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a) \(\left(x-9\right)^4=\left(x-9\right)^7\)
\(\Rightarrow\left[{}\begin{matrix}x-9=1\\x-9=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=10\\x=9\end{matrix}\right.\)
b) \(\left(3x-15\right)^{10}=\left(3x-15\right)^{15}\)
\(\Rightarrow\left[{}\begin{matrix}3x-15=0\\3x-15=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{3}\\x=\dfrac{16}{3}\end{matrix}\right.\)
c) \(\left(x-8\right)^3=\left(x-8\right)^6\)
\(\Rightarrow\left[{}\begin{matrix}x-8=0\\x-8=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=8\\x=9\end{matrix}\right.\)
a, `(x-9)^4=(x-9)^7`
`(x-9)^4-(x-9)^7=0`
`(x-9)^4 . [(1-(x-9)^3]=0`
TH1: `(x-9)^4=0`
`x-9=0`
`x=9`
TH2: `1-(x-9)^3=0`
`(x-9)^3=1^3`
`x-9=1`
`x=10`
b, `(3x-15)^10=(3x-15)^15`
`(3x-15)^10 . [1-(3x-15)^5]=0`
TH1: `(3x-15)^10=0`
`3x-15=0`
`x=5`
TH2: `1-(3x-15)^5=0`
`(3x-15)^5=1^5`
`3x-15=1`
`x=16/3` (Loại)
c, `(x-8)^3=(x-8)^6`
`(x-8)^3 .[1-(x-8)^3]=0`
TH1: `(x-8)^3=0`
`x=8`
TH2: `1-(x-8)^3=0`
`x-8=1`
`x=9`
\(\frac{1}{x+2}-\frac{1}{x+5}+...+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\frac{1}{x+2}-\frac{1}{x+7}=\frac{x}{\left(x+2\right)\left(x+7\right)}\)
\(\Rightarrow x=1\)
Áp dụng BĐT giá trị tuyệt đối:
\(\left|x-1\right|+\left|x-5\right|=\left|x-1\right|+\left|5-x\right|\ge\left|x-1+5-x\right|=\left|4\right|=4\)
Dấu "=" xảy ra khi \(\left(x-1\right)\left(5-x\right)\ge0\)
\(\Rightarrow x-1,5-x\) cùng dấu
Nếu \(x-1,5-x\ge0\) thì \(x\ge1;5\ge x\Leftrightarrow1\le x\le5\)
Nếu \(x-1,5-x\le0\) thì \(x\le1;x\ge5\)(loại)
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
\(\left|x-1\right|+\left|x-3\right|+\left|x-5\right|+\left|x-7\right|=\left(\left|x-1\right|+\left|x-7\right|\right)+\left(\left|x-3\right|+\left|x-5\right|\right)\\ \)
\(=\left(\left|x-1\right|+\left|7-x\right|\right)+\left(\left|x-3\right|+\left|5-x\right|\right)\)
\(\ge\left|x-1+7-x\right|+\left|x-3+5-x\right|=\left|6\right|+\left|2\right|=8\)
\(\left|x+1\right|+\left|x+3\right|+\left|x+5\right|=\left(\left|x+1\right|+\left|x+3\right|\right)+\left|x+5\right|=\left(\left|x+1\right|+\left|3-x\right|\right)+\left|x+5\right|\)
\(\ge\left|x+1+3-x\right|+\left|x+5\right|=\left|4\right|+\left|x+5\right|=4+\left|x+5\right|\ge4\)
\(\left|x-1\right|+2\left|x-3\right|+\left|x-5\right|=\left(\left|x-1\right|+\left|x-5\right|\right)+2\left|x-3\right|=\left(\left|x-1\right|+\left|5-x\right|\right)+2\left|x-3\right|\)
\(\ge\left|x-1+5-x\right|+2\left|x-3\right|=\left|4\right|+2\left|x-3\right|=4+2\left|x-3\right|\ge4\)