Tìm x
6060 : [( 50x + 20 ) : x ] = 101
tìm x
6060*50x+20/x=101
\(6060.\frac{50x+20}{x}=101\)
<=> \(\frac{50x+20}{x}=\frac{1}{60}\)
<=> \(3000x+1200=x\)
<=> \(2999x+1200=0\)
<=> \(x=-\frac{1200}{2999}\)
6060÷[(50x+20)÷x]=101
\(6060:\left[\left(50x+20\right):x\right]=101\)
\(\left(50x+20\right):x=60\)
\(50x+20=60x\)
\(60x-50x=20\)
\(10x=20\)
\(x=2\)
x-y 6060 : [(50x+20):x]=101
Tìm x
6060 . \(\frac{50x+20}{x}\)= 101
6060/50x+20 phan x =101 hay tim con x do
6060\(\frac{50x+20}{x}\)=101
\(6060\frac{50x+20}{x}=101\)
\(\Rightarrow\frac{6060x+50x+20}{x}=101\)
\(\Rightarrow\frac{6110x+20}{x}=101\)
\(\Rightarrow6110x+20=101x\)
\(\Rightarrow6110x-101x=-20\)
\(\Rightarrow6009x=-20\)
\(\Rightarrow x=-\frac{20}{6009}\)
Tìm x , biết :
a)\(|x+\frac{1}{101}|+|x+\frac{2}{101}|+|x+\frac{3}{101}|+...+|x+\frac{100}{101}|=101x\)
b) \(|x+\frac{1}{1\times3}|+|x+\frac{1}{3\times5}|+|x+\frac{1}{9\times13}|+...+|x+\frac{1}{97\times99}=50x\)
\(|A\left(x\right)|+|B\left(x\right)|+|C\left(x\right)|=D\left(x\right)\))
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:
\(\Leftrightarrow100x+\frac{5050}{101}=101x\Leftrightarrow100x+50=101x\Leftrightarrow201x=50\Leftrightarrow x=\frac{50}{201}\)
\(\text{Tìm x , biết : }\)
\(a.\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(b.x-\left(\frac{50x}{100}+\frac{25x}{200}\right)=\frac{45}{4}\)
\(c.1+2+3+4+...+x=820\)
Ai làm nhanh mk cho 20 tích
\(a)\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x(x+3)}=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}\left[(\frac{1}{5}-\frac{1}{8})+(\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3})\right]=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}\left[\frac{1}{5}-\frac{1}{x+3}\right]=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{5}{1540}=\frac{1}{308}\)
\(\Rightarrow x+3=308\Rightarrow x=305\)
\(b)x-(\frac{50x}{100}-\frac{25x}{200})=\frac{45}{4}\)
\(\Rightarrow x-(\frac{100x}{200}-\frac{25x}{200})=\frac{45}{4}\)
\(\Rightarrow x-\frac{5x}{8}=\frac{45}{4}\)
\(\Rightarrow\frac{3x}{8}=\frac{45}{4}\)
\(\Rightarrow3x=\frac{45}{4}\cdot8\)
\(\Rightarrow3x=90\Rightarrow x=30\)
\(c)1+2+3+4+...+x=820\)
Ta có : \(1+2+3+4+...+x=\frac{(1+x)\cdot x}{2}\)
Do đó : \(\frac{(1+x)\cdot x}{2}=820\)
\(\Rightarrow(1+x)\cdot x=820\cdot2\)
\(\Rightarrow(1+x)\cdot x=1640\)
\(\Rightarrow(1+x)\cdot x=40\cdot41\)
Vì x và x + 1 là hai số tự nhiên liên tiếp nên => x = 40
Chúc bạn học tốt :3
100x + x - 50x + 100 - 50x = 100 . tìm x
( 100x - 50x - 50x ) + x + 100 = 100
=> x + 100 = 100
=> x = 0
Ta có: 100x + x - 50x + 100 - 50x = 100
=> (100x + x - 50 - 50x) + 100 = 100
=> x + 100 = 100
=> x = 100 - 100
=> x = 0
100x + x - 50x + 100 - 50x=100
<=>(100x-50x-50x)+x=100-100
<=>x=0
Vay x=0