5. Tìm x biết:
a, \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+...+\left|x+10\right|=11x+1\)
b, \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)
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Nhận xét :
\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)
Vì \(x\ge0\) nên pt a) tương đương với : \(100x+\frac{1+2+3+...+100}{101}=101x\)
\(\Leftrightarrow x=\frac{100.101}{2.101}=50\)
b)
Tương tự câu a) , phương trình tương đương với :
\(49x+\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{...1}{97.99}=50x\)
\(\Rightarrow x=\frac{97}{195}\)
Ta có: \(\left|x+\frac{1}{101}\right|\ge0\); \(\left|x+\frac{2}{101}\right|\) \(\ge0\); ...; \(\left|x+\frac{100}{101}\right|\ge0\)
\(\Rightarrow101x\ge0\)
và \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|\ge0\)
\(\Rightarrow\left|x+\frac{1}{101}\right|=x+\frac{1}{101}\); \(\left|x+\frac{2}{101}\right|=x+\frac{2}{101}\); ...; \(\left|x+\frac{100}{101}\right|=x+\frac{100}{101}\)
Thay vào đề bài ta đc:
\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}\right)=101x\)
\(\Rightarrow\) \(100x\) + \(\left(\frac{1+2+...+101}{101}\right)=101x\)
\(\Rightarrow100x+101=101x\)
\(\Rightarrow x=101\)
Vậy \(x=101.\)
\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+\left|x+\frac{3}{101}\right|+....+\left|x+\frac{100}{101}\right|\)=101x (1)
điều kiện:101x\(\ge\) 0 \(\Rightarrow\) x\(\ge\) 0
từ (1) \(\Rightarrow\) \(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}\)=101x
\(\Rightarrow\) 100x+(\(\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}\))=101x
\(\Rightarrow\) 100x+\(\frac{5050}{101}\)=101x
\(\Rightarrow\) \(\frac{5050}{101}\)=101x-100x
\(\Rightarrow\) x=50
k bt mk lm sai hay lm đúng nữa
nếu mk lm sai thì thôi nha!
Vì \(\left|x+\frac{1}{101}\right|\ge0;\left|x+\frac{2}{101}\right|\ge0;...;\left|x+\frac{100}{101}\right|\ge0\forall x\)
\(\Rightarrow\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|\ge0\forall x\)
\(\Rightarrow101x\ge0\)
\(\Rightarrow x\ge0\)
Từ điều kiện trên ta có :
\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)
\(100x+\frac{1+2+...+100}{101}=101x\)
\(101x-100x=\frac{5050}{101}\)
\(x=50\)
Vậy x = 50
\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+....+\left|x+\frac{100}{101}\right|=101x\)
\(KĐ:101x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)
\(x+\frac{1}{101}+x+\frac{2}{101}+....+x+\frac{100}{101}=101x\)
\(100x+\left(\frac{1}{101}+\frac{2}{101}+....+\frac{100}{101}\right)=101x\)
\(\Rightarrow101-100x=\frac{1+2+....+100}{101}\)
\(x=\frac{\left(1+100\right)\left(100-1+1\right):2}{101}\)
\(x=\frac{101.100:2}{101}\)
\(x=50\)
Vì \(\left|x+\frac{1}{101}\right|+\left|x+\frac{1}{102}\right|+....+\left|x+\frac{100}{101}\right|>0\)
\(\Rightarrow101x>0\)
\(\Rightarrow x>0\)
\(\Rightarrow\left(x+\frac{1}{101}\right)+.....+\left(x+\frac{100}{101}\right)=101x\)
\(\Rightarrow100x+\left(\frac{1}{101}+\frac{2}{101}+....+\frac{100}{101}\right)=101x\)
\(\Rightarrow x=\frac{\left(100+1\right)100:2}{101}\)
\(\Rightarrow x=\frac{50.101}{101}\)
\(\Rightarrow x=50\)
Vậy x = 50
Do \(\left|x+\frac{1}{101}\right|\ge0;\left|x+\frac{2}{101}\right|\ge0;\left|x+\frac{3}{101}\right|\ge0;...;\left|x+\frac{100}{101}\right|\ge0\)
=> \(101x\ge0\)
=> \(x\ge0\)
=> \(\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+\left(x+\frac{3}{101}\right)+...+\left(x+\frac{100}{101}\right)=101x\)
=> \(\left(x+x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+\frac{3}{101}+...+\frac{100}{101}\right)=101x\)
100 số x 100 phân số
=> \(100x+\frac{\left(1+100\right).100:2}{101}=101x\)
=> \(\frac{101.50}{101}=101x-100x\)
=> \(x=50\)
a, Vào câu hỏi tương tự nhé
b, Vì \(\hept{\begin{cases}\left|x+3\right|\ge0\\\left|x+1\right|\ge0\end{cases}\Rightarrow\left|x+3\right|+\left|x+1\right|\ge0\Rightarrow3x\ge0\Rightarrow x\ge0}\)
=> x+3+x+1=3x
=> 2x+4=3x
=>x=4
c, \(\left|x-4\right|+\left|x-10\right|+\left|x+101\right|+\left|x+990\right|+\left|x+1000\right|=\left|4-x\right|+\left|10-x\right|+\left|x+101\right|+\left|x+990\right|+\left|x+1000\right|\)
Có \(\left|4-x\right|\ge4-x;\left|10-x\right|\ge10-x;\left|x+990\right|\ge x+990;\left|x+1000\right|\ge x+1000\)
=>\(\left|4-x\right|+\left|10-x\right|+\left|x+101\right|+\left|x+990\right|+\left|x+1000\right|\)
=> \(2005\ge4-x+10-x+x+990+x+1000+\left|x+101\right|\)
=> \(2005\ge\left|x+101\right|+2004\)
=> \(\left|x+101\right|\le1\)
=> \(x+101\in\left\{-1;0;1\right\}\Rightarrow x\in\left\{-102;-101;-100\right\}\)
d, tương tự b
Do \(\left|a\right|\ge0\) nên:
a) \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\ge0\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}\right)=101x\) (100 số hạng x)
\(\Leftrightarrow100x+5050=101x\Leftrightarrow201x=5050\Leftrightarrow x=\frac{5050}{201}\)
b) Đề sai nhé!
a ) \(3-4.\left|5-6x\right|=7\)
\(\Leftrightarrow4.\left|5-6x\right|=-4\)
\(\Leftrightarrow\left|5-6x\right|=-1\)
\(\Leftrightarrow\) Không thõa mãn ( vì \(x\ge0\) )
b) Do \(\left|x+2\right|\ge0;\left|x+\frac{3}{5}\right|\ge0;\left|x+\frac{1}{2}\right|\ge0\)
=> \(4x\ge0\)
=> \(x\ge0\)
Lúc này ta có: \(\left(x+2\right)+\left(x+\frac{3}{5}\right)+\left(x+\frac{1}{2}\right)=4x\)
=> \(\left(x+x+x\right)+\left(2+\frac{3}{5}+\frac{1}{2}\right)=4x\)
=> \(3x+\frac{31}{10}=4x\)
=> \(4x-3x=\frac{31}{10}\)
=> \(x=\frac{31}{10}\)
Vậy \(x=\frac{31}{10}\)
c) Do \(\left|x+\frac{1}{101}\right|\ge0;\left|x+\frac{2}{101}\right|\ge0;\left|x+\frac{3}{101}\right|\ge0;...;\left|x+\frac{100}{101}\right|\ge0\)
=> \(101x\ge0\)
=> \(x\ge0\)
Lúc này ta có: \(\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+\left(x+\frac{3}{101}\right)+...+\left(x+\frac{100}{101}\right)=101x\)
=> \(\left(x+x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+\frac{3}{101}+...+\frac{100}{101}\right)=101x\)
100 số x
=> \(100x+\frac{\left(1+100\right).100:2}{101}=101x\)
=> \(\frac{101.50}{101}=101x-100x\)
=> \(x=50\)
Vậy x = 50