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20 . 2^x + 1 = 10.4^2 + 1
20 . 2^x + 1 = 10 . 16 + 1
20 . 2^x + 1 = 161
20 . 2^x = 161 - 1
20 . 2^x = 160
2^x = 8
2^x = 2^3
=> x = 3
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(x+1)+(x-3)=x-2
<=> x+1+x-3=x-2
<=> x+1+x-3-(x-2)=0
<=> x+1+x-3-x+2=0
<=> (x+x-x)+(1-3+2)=0 <=> x=0
Đúng thì tick nha ơn nhìu
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x(x-2)-3(x+1)=x(x-6)
=>x2-2x-3x-3=x2-6x
=>x2-2x-3x-3-x2+6x=0
=>(x2-x2)+(-2x-3x+6x)-3=0
=>0+x-3=0
=>x=3
x(x-2)-3(x+1)=x(x-6)
=>x2-2x-3x-1=x2-6x
=>x2-2x-3x-1-x2+6x=0
=>(x2-x2)+(-2x-3x+6x)-1=0
=>0+x-1=0
=>x=1
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Nguyễn Thị Quỳnh Linh
Ta có :
\(\left|x+1\right|\ge0\)
\(\left|x+2\right|\ge0\)
\(\left|x+3\right|\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+3\right|\ge0\)
\(\Rightarrow4x\le0\)
Mà 4 > 0
=> x > 0
=> x + 1 + x + 2 + x + 3 = 4x (phá trị uyệt đối vì x dương)
=> 3x + 6 = 4x
=> 4x - 3x = 6
=> x = 6
\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|=4x\)
\(3x+6=4x\Rightarrow x=6\)
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x + 1/2 = x - 2/3
x - x = 1/2 + 2/3
0 = 7/6 (vl)
Vậy không có x thỏa mãn