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\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow\frac{x}{\left(x+2\right)\left(x+14\right)}=\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}-\frac{1}{x+8}-\frac{1}{x+14}\)
\(\Leftrightarrow\frac{x}{\left(x+2\right)\left(x+14\right)}=\frac{1}{x+2}-\frac{1}{x+14}\)
\(\Leftrightarrow\frac{x}{\left(x+2\right)\left(x+14\right)}=\frac{\left(x+14\right)-\left(x+2\right)}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow x=\left(x+14\right)-\left(x+2\right)\)
\(\Leftrightarrow x=x+14-x-2\)
\(\Leftrightarrow x=\left(x-x\right)+\left(14-2\right)\)
\(\Leftrightarrow x=0+12\)
\(\Leftrightarrow x=12\)
1. a, 3x + |x - 2| = 8
<=> |x - 2| = 8 - 3x
Xét 2 TH :
TH1: x - 2 = 8 - 3x
<=> x + 3x = 8 + 2
<=> 4x = 10
<=> x = \(\dfrac{5}{2}\) (thỏa mãn)
TH2: x - 2 = -(8 - 3x)
<=> x - 2 = -8 + 3x
<=> -2 + 8 = 3x - x
<=> 6 = 2x
<=> x = 3 (thỏa mãn)
b, 5 - |x - 1| = 4
<=> |x - 1| = 1
<=> \(\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\) (thỏa mãn)
@Nguyễn Hoàng Vũ
2. 5.(x - 2) - 4.(1 - 3x) = |3 - 7| + 2.(1 + 2x)
<=> 5x - 10 - 4 + 12x = 4 + 2 + 4x
<=> 17x - 14 = 6 + 4x
<=> 17x - 4x = 6 + 14
<=> 13x = 20
<=> x = \(\dfrac{20}{13}\) (thỏa mãn)
@Nguyễn Hoàng Vũ
a ) Ta có : 4(x - 5) - 3(x + 7) = -19
<=> 4x - 20 - 3x - 21 = -19
=> x - 41 = -19
=> x = -19 + 41
=> x = 22
b) Ta có " 7(x - 3) - 5(3 - x) = 11x - 5
<=> 7x - 21 - 15 + 5x = 11x - 5
<=> 12x - 36 = 11x - 5
=> 12x - 11x = -5 + 36
=> x = 31
=> \(\frac{1}{x+2}-\frac{1}{x+14}=\frac{1}{16}\)
=> \(\frac{x+14}{\left(x+2\right).\left(x+14\right)}-\frac{x+2}{\left(x+2\right).\left(x+14\right)}=\frac{1}{16}\)
=> \(\frac{\left(x+14\right)-\left(x+2\right)}{\left(x+2\right).\left(x+14\right)}=\frac{1}{16}\)
=> \(\frac{x+14-x-2}{x\left(2+14\right)}=\frac{1}{16}\)
=> \(\frac{12}{16x}=\frac{1}{16}\)
=> x = 12
a) \(2\left(x-5\right)-3\left(x+7\right)=14\)
\(\Leftrightarrow2x-10-3x-21=14\)
\(\Leftrightarrow-x-31=14\)
\(\Leftrightarrow-x=45\Leftrightarrow x=-45\)
b) \(5\left(x-6\right)-2\left(x+3\right)=12\)
\(\Leftrightarrow5x-30-2x-6=12\)
\(\Leftrightarrow3x-36=12\)
\(\Leftrightarrow3x=48\Leftrightarrow x=16\)
c) \(3\left(x-4\right)-\left(8-x\right)=12\)
\(\Leftrightarrow3x-12-8+x=12\)
\(\Leftrightarrow4x-20=12\)
\(\Leftrightarrow4x=32\Leftrightarrow x=8\)
d) \(-7\left(3x-5\right)+2\left(7x-14\right)=28\)
\(\Leftrightarrow-21x+35+14x-28=28\)
\(\Leftrightarrow-7x+35=0\Leftrightarrow x=5\)
(x-2)^4-(x-2)^2=0
(x-2)^2.(x-2)^2-(x-2)^2.1=0
(x-2)^2.[(x-2)^2-1]=0
Nếu (x-2)^2=0
(x-2)^2=0^2
Suy ra : x-2=0
x =0+2
x =2
Nếu (x-2)^2-1=0
(x-2)^2 =0+1
(x-2)^2 =1
(x-2)^2 =1^2
Suy ra : x-2=1
x = 1+2
x = 3
Vậy x thuộc { 2;3 }
\(\left(x-2\right)^2=\left(x-2\right)^4\)
\(\left(x-2\right)^2-\left(x-2\right)^4=0\)
\(\left(x-2\right)^2.\left[1-\left(x-2\right)^2\right]=0\)
TH1: TH2:
\(\left(x-2\right)^2=0\) \(1-\left(x-2\right)^2=0\)
\(x-2=0\) \(\left(x-2\right)^2=1\)
\(x=0+2\) * x-2=1 * x-2=-1
\(x=2\) x=1+2 x=-1+2
x=3 x=1
vậy x=2 hoặc x=3 hoặc x=1