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Với x > 0
ta có
x + 1/101 + x + 2/101 + ... + x + 100/ 101 = 101x
=> 100x + ( 1 + 2 + 3 + ... + 100)/101 = 101x
=> 5050/101 = 101 x - 100x
=> x = 50
x < 0 ta có :
-x - 1/101 - x - 2/101 - ... - x - 100/101 = 101x
=> - 100x - ( 1 + 2 + .. + 100)/101 = 101x
=> 5050/101 = -100x - 101x
=> 50 = -201x
=> x =
thang Tran trả lời sai, x chỉ có thể lớn hơn 0 thôi, ta có : VT= |x+1/101|+|x+2/101|+|x+3/101|+...+|x+100/101| >= 0
Mà VT=VP =)) VP= 101x >= (lớn hơn hoặc bằng) 0 mà 101 >= 0 =)) x >= 0
<sau đó mới làm giống TH x>0 của bn í>
SAi vậy mà bn vẫn ak???
\(\left|x+\dfrac{1}{101}\right|+\left|x+\dfrac{2}{101}\right|+.....+\left|x+\dfrac{100}{101}\right|=101x\left(1\right)\)
VT(1) \(\ge0\) \(\Rightarrow VP\left(1\right)\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left|x+\dfrac{1}{101}\right|+\left|x+\dfrac{2}{101}\right|+...+\left|x+\dfrac{100}{101}\right|=100x+\dfrac{5050}{101}=101x\\ \Rightarrow x=50\)
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
=> \(100x+\frac{\left(1+100\right).100:2}{101}=101x\)
=> \(\frac{101.50}{101}=101x-100x\)
=> \(x=50\)
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 \(VT\ge0\Leftrightarrow VP\ge0\Leftrightarrow101x\ge0\Leftrightarrow x\ge0\)
=>\(\left|x+\frac{1}{101}\right|=x+\frac{1}{101};\left|x+\frac{2}{101}\right|=x+\frac{2}{101};...;\left|x+\frac{100}{101}\right|=x+\frac{100}{101}\)
=>\(x+\frac{1}{101}+x+\frac{2}{101}+x+\frac{3}{101}+...+x+\frac{100}{101}=101x\)
=>\(\left(x+x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+\frac{3}{101}+...+\frac{100}{101}\right)=100x\)
=>\(100x+50=101x\)
=> x = 50
bài này khó quá ta
mk chịu mất rùi
chúc bn học gioi!
và các bn khác giúp bn này nha'
hihi@@@
\(\left|x+\dfrac{1}{101}\right|+\left|x+\dfrac{2}{101}\right|+\left|x+\dfrac{3}{101}\right|+...+\left|x+\dfrac{100}{101}\right|=101x\)
Ta có : \(\left\{{}\begin{matrix}\left|x+\dfrac{1}{101}\right|\ge0\\\left|x+\dfrac{1}{102}\right|\ge0\\....\\\left|x+\dfrac{100}{101}\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x+\dfrac{1}{101}\right|+\left|x+\dfrac{2}{101}\right|+\left|x+\dfrac{3}{101}\right|+...+\left|x+\dfrac{100}{101}\right|\ge0\)
\(\Rightarrow101x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{1}{101}\right|=x+\dfrac{1}{101}\\\left|x+\dfrac{2}{101}\right|=x+\dfrac{2}{101}\\....\\\left|x+\dfrac{100}{101}\right|=x+\dfrac{100}{101}\end{matrix}\right.\)
\(\Rightarrow x+\dfrac{1}{101}+x+\dfrac{2}{101}+x+\dfrac{3}{101}+...+x+\dfrac{100}{101}=101x\)
\(\Rightarrow100x+\dfrac{1+2+3+...+100}{101}=101x\)
\(\Rightarrow100x+\dfrac{5050}{101}=101x\)
\(\Rightarrow100x+50=101x\)
\(\Rightarrow101x-100x=50\)
\(\Rightarrow x=50\)
Vậy \(x=50\)
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!
ko biet