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\(\frac{\frac{41}{10}}{\frac{9}{4}}=\frac{41}{10}\div\frac{9}{4}=\frac{41}{10}\times\frac{4}{9}=\frac{82}{45}\Rightarrow\frac{82}{45}=\frac{x}{7,3}\)
Đến đấy thôi
a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\)
\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)
Nên x + 1 = 0
=> x = -1
a) \(\frac{1}{4}+\frac{1}{3}:2x=-5\)
\(\frac{1}{3}:2x=\frac{-21}{4}\)
\(2x=\frac{-4}{63}\)
\(x=\frac{2}{63}\)
b) \(\left(3x-\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-1}{2}\end{cases}}\)
Vậy.........
a) \(\frac{1}{2}-|\frac{5}{4}-2x|=\frac{1}{3}\Leftrightarrow|\frac{5}{4}-2x|=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{4}-2x=\frac{1}{6}\\\frac{5}{4}-2x=-\frac{1}{6}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{5}{4}-\frac{1}{6}=\frac{13}{12}\\2x=\frac{5}{4}+\frac{1}{6}=\frac{17}{12}\end{cases}}}\)
Tự làm nốt và kết luận
b) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)=0\)
Vì \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)\ne0\forall x\Rightarrow x+1=0\Leftrightarrow x=-1\)
Vậy ....
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
\(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
\(\Leftrightarrow\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)
\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)
\(\Leftrightarrow x+1=0\)( \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\))
\(\Leftrightarrow x=-1\)
Vậy x=-1
mỗi phân số + 1 thì sẽ có tử chung là x + 1
chuyển vế có \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)\)) =0
mà tổng các phân số kia khác 0 nên x+1 bằng 0
=> x=-1
/X/ = \(\frac{10}{3}\)
suy ra: X = \(\frac{10}{3}\)hay X = \(\frac{-10}{3}\)
/X - \(\frac{5}{6}\)/ = \(\frac{1}{2}\)
suy ra : X - \(\frac{5}{6}\)= \(\frac{1}{2}\)hay X - \(\frac{5}{6}\)= \(\frac{-1}{2}\)
X = \(\frac{1}{2}+\frac{5}{6}\) hay X = \(\frac{-1}{2}+\frac{5}{6}\)
X = \(\frac{4}{3}\) hay X = \(\frac{1}{3}\)
/ X + \(\frac{4}{9}\)/ - \(\frac{1}{2}\)= \(\frac{3}{2}\)
/ X + \(\frac{4}{9}\)/ = \(\frac{3}{2}+\frac{1}{2}\)
/ X + \(\frac{4}{9}\)/ = \(2\)
suy ra : X + \(\frac{4}{9}\)= \(2\)hay X + \(\frac{4}{9}\)= \(-2\)
X = \(2-\frac{4}{9}\)hay X = \(\left(-2\right)-\frac{4}{9}\)
X = \(\frac{14}{9}\) hay X = \(\frac{-22}{9}\)
a. |x| = \(10/3\)
=> x = 10/3 hoặc x = -10/3
b/ |x - 5/6| = 1/2
*TH1: x - 5/6 = 1/2
=> x = 1/2 + 5/6
=> x = 4/3
*TH2: -x + 5/6 = 1/2
=> -x = 1/2 - 5/6
=> -x = -1/3
=> x = 1/3
Vậy...
Phần c mk không chắc nên để lại ạ
\(\frac{\frac{41}{10}}{\frac{9}{4}}=\frac{x}{7,3}\)
\(\Rightarrow x.\frac{9}{4}=\frac{41}{10}.7,3\)
\(\Rightarrow x.\frac{9}{4}=\frac{2993}{100}\)
\(\Rightarrow x=\frac{2993}{100}:\frac{9}{4}\)
\(\Rightarrow x=\frac{2993}{225}\)
Vậy \(x=\frac{2993}{225}.\)
Chúc bạn học tốt!
\(\frac{\frac{41}{10}}{\frac{9}{4}}=\frac{x}{7,3}\\ \Rightarrow\frac{9}{4}x=\frac{41}{10}\cdot7,3\left(\text{tính chất của tỉ lệ thức}\right)\\ x=\frac{\frac{41}{10}\cdot7,3}{\frac{9}{4}}=\frac{2993}{225}\)
Vậy \(x=\frac{2993}{225}\)