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a) 2(x-2)+1=2x
=>2x-4+1=2x
=>2x-3-2x=0
=>-3=0 vô lý
b)x/-6=6/5
=>5x=-36
=>x=-36/5
c)6/-x=9/3
=>6=-3x
=>x=-2
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2x^2+3x+2=2x(x+1)+2(x+1)=(x+1)(2x+1) => luôn chia hết cho x+1
kl: Đúng với mọi x khác -1
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Ta có -2-(x-17)=34-(-x+25)
-2-x+17=34+x+25
-2 +(-x)+17+(-x)=34+25
[(-x)+(-x)]+[(-2)+17]=29
(-x).2+15=29
(-x).2=29-15
(-x).2=14
(-x)=14:2
(-x)=7
x=-7
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Tu de bai ra ta co:
2x+3va y+1 thuoc U(24)={1;2;3;4;6;8;12;24}
Ta co bang gia tri:
2x+3 1 2 3 4 6 8 12 24
y+1 24 12 8 6 4 3 2 1
x -1 -0,5 0 0,5 1,5 2,5 4,5 10,5
y 23 11 7 5 3 2 1 0
chon/loai loai loai chon loai loai loai loai loai
Vay cac cap (x,y thoa man de bai la:(o,7) vay x=0,y=7 se thoa man
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\(\Rightarrow\)x+2\(\in\)Ư(9)
Ư(9)={\(\pm1\); \(\pm3\); \(\pm9\)}
\(\Rightarrow\)x+2\(\in\left\{\pm1;\pm3;\pm9\right\}\)
\(\Rightarrow\)x\(\in\left\{\pm1;-3;-5;-11;7\right\}\)
Vậy x\(\in\left\{\pm1;-3;-5;-11;7\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét vế trái của biểu thức (VT)
\(VT=1+2^2\left(1+1\right)+3^2\left(2+1\right)+4^2\left(3+1\right)+...+10^2\left(9+1\right)\)
\(VT=1+1.2^2+2^2+2.3^2+3^2+3.4^2+4^2+...+9.10^2+10^2\)
\(VT=\left(1+2^2+3^2+4^2+...+10^2\right)+\left(1.2^2+2.3^2+3.4^2+...+9.10^2\right)\)
Đặt \(A=1+2^2+3^2+4^2+...+10^2\)
\(A=1+2\left(1+1\right)+3\left(2+1\right)+4\left(3+1\right)+...+10\left(9+1\right)\)
\(A=1+1.2+2+2.3+3+3.4+4+...+9.10+10\)
\(A=\left(1+2+3+4+...+10\right)+\left(1.2+2.3+3.4+...+9.10\right)\)
Đặt \(B=1.2^2+2.3^2+3.4^2+...+9.10^2\)
\(B=1.2\left(3-1\right)+2.3\left(4-1\right)+3.4\left(5-1\right)+...+9.10\left(11-1\right)\)
\(B=1.2.3-1.2+2.3.4-2.3+3.4.5-3.4+...+9.10.11-9.10\)
\(B=\left(1.2.3+2.3.4+3.4.5+...+9.10.11\right)-\left(1.2+2.3+3.4+...+9.10\right)\)
\(VT=A+B=\left(1+2+3+4+...+10\right)+\left(1.2.3+2.3.4+3.4.5+...+9.10.11\right)\)
Đặt \(C=1+2+3+4+...+10=55\) (Tổng cấp số cộng)
Đặt \(D=1.2.3+2.3.4+3.4.5+...+9.10.11\)
\(4D=1.2.3.4+2.3.4.4+3.4.5.4+...+9.10.11.4\)
\(4D=1.2.3.4+2.3.4\left(5-1\right)+3.4.5\left(6-2\right)+...+9.10.11\left(12-8\right)\)
\(4D=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-...-8.9.10.11+9.10.11.12\)
\(4D=9.10.11.12\Rightarrow D=3.9.10.11\)
\(\Rightarrow VT=C+D=55+3.9.10.11=3025=55^2\)
\(\Rightarrow55^2=\left(x+1\right)^2\Rightarrow55=x+1\Rightarrow x=54\)
Ta có: (2x +1)3 =125
(2x +1)3 = 53
2x + 1 = 5
2x = 4
x = 2
( 2x + 1 ) \(^3\)= 125
( 2x + 1 ) = \(\sqrt[3]{125}\)
( 2x + 1 ) = 5
2x = 5 - 1
2x = 4
x = 4 : 2
x = 2