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(d,\left(x-3\right)^{10}=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
\(\left(x-3\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\2x+4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
\(\left(x^2+3\right)\left(2x^2-50\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+3=0\\2x^2-50=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=-3\left(loại\right)\\x^2=25\end{cases}}\)
\(\Rightarrow x=\pm5\)<=>x=-5 hoặc x=5
\(A=2^2+2^2+2^3+2^4+...+2^{2006}\)
\(2A=2^3+2^3+2^4+2^5+...+2^{2007}\)
\(2A-A=2^{2007}+2^3-\left(2^2+2^2\right)\)
\(A=2^{2007}+8-8\)
\(A=2^{2007}\)
\(\Rightarrow\text{ }2A=2^{2008}=2^{2\cdot1004}=\left(2^2\right)^{1004}=4^{1004}\)
\(\Rightarrow\text{ }x=1004\)
Đặt B=2^2+2^3....+2^2006
2B=2^3+2^4+....+2^2007
=>2B-B=(2^3+2^4+...+2^2007)-(2^2+2^3+....+2^2006)
B=2^2007-2^2
=>A=2^2007-2^2+2^2
A=2^2007
=>2A=2^2008
=>2A=4^1004
Vậy x=1004
+) \(A=3\left(x-4\right)^4-4\ge-4\)
Min A = -4 \(\Leftrightarrow x-4=0\Leftrightarrow x=4\)
+) \(B=5+2\left(x-2019\right)^{2020}\ge5\)
Min B = 5 \(\Leftrightarrow x-2019=0\Leftrightarrow x=2019\)
+) \(C=5+2018\left(2020-x\right)^2\)
Min C = 5 \(\Leftrightarrow2020-x=0\Leftrightarrow x=2020\)
+) \(D=\left(x-1\right)^{2020}+\left(y+x\right)-1\ge-1\)
Min D = -1 \(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-x\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=-1\end{cases}}}\)
+) \(E=2\left(x-1\right)^2+3\left(2x-y\right)^4-2\ge-2\)
Min E = -2 \(\Leftrightarrow\hept{\begin{cases}x-1=0\\2x-y=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\2x=y\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=2\end{cases}}}\)
\(3^{2x+2}=9^{x+3}\)
\(\Rightarrow3^{2x+2}=3^{2x+6}\)
\(\Rightarrow2x+6=2x+2\)
\(\Rightarrow\left(2x-2x\right)+\left(6-2\right)=0\)
\(\Rightarrow0x=-4\left(loại\right)\)
\(b,\left(x-3\right)^4=\left(x-3\right)^6\)
\(\Rightarrow\left(x-3\right)^4-\left(x-3\right)^4.\left(x-3\right)^2=0\)
\(\Rightarrow\left(x-3\right)^4.\left[1-\left(x-3\right)^2\right]=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-3\right)^4=0\\1-\left(x-3\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x-3\in\left\{\pm1\right\}\end{cases}\Rightarrow}\hept{\begin{cases}x=3\\x\in\left\{4;2\right\}\end{cases}\Rightarrow}x\in\left\{2;3;4\right\}}\)
a, => 32x+2 =32.(x+3)
2x+2=2.(x+3)
2(x+1)=2(x+3)
x+1=x+3
=> x= rỗng
vậy............
c, x15-x2=0
x2(x13-1)=0
\(\orbr{\begin{cases}x^2=0\\x^{13}=0\end{cases}}< =>\orbr{\begin{cases}x=0\\x^{13}=1^{13}\end{cases}}< =>\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
vậy.........
\(2^{10}.2^{x+4}=64^5\)
\(\Leftrightarrow2^{x+14}=2^{30}\)
\(\Leftrightarrow x+14=30\)
\(\Leftrightarrow x=16\)
\(5^x+5^{x+3}=630\)
\(\Rightarrow5^x.1+5^x.125=630\)
\(\Rightarrow5^x.126=630\)
\(\Rightarrow5^x=5\)
\(\Rightarrow x=1\)
\(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+..............+\left(x+100\right)=7450\)
\(\Rightarrow\left(x+x+x+x+.........+x\right)+\left(1+2+3+..........+100\right)=7450\)
\(101x+5050=7450\)
Đến đây tự tính
a: x=15
c: \(\Leftrightarrow2x+1=3\)
hay x=1
a) \(\Rightarrow x=2021-2006=15\)
b) \(\Rightarrow2x-2016=64\Rightarrow2x=2016+64=2080\Rightarrow x=1040\)
c) \(\Rightarrow\left(2x+1\right)^3=81:3=27\Rightarrow2x+1=3\)
\(\Rightarrow2x=3-1=2\Rightarrow x=1\)