Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)<=>3(x+3)=1(-15)
=>3x+9=-15
=>3x=-15-9
=>3x=-24
=>x=-8
b)<=>4,5-2x=\(\frac{11}{14}\div\frac{-11}{7}\)
\(\Rightarrow4\frac{1}{2}-2x=-\frac{1}{2}\)
\(\Leftrightarrow2x=\frac{9}{2}-\frac{-1}{2}\)
\(\Rightarrow2x=5\)
\(\Rightarrow x=\frac{5}{2}\)
A/\(\left(2,8x-32\right):\frac{2}{3}=-90\)
\(\left(\frac{28}{10}x-32\right)=\frac{-90}{1}.\frac{2}{3}\)
\(\left(\frac{14}{5}x-32\right)=\frac{-30}{1}.\frac{2}{1}\)
\(\left(\frac{14}{5}x-32\right)=-60\)
\(\frac{14}{5}x=-60+32\)
\(\frac{14}{5}x=-28\)
\(x=\frac{-28}{1}:\frac{14}{5}\)
\(x=\frac{-28}{1}.\frac{5}{14}\)
\(x=\frac{-2}{1}.\frac{5}{1}=-10\)
B/\(\left(4,5-2x\right).1\frac{4}{7}=\frac{11}{14}\)
\(\left(\frac{45}{10}-2x\right).\frac{11}{7}=\frac{11}{14}\)
\(\left(\frac{9}{2}-2x\right)=\frac{11}{14}:\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)=\frac{11}{14}.\frac{7}{11}\)
\(\left(\frac{9}{2}-2x\right)=\frac{1}{2}.\frac{1}{1}=\frac{1}{2}\)
\(2x=\frac{9}{2}-\frac{1}{2}\)
\(2x=\frac{8}{2}\)
\(x=\frac{8}{2}:\frac{2}{1}=\frac{8}{2}.\frac{1}{2}\)
\(x=\frac{4}{2}.\frac{1}{1}=\frac{4}{2}=2\)
Giải:
a) \(\left(4,5-2x\right).\left(-1\dfrac{4}{7}\right)=\dfrac{11}{14}\)
\(\Leftrightarrow\left(4,5-2x\right).\left(-\dfrac{3}{7}\right)=\dfrac{11}{14}\)
\(\Leftrightarrow4,5-2x=\dfrac{11}{14}:\left(-\dfrac{3}{7}\right)=-\dfrac{11}{6}\)
\(\Leftrightarrow2x=4,5-\left(-\dfrac{11}{6}\right)\)
\(\Leftrightarrow2x=\dfrac{19}{3}\)
\(\Leftrightarrow x=\dfrac{19}{3}:2=\dfrac{19}{6}\)
Vậy ...
b) \(\dfrac{4}{9}x=\dfrac{9}{8}-0,125\)
\(\Leftrightarrow\dfrac{4}{9}x=\dfrac{9}{8}-\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{4}{9}x=1\)
\(\Leftrightarrow x=1:\dfrac{4}{9}=\dfrac{9}{4}\)
Vậy ...
Các câu còn lại làm tương tự.
\(a)x+30\%x=-1,31\)
\(\Leftrightarrow x+\frac{3x}{10}=-1,31\)
\(\Leftrightarrow10x+3x=-13,1\)
\(\Leftrightarrow13x=-13,1\Leftrightarrow x=-\frac{131}{130}\)
\(b)\left(x-\frac{1}{2}\right):\frac{1}{3}+\frac{5}{7}=9\frac{5}{7}\)
\(\Leftrightarrow\frac{2x-1}{2}.3+\frac{5}{7}=\frac{68}{7}\)
\(\Leftrightarrow\frac{6x-3}{2}=\frac{63}{7}\)
\(\Leftrightarrow\frac{6x-3}{2}=9\)
\(\Leftrightarrow6x-3=18\)
\(\Leftrightarrow x=\frac{7}{2}\)
a) \(\left(4,5-2x\right)\cdot1\frac{4}{7}=\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)\cdot\frac{11}{7}=\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)=\frac{11}{7}\div\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)=1\)
\(2x=\frac{9}{2}-1\)
\(x=\frac{7}{2}\div2\)
\(x=\frac{7}{4}\)
b) \(|\frac{3}{4}\cdot x-\frac{1}{2}|-1=\frac{1}{4}\)
\(|\frac{3}{4}\cdot x-\frac{1}{2}|=\frac{1}{4}+1\)
\(|\frac{3}{4}\cdot x|=\frac{5}{4}+\frac{1}{2}\)
\(x=\frac{7}{4}\div\frac{3}{4}\)
\(x=\frac{7}{3}\)
c) \(\frac{1}{4}-|3-x|=-\frac{3}{4}\)
\(|3-x|=\frac{1}{4}-\left(-\frac{3}{4}\right)\)
\(|3-x|=1\)
\(x=3-1\)
\(\Rightarrow x=2\)
d) \(4\cdot\left(x-\frac{6}{7}\right)-\frac{3}{5}=1,4\)
\(4\cdot\left(x-\frac{6}{7}\right)-\frac{3}{5}=\frac{7}{5}\)
\(4\cdot\left(x-\frac{6}{7}\right)=\frac{7}{5}+\frac{3}{5}\)
\(4\cdot\left(x-\frac{6}{7}\right)=2\)
\(\left(x-\frac{6}{7}\right)=2\div4\)
\(x=\frac{1}{2}+\frac{6}{7}\)
\(x=\frac{19}{14}\)
\(\)
1, Tính tổng:
\(C=\frac{5}{7}\cdot\frac{5}{11}+\frac{5}{7}\cdot\frac{2}{11}-\frac{5}{7}\cdot\frac{14}{11}\)
\(=\frac{5}{7}\cdot\left(\frac{5}{11}+\frac{2}{11}-\frac{14}{11}\right)=\frac{5}{7}\cdot\frac{-7}{11}=\frac{-5}{11}\)
2, Tìm x:
\(x+\frac{5}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}=\frac{-37}{45}\)
\(\Rightarrow x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)
\(\Rightarrow x+\frac{1}{5}-\frac{1}{45}=\frac{-37}{45}\Rightarrow x+\frac{9}{45}-\frac{1}{45}=\frac{-37}{45}\)
\(\Rightarrow x+\frac{8}{45}=\frac{-37}{45}\Rightarrow x=\frac{-37}{45}-\frac{8}{45}=\frac{-45}{45}=-1\)
- Các bài tìm x còn lại bạn cứ theo trình tự thực hiện phép tính mà làm nhé!
\(C=\frac{5}{7}\cdot\frac{5}{11}+\frac{5}{7}\cdot\frac{2}{11}-\frac{5}{7}\cdot\frac{14}{11}\)
\(=\frac{5}{7}\cdot\left(\frac{5}{11}+\frac{2}{11}-\frac{14}{11}\right)\)
\(=\frac{5}{7}\cdot-\frac{7}{11}\)
\(=-\frac{5}{11}\)
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)