Tìm x biết: \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+\left|x+\frac{1}{20}\right|+...+\left|x+\frac{1}{110}\right|=11x\) = 11x
Help tìm x: \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+\left|x+\frac{1}{20}\right|+...+\left|x+\frac{1}{110}\right|=11x\)
Ta có:
\(\left\{{}\begin{matrix}\left|x+\frac{1}{2}\right|\ge0\\\left|x+\frac{1}{6}\right|\ge0\\...\\\left|x+\frac{1}{110}\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{110}\right|\ge0\)
\(\Rightarrow11x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{110}\right|\)
=\(x+\frac{1}{2}+x+\frac{1}{6}+...+x+\frac{1}{110}\)
\(=10x+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)
Đặt \(A=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\)
\(\Rightarrow A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{11-10}{10.11}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)
\(\Rightarrow A=1-\frac{1}{11}=\frac{10}{11}\)
\(\Rightarrow10x+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)=10x+A=10x+\frac{10}{11}=11x\)
\(\Rightarrow\frac{10}{11}=11x-10x\)
\(\Rightarrow x=\frac{10}{11}\)
Tìm x : \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+\left|x+\frac{1}{20}\right|+...+\left|x+\frac{1}{110}\right|=11x\)
Vì \(\left|x+\frac{1}{2}\right|\ge0;\left|x+\frac{1}{6}\right|\ge0;\left|x+\frac{1}{12}\right|\ge0;...;\left|x+\frac{1}{110}\right|\ge0\)
\(\Rightarrow11x\ge0\)
\(\Rightarrow x\ge0\)
Với \(x\ge0\) ta có:
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+\left(x+\frac{1}{12}\right)+...+\left(x+\frac{1}{110}\right)=11x\)
\(\Rightarrow\left(x+\frac{1}{1.2}\right)+\left(x+\frac{1}{2.3}\right)+\left(x+\frac{1}{3.4}\right)+...+\left(x+\frac{1}{10.11}\right)=11x\)
\(\Rightarrow\left(x+x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=11x\)
10 số x
\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=11x\)
\(\Rightarrow1-\frac{1}{11}=11x-10x\)
\(\Rightarrow x=\frac{10}{11}\)
Vậy \(x=\frac{10}{11}\)
Tìm x biết: \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+...+\left|x+\frac{1}{110}\right|=11x\)
Ta có:
\(\left|x+\frac{1}{2}\right|\ge0;\left|x+\frac{1}{6}\right|\ge0;....;\left|x+\frac{1}{110}\right|\ge0\)
\(\Rightarrow VT\ge0\Rightarrow VP\ge0\)
\(\Rightarrow x+\frac{1}{2}+x+\frac{1}{6}+...+\frac{1}{110}=11x\)
\(\Rightarrow\left(x+...+x\right)+\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{110}\right)=11x\)
\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10\cdot11}\right)=11x\)
\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=11x\)
\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=11x\)
\(\Rightarrow10x+\frac{10}{11}=11x\)
\(\Rightarrow x=\frac{10}{11}\)
mk viết vội nên nhầm dòng thứ 4 từ trên xuống, bn sửa 1/3 thành 1/6 nhé
kq vẫn đúng đấy
Tim x biết \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+...+\left|x+\frac{1}{110}\right|=11x\)
Ta có: \(\left|x+\frac{1}{2}\right|\ge0;\left|x+\frac{1}{6}\right|\ge0;\left|x+\frac{1}{12}\right|\ge0;...;\left|x+\frac{1}{110}\right|\ge0\)
=> VT \(\ge\)0
=>VP \(\ge\)0 => 11x \(\ge\)0 => x \(\ge\)0.
=> \(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{6}\right|=x+\frac{1}{6};\left|x+\frac{1}{12}\right|=x+\frac{1}{12};...;\left|x+\frac{1}{110}\right|=x+\frac{1}{110}\)
Phương trình <=> \(x+\frac{1}{2}+x+\frac{1}{6}+x+\frac{1}{12}+...+x+\frac{1}{110}=11x\)
<=> \(\left(x+x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=11x\)
<=> \(10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=11x\)
<=> \(1-\frac{1}{11}=11x-10x\)
<=> \(\frac{10}{11}=x\)
<=> \(x=\frac{10}{11}\left(tm\right)\)
Bởi vì
\(\frac{1}{2}=\frac{1}{1.2};\frac{1}{6}=\frac{1}{2.3};...;\frac{1}{110}=\frac{1}{10.11}\)
nên từ \(\frac{1}{2}\)đến \(\frac{1}{110}\)chỉ có 10 số
nên chỉ có 10 x
Ta có : \(|x+\frac{1}{2}|\ge0;|x+\frac{1}{6}|\ge0;....;|x+\frac{1}{110}|\ge0\)
\(\Rightarrow|x+\frac{1}{2}|+|x+\frac{1}{6}|+...+|x+\frac{1}{110}|\ge0\)
\(\Rightarrow11x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow x+\frac{1}{2}>0;x+\frac{1}{6}>0;....;x+\frac{1}{110}>0\)
\(\Rightarrow|x+\frac{1}{2}|=x+\frac{1}{2};|x+\frac{1}{6}|=x+\frac{1}{6};....;|x+\frac{1}{110}|=x+\frac{1}{110}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{110}\right)=11x\)
\(\Rightarrow10x+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)=11x\)
\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=11x\)
\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{10}-\frac{1}{11}\right)=11x\)
\(\Rightarrow10x+\frac{1}{10}=11x\)
\(\Rightarrow x=\frac{10}{11}\)
Vậy.....
Tim x biết
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+...+\left|x+\frac{1}{110}=11x\right|\)
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+...+\left|x+\frac{1}{110}\right|=11x\)
Với mọi x ta có:
+) \(\left\{{}\begin{matrix}\left|x+\frac{1}{2}\right|\ge0\\\left|x+\frac{1}{6}\right|\ge0\\.........\\\left|x+\frac{1}{110}\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{110}\right|\ge0\) \(\forall x.\)
Mà \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{110}\right|=11x\)
\(\Rightarrow11x\ge0\)
\(\Rightarrow x\ge0.\)
Với \(x\ge0\) thì:
\(\left\{{}\begin{matrix}\left|x+\frac{1}{2}\right|=x+\frac{1}{2}\\\left|x+\frac{1}{6}\right|=x+\frac{1}{6}\\..........\\\left|x+\frac{1}{110}\right|=x+\frac{1}{110}\end{matrix}\right.\)
\(\Rightarrow x+\frac{1}{2}+x+\frac{1}{6}+...+x+\frac{1}{110}=11x\)
\(\Rightarrow11x+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)=11x\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)=11x-11x\)
\(\Rightarrow\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}=0x\) (vô lí).
\(\Rightarrow x\in\varnothing.\)
Vậy không tồn tại giá trị của x thỏa mãn yêu cầu đề bài.
Chúc bạn học tốt!
mình sửa đề chút nhé + \(\left|x+\frac{1}{110}\right|=11x\)
tim x biet:
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+...+\left|x+\frac{1}{110}\right|=11x\)11x
Trinh bay cach lam ho minh nhe
Vế trái lớn hơn hoặc bằng 0 nên 11x lớn hơn hoặc bằng 0.
\(\Rightarrow x\ge0\)
Do vậy chỉ cần bỏ dấu giá trị tuyệt đối là tính được.
Kết quả cuối cùng được \(x=\frac{10}{11}\)
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{8}\right|+......+\left|x+\frac{1}{110}\right|=11x\)
GIẢI PHƯƠNG TRÌNH
Ta có vế trái của pt luôn \(\ge0\)
Do đó : \(11x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\hept{\begin{cases}\left|x+\frac{1}{2}\right|=x+\frac{1}{2}\\...\\\left|x+\frac{1}{110}\right|=x+\frac{1}{110}\end{cases}}\)
Khi đó pt trở thành :
\(x+\frac{1}{2}+x+\frac{1}{6}+...+x+\frac{1}{110}=11x\)
\(\Leftrightarrow10x+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}=11x\)
\(\Leftrightarrow x=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)
\(\Leftrightarrow x=1-\frac{1}{11}=\frac{10}{11}\) ( thỏa mãn )
Vậy : pt đã cho có nghiệm \(S=\left\{\frac{10}{11}\right\}\)
Dễ thấy \(VT>0\forall x\)
\(\Rightarrow11x>0\Rightarrow x>0\)
Phương trình trở thành \(10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=11x\)
\(\Rightarrow x=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)
\(\Rightarrow x=1-\frac{1}{11}=\frac{10}{11}\)
Vậy \(x=\frac{10}{11}\)
Tìm x biết
a)\(\left|2x-3\right|-x=\)\(\left|2-x\right|\)
b)\(\left|x+3\right|+\left|x+1\right|=\)\(3x\)
c)\(2005=\left|x-4\right|+\left|x-10\right|+\left|x+101\right|+\left|x+990\right|+\left|x+1000\right|\)
d)\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{20}\right|+..+\left|x+\frac{1}{110}\right|=11x\)
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
Giaỉ các phương trình sau:
a) \(\left(x^2+11x+12\right)\left(x^2+9x+20\right)\left(x^2+13x+42\right)=36\left(x^2+11x+30\right)\left(x^2+11x+31\right)\)
b) \(20\left(\frac{x-2}{x+1}\right)^2-5\left(\frac{x+2}{x-1}\right)^2+48\cdot\frac{x^2-4}{x^2-1}=0\)