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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
\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+....+\left|x+\frac{100}{101}\right|=101x\)
\(KĐ:101x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)
\(x+\frac{1}{101}+x+\frac{2}{101}+....+x+\frac{100}{101}=101x\)
\(100x+\left(\frac{1}{101}+\frac{2}{101}+....+\frac{100}{101}\right)=101x\)
\(\Rightarrow101-100x=\frac{1+2+....+100}{101}\)
\(x=\frac{\left(1+100\right)\left(100-1+1\right):2}{101}\)
\(x=\frac{101.100:2}{101}\)
\(x=50\)
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@@@
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
=> \(100x+\frac{\left(1+100\right).100:2}{101}=101x\)
=> \(\frac{101.50}{101}=101x-100x\)
=> \(x=50\)
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: \(\left|x+\frac{1}{1.5}\right|+\left|x+\frac{1}{5.9}\right|+\left|x+\frac{1}{9.13}\right|+...+\left|x+\frac{1}{397.401}\right|\ge0\)
\(\Rightarrow101x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow x+\frac{1}{1.5}+x+\frac{1}{5.9}+...+x+\frac{1}{397.401}=101x\)
\(\Rightarrow100x+\left(\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{397.401}\right)=101x\)
\(\Rightarrow\frac{1}{4}\left(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{397.401}\right)=x\)
\(\Rightarrow x=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{397}-\frac{1}{401}\right)\)
\(\Rightarrow x=\frac{1}{4}\left(1-\frac{1}{401}\right)\)
\(\Rightarrow x=\frac{1}{4}.\frac{400}{401}\)
\(\Rightarrow x=\frac{100}{401}\)
Vậy...
\(a,\frac{x+1}{65}+\frac{x+2}{64}=\frac{x+3}{63}+\frac{x+4}{62}\)
\(\Rightarrow\left[\frac{x+1}{65}+1\right]+\left[\frac{x+2}{64}+1\right]=\left[\frac{x+3}{63}+1\right]+\left[\frac{x+4}{62}+1\right]\)
\(\Rightarrow\frac{x+1+65}{65}+\frac{x+2+64}{64}=\frac{x+3+63}{63}+\frac{x+4+62}{62}\)
\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}\)
\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}=0\)
\(\Rightarrow\left[x+66\right]\left[\frac{1}{65}+\frac{1}{64}-\frac{1}{63}+\frac{1}{62}\right]=0\)
Mà \(\frac{1}{65}+\frac{1}{64}-\frac{1}{63}+\frac{1}{62}\ne0\)
\(\Rightarrow x+66=0\)
\(\Rightarrow x=0-66=-66\)
Auto làm nốt câu b
a, Cộng cả 2 vế với 2
Ta có \(\frac{x+1}{64}+\frac{x+2}{63}+2=\frac{x+3}{62}+\frac{x+4}{61}+2\)
\(\left(\frac{x+1}{64}+\frac{64}{64}\right)+\left(\frac{x+2}{63}+\frac{63}{63}\right)=\left(\frac{x+3}{62}+\frac{62}{62}\right)+\left(\frac{x+4}{61}+\frac{61}{61}\right)\)
=> \(\frac{x+65}{64}+\frac{x+65}{63}=\frac{x+65}{62}+\frac{x+65}{61}\)\(\)
=> \(\frac{x+65}{64}+\frac{x+65}{63}-\frac{x+65}{62}-\frac{x+65}{61}=0\)
=> \(\left(x+65\right)\left(\frac{1}{64}+\frac{1}{63}-\frac{1}{62}-\frac{1}{61}\right)=0\)
Do \(\frac{1}{64}+\frac{1}{63}-\frac{1}{62}-\frac{1}{61}\ne0\)=> \(x+65=0\)
=> \(x=-65\)
b , Lm tương tự như Câu a
Chúc bn hok tốt
ĐKXĐ : 101x \(\ge\)0 nên x \(\ge\)0
khi đó : \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)
\(\Leftrightarrow\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+...+\left(x+\frac{100}{101}\right)=101x\)
\(\Leftrightarrow100x+\frac{5050}{101}=101x\Leftrightarrow x=50\)
*ĐK : 101x\(\ge\) 0 => x\(\ge\)0
=> \(x+\frac{1}{101}\ge\frac{1}{101}>0\Rightarrow\left|x+\frac{1}{101}\right|=x+\frac{1}{101}\)
\(x+\frac{2}{101}\ge\frac{2}{101}>0\Rightarrow\left|x+\frac{2}{101}\right|=x+\frac{2}{101}\)
...
\(x+\frac{100}{101}\ge\frac{100}{101}>0\Rightarrow\left|x+\frac{100}{101}\right|=x+\frac{100}{101}\)
Ta có :
\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)
<=> \(100x+\left(\frac{1+2+...+100}{101}\right)=101x\)
<=> \(100x+\frac{5050}{101}=101x\)
<=> \(100x-101x=\frac{-5050}{101}\)
<=> -x = -50
=> x = 50 ( t/m x >/ 0)
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!
Chết,nhầm ở câu cuối cùng của câu a) . Mình là ẩu thật :v. Sửa lại nhé:
\(\Leftrightarrow100x+\frac{5050}{101}=101x\Leftrightarrow100x+50=101x\Leftrightarrow201x=50\Leftrightarrow x=\frac{50}{201}\)