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4A=1+1/4+1/42+......+1/498
4A - A = ( 1+1/4+1/42+..........+1/498) - ( 1/4+1/42+1/43+.......+1/499)
3A= 1-1/499
A= 1/3 - 1/499 : 3
Mà 1/499 : 3 > 0 => 1/3 - 1/499 : 3 < 1/3
Hay A < 1/3
a/ Rút gọn:
\(A=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{99}}.\)
=> \(4A=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{98}}\)
=> \(4A=1+\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{98}}+\frac{1}{4^{99}}\right)-\frac{1}{4^{99}}\)
<=> \(4A=1+A-\frac{1}{4^{99}}\)
=> \(3A=1-\frac{1}{4^{99}}\)
=> \(A=\frac{1}{3}-\frac{1}{3.4^{99}}\)
b/ Ta có: \(A=\frac{1}{3}-\frac{1}{3.4^{99}}< \frac{1}{3}\)
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\(\left(3x-1\right)^2+2\left(9x^2-1\right)+\left(3x+1\right)^2\)
\(=9x^2-6x+1+18x^2+2+9x^2+6x+1\)
\(=36x^2+4\)
\(\left(x^2-1\right)\left(x+3\right)-\left(x-3\right)\left(x^3+3x+9\right)\)
\(=x^3+3x^2-x+3-\left(x^4+3x^2+9x-3x^3-9x-27\right)\)
\(=x^3+3x^2-x+3-x^4-3x^2-9x+3x^3+9x-27\)
\(=\left(3x^2-3x^2\right)+\left(9x-9x\right)-x-\left(27-3\right)+x^3-x^4+3x^3\)
\(=-x-24+x^3-x^4+3x^3\)
\(\left(x+4\right)\left(x-4\right)-\left(x-4\right)^2\)
\(=x^2-16-\left(x-4\right)^2\)
\(=x^2-16-x^2+8x-16\)
\(=8x-32\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(\left(x+1\right)^2-\left(x-1\right)^2\right)-3\left(x^2-1\right)\)
\(=4x-3x^2+1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=3\left(x-1\right)-2\left|x-3\right|=3x-3-2\left|x-3\right|\)
Trường hợp 1: x>=3
A=3x-3-2(x-3)=3x-3-2x+6=x+3
TRường hợp 2: x<=-3
A=3x-3-2(3-x)=3x-3-6+2x=5x-9
![](https://rs.olm.vn/images/avt/0.png?1311)
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Đặt \(A=2\left|2x-1\right|-3\left|2x+3\right|\)
TH1 : \(x< -\frac{3}{2};\) ta có :
\(A=2.\left(1-2x\right)-3.\left[-\left(2x+3\right)\right]\)
\(=2-4x+6x+9\)
\(=2x+11\)
TH2 : \(-\frac{3}{2}\le x< \frac{1}{2};\)ta có :
\(A=2\left(1-2x\right)-3\left(2x+3\right)\)
\(=2-4x-6x-9\)
\(=-10x-7\)
TH3 : \(x\ge\frac{1}{2};\)ta có :
\(A=2\left(2x-1\right)-3\left(2x+3\right)\)
\(=4x-2-6x-9\)
\(=-2x-11\)
+ Với \(x< -\frac{3}{2}\) thì |2x - 1| = 1 - 2x; |2x + 3| = -(2x + 3) = -2x - 3
Ta có: 2.|2x - 1| - 3.|2x - 3| = 2.(1 - 2x) - 3.(-2x - 3)
= 2 - 4x + 6x + 9
= 2x + 11
+ Với \(-\frac{3}{2}\le x< \frac{1}{2}\) thì |2x - 1| = 1 - 2x; |2x + 3| = 2x + 3
Ta có: 2.|2x - 1| - 3.|2x + 3| = 2.(1 - 2x) - 3.(2x + 3)
= 2 - 4x - 6x - 9
= -10x - 7
+ Với \(x>\frac{1}{2}\) thì |2x - 1| = 2x - 1; |2x + 3| = 2x + 3
Ta có: 2.|2x - 1| - 3.|2x + 3| = 2.(2x - 1) - 3.(2x + 3)
= 4x - 2 - 6x - 9
= -2x - 11
Đúng thì thôi c` sai thì cấm chửi
P= (1-1/1+2).(1-1/1+2+3).(1-1/1+2+3+4).....(1/1+2+3+4+...+100)
P=0
P=0
NHẤN MÁY TÍNH LÀ RA THÔI