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a) Với \(2x+1\ge0\Rightarrow x\ge\frac{-1}{2}\) thì \(2x+1-2=1\Rightarrow2x+1=3\Rightarrow2x=2\Rightarrow x=1\) (nhận)
Với 2x + 1 <0 => x < -1/2 thì \(-\left(2x+1\right)-2=1\Rightarrow-2x-1-2=1\Rightarrow-2x=4\Rightarrow x=-2\) (nhận)
Vậy x = {-2;1}
b) Với \(3x-1\ge0\Rightarrow x\ge\frac{1}{3}\) thì \(3x-1=1-3x\Rightarrow6x=2\Rightarrow x=\frac{1}{3}\) (nhận)
Với 3x - 1 < 0 => x < 1/3 thì \(-\left(3x-1\right)=1-3x\Rightarrow-3x+1=1-3x\Rightarrow0x=0\) (nhận)
Vậy x = 1/3 hoặc x vô số nghiệm với x < 1/3
a) | 2x + 1| - 2 = 1
| 2x + 1| = 3
TH1: 2x + 1 = 3
2x = 2
x = 1
TH2: 2x + 1 = -3
2x = -4
x = -2
KL: x = 1 hoặc x = - 2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{8^5.\left(-5\right)^8+\left(-2\right)^5.10^9}{2^{16}.5^7+20^8}\)
\(=\frac{2^{15}.5^8+\left(-2\right)^5.10^9}{2^{16}.5^7+2.10^8}\)
\(=\frac{5-2^4.10}{2}\)
\(=5-8.10\)
\(=5-80\)
\(=-75\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Mk lm đc câu nào hay câu đó nha
A = 3/8 + [ 11/17 + (-3/8 ) ]
= [ 3/8 + ( -3 /8 ) ] + 11/17
= 0 + 11/17
= 11/17
C = 1/3 . 17/21 + 4/21 . 1/3 - 5/12
= 1/3 ( 17/21 + 4/21 ) - 5/12
= 1/3 - 5/12
= -1/12
D = 1/7 + 4/9 + 6/7 + 25/45
= ( 1/7 + 6/7 ) + ( 4/9 + 25/45 )
= 1 +1
= 2
Mk chỉ lm đc thế này thôi mong thông cảm hjhj
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
\(A=\frac{8^5(-5)^8+(-2)^5.10^9}{2^{16}.5^7+20^8}\) \(=\frac{(2^3)^5(-5)^8+(-2)^5.2^9.5^9}{2^{16}.5^7+(2^2.5)^8}\)
\(=\frac{2^{15}.5^8-2^5.2^9.5^9}{2^{16}.5^7+2^{16}.5^8}\)
\(=\frac{2^{14}.5^8(2-5)}{2^{16}.5^7(1+5)}\)
\(=\frac{5(-3)}{2^2.6}=\frac{-5}{8}\)
Bài 3:
Đặt \(\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt\)
Thay vào:
\(\frac{5a+3b}{5a-3b}=\frac{5bt+3b}{5bt-3b}=\frac{b(5t+3)}{b(5t-3)}=\frac{5t+3}{5t-3}\)
\(\frac{5c+3d}{5c-3d}=\frac{5dt+3d}{5dt-3d}=\frac{d(5t+3)}{d(5t-3)}=\frac{5t+3}{5t-3}\)
Do đó: \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\) (đpcm)
Bài 4:
Ta có:
\(A=3+3^2+3^3+3^4+...+3^{100}\)
\(=(3+3^2+3^3+3^4)+(3^5+3^6+3^7+3^8)+....+(3^{97}+3^{98}+3^{99}+3^{100})\)
\(=3(1+3+3^2+3^3)+3^5(1+3+3^2+3^3)+...+3^{97}(1+3+3^2+3^3)\)
\(=3.40+3^5.40+....+3^{97}.40\)
\(=120(1+3^4+....+3^{96})\vdots 120\)
Ta có đpcm.
a) A=6.10^7+5.10^5+4.10^3+2.10
A= 6.10000000+5.100000+4.1000+2.10
A= 60000000+500000+4000+20
A= 60504020
b) B=12.10^8+17.10^7+5.10^4+3
B= 12.100000000+17.10000000+5.10000+3
B=1200000000+170000000+50000+3
B=1370050003