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\(C=\left|x+7\right|+\left|x-5\right|+\left|x-1\right|=\left(\left|x+7\right|+\left|5-x\right|\right)+\left|x-1\right|\)
Ta có: \(\left|x+7\right|+\left|5-x\right|\ge\left|x+7+5-x\right|=8\)
Mà \(\left|x-1\right|\ge0\)
\(\Rightarrow C=\left(\left|x+7\right|+\left|5-x\right|\right)+\left|x-1\right|\ge12+0=12\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(x+7\right)\left(5-x\right)\ge0\\\left|x-1\right|=0\end{cases}\Rightarrow\hept{\begin{cases}-7\le x\le5\\x=1\end{cases}}\Rightarrow x=1}\)
Vậy Cmin = 12 khi x = 1
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a)10-Ix+1I=5
Ix+1I=10-5
Ix+1I=5=>x+1=+-5
TH1:x+1=5 TH2:x+1=-5
=>x=4 =>x=-6
b)(2.x+7)+135=0
2.x+7=0-135
2.x+7=-135
2.x=-135-7
2.x=-142
x=-142:2
x=-71
c)1/2x-2/5=1/5
1/2x=1/5+2/5
1/2x=3/5
x=3/5:1/2
x=6/5
d)1/2x+150%x=2014
1/2x+15/10x=2014
x.(1/2+15/10)=2014
x.2=2014
x=2014:2
x=1007
mình trả lời cho cậu rồi nhớ nha
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a) (2x-6)3 = (2x-6)2018
=> (2x-6)3 - (2x-6)2018 = 0
(2x-6)3.[1-(2x-6)2015 ] = 0
=> (2x-6)3 = 0 =>...
1 - (2x-6)2015 = 0 => (2x-6)2015 = 1 => ...
b) (2x-1)3 = 27 = 33
=> 2x - 1 = 3
=> ...
c) (x + 1) + (x+2) + (x+3) + ...+ (x+100) = 5750
x.100 + (1+2+3+...+100) = 5750
x.100 + [(1+100).100:2] = 5750
x.100 + 5050 = 5750
x.100 = 700
x = 7
x . x + 2x = 63
x ( x + 2 ) = 63 = 1 . 63 = 3 . 21 = 7 . 9 = ... ( các t/h âm )
=> lập bảng tìm x
x*x+2x=63
x*x+2x+1=63+1
(x*x+x)+(x+1)=64
x(x+1)+(x+1)=64
(x+1)(x+1)=64
(x+1)2 =82=(-8)2
=>x+1=8 hoặc x+1=-8
=>x=7 hoặc x=-9
Vậy x=7 hoặc x=-9