[4mux2 -3 mũ2]x+ơ4mũ2 +3mũ2]x =2 n\mũ7
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A=1+3+3^2+3^3+3^4+3^5+3^6
3A=3+3^2+3^3+3^4+3^5+3^6+3^7
3A-A=(3+3^2+3^3+3^4+3^5+3^6+3^7)-(1+3+3^2+3^3+3^4+3^5+3^6)
A=3^7-1
Vì A =3^7-1 ; B =3^7-1
=> A=B
Sửa đề:
\(A=1+3+3^2+3^3+3^4+3^5+3^6\)
\(3A=3+3^2+...+3^7\)
\(3A-A=\left(3+3^2+3^3+...+3^7\right)-\left(1+3+3^2+...+3^6\right)\)
\(2A=3^7-1\)
\(\Rightarrow A=\frac{3^7-1}{2}< 3^7-1=B\)
Vậy \(A< B\)
S = 1 + 2 + 22 + 23 + 24 + 25 + 26 + 27
= (1 + 2) + (22 + 23) + (24 + 25) + (26 + 27)
= (1 + 2) + 22(1 + 2) + 24(1 + 2) + 26(1 + 2)
= (1 + 2)(1 + 22 + 24 + 26)
= 3(1 + 22 + 24 + 26) \(⋮3\)(ĐPCM)
2S = 1 + 2 + 22 + 23 + 24 + 25 + 26 + 27
S = (1+2 ) + (22 + 23 ) + (24 + 25 ) + (26 +27)
S = 3 + 22(1+2) + 24(1+2) + 26(1+2)
S = 3+22.3 + 24.3 + 26 .3
S = 3(1+22 + 24 + 26 ) \(⋮\) 3
=> đpcm
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)(ĐPCM)
a)ta có:g(x)=(x-3).(16-4x)=0
Th1:x-3=0
=>x=3
Th2:16-4x=0
=>4x=16
=>x=4
1. (-2x - 1)(x2 - x - 3) - (x + 2)(x + 1)2
= -2x3 + 2x2 + 6x - x2 + x + 3 - (x + 2)(x2 + 2x + 1)
= -2x3 + x2 + 7x + 3 - x3 - 2x2 - x - 2x2 - 2x - 2
= -3x3 - 3x2 + 4x + 1
2. (x + 2)(x - 1) - (x - 3)(x + 2) = 3
=> (x + 2)(x - 1 - x + 3) = 3
=> (x + 2).0 = 3
...(xem lại đề)
\(\left(x+2\right)\left(x-1\right)-\left(x-3\right)\left(x+2\right)=3\)
\(\Leftrightarrow\left(x+2\right)\left(x-1-x+3\right)=3\)
\(\Leftrightarrow2\left(x+2\right)=3\)
\(\Leftrightarrow x+2=\frac{3}{2}\)
\(\Leftrightarrow x=\frac{3}{2}-2\)
\(\Leftrightarrow x=-\frac{1}{2}\)