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!

\(C=\frac{5}{7}\cdot\frac{5}{11}+\frac{5}{7}\cdot\frac{2}{11}-\frac{5}{7}\cdot\frac{14}{11}\)

\(=\frac{5}{7}\cdot\left(\frac{5}{11}+\frac{2}{11}-\frac{14}{11}\right)\)

\(=\frac{5}{7}\cdot-\frac{7}{11}\)

\(=-\frac{5}{11}\)

 \(\frac{5}{11}.\frac{3}{7}+\frac{5}{7}.\frac{4}{11}-\frac{3}{11}\)

\(=\frac{3}{11}.\frac{5}{3}.\frac{3}{7}+\frac{5}{7}.\frac{4}{3}.\frac{3}{11}-\frac{3}{11}\)

\(=\frac{3}{11}.\left(\frac{5}{7}+\frac{20}{21}-1\right)\)

\(=\frac{3}{11}.\left(\frac{15+20}{21}-1\right)\)

\(=\frac{3}{11}.\left(\frac{35}{21}-1\right)\)

\(=\frac{3}{11}.\frac{14}{21}\)

\(=\frac{14}{77}\)

5/11 x 3/7 + 5/7 x 4/11 - 3/11

= 5/11 x 3/7 + 5/11 x 4/7 - 3/11

= 5/11 x ( 3/7 + 4/7 ) - 3/11

= 5/11 x 1 - 3/11

= 5/11 - 3/11

= 2/11

Tk nha !!

ta có

\(2.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(5.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}+\dfrac{1}{11}\right)\)

_______________________ X ________________________

\(4.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(9.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}\dfrac{1}{11}\right)\)

= \(\dfrac{2}{4}X\dfrac{5}{9}\)= \(\dfrac{10}{36}\)= \(\dfrac{5}{18}\)

B=\(6\frac{4}{9}-4\frac{4}{9}+3\frac{7}{11}\)

B=\(2+3\frac{7}{11}\)

B=\(5\frac{7}{11}\)

B = \(5\frac{7}{11}=\frac{62}{11}\)

C = 1

D = \(\frac{5}{2}=2\frac{1}{2}\)

a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)

\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)

=>x=10

b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)

\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)

\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)

hay \(x\in\left\{0;1;2\right\}\)

c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)

\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)

\(\Leftrightarrow6-x=0\)

hay x=6

Câu a:

\(\frac{x+5}{x+9}\) = \(\frac{x+4}{x+3}\)

(\(x\) + 5)(\(x+3\)) = (\(x+4\))(\(x+9\))

\(x^2\) + 3\(x\) + 5\(x\) + 15 = \(x^2\) + 9\(x\) + 4\(x\) + 36

\(x^2+3x+5x\) - \(x^2-9x\) - 4\(x\) = 36 - 15

(\(x^2-x^2\)) + (\(3x+5x\) - 9\(x-4x\)) = 21

0 + (8\(x\) - 9\(x\) - 4\(x\)) = 21

-\(x-4x\) = 21

-5\(x\) = 21

\(x\) = 21 : (-5)

\(x\) = - \(\frac{21}{5}\)

Vậy \(x=-\frac{21}{5}\)

\(a;\dfrac{3}{2}x-\dfrac{2}{3}=\dfrac{2}{3}:\dfrac{3}{2}\\ \dfrac{3}{2}x-\dfrac{2}{3}=\dfrac{4}{9}\\ \dfrac{3}{2}x=\dfrac{4}{9}+\dfrac{2}{3}=\dfrac{10}{9}\\ x=\dfrac{10}{9}:\dfrac{3}{2}=\dfrac{20}{27}\\ b;\left(\dfrac{9}{11}-x\right):\left(-\dfrac{10}{11}\right)=1-\dfrac{4}{5}\\ \left(\dfrac{9}{11}-x\right):\left(-\dfrac{10}{11}\right)=\dfrac{1}{5}\\ \dfrac{9}{11}-x=\dfrac{1}{5}\cdot\left(-\dfrac{10}{11}\right)\\ \dfrac{9}{11}-x=-\dfrac{2}{11}\\ x=\dfrac{9}{11}-\left(-\dfrac{2}{11}\right)=\dfrac{9}{11}+\dfrac{2}{11}\\ x=1\\ c;-\dfrac{11}{12}x+\dfrac{3}{4}=-\dfrac{1}{6}\\ -\dfrac{11}{12}x=-\dfrac{1}{6}-\dfrac{3}{4}\\ -\dfrac{11}{12}x=-\dfrac{11}{12}\\ x=\left(-\dfrac{11}{12}\right):\left(-\dfrac{11}{12}\right)=1\)

\(d;-\dfrac{5}{4}-\left(1\dfrac{1}{2}+x\right)=4,5\\ \Leftrightarrow-\dfrac{5}{4}-\left(\dfrac{3}{2}+x\right)=4,5\\\dfrac{3}{2}+x=-\dfrac{5}{4}-4,5\\ \dfrac{3}{2}+x=-\dfrac{23}{4}\\ x=-\dfrac{23}{4}-\dfrac{3}{2}\\ x=-\dfrac{29}{4}\\ đ;\left(\dfrac{3}{4}-x:\dfrac{2}{15}\right)\cdot\dfrac{1}{5}=-2,6\\ \dfrac{3}{4}-x:\dfrac{2}{15}=-2,6:\dfrac{1}{5}\\ \dfrac{3}{4}-x:\dfrac{2}{15}=-13\\ x:\dfrac{2}{15}=\dfrac{3}{4}-\left(-13\right)\\ x:\dfrac{2}{15}=\dfrac{55}{4}\\ x=\dfrac{55}{4}\cdot\dfrac{2}{15}=\dfrac{11}{6}\\ e;3-\left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=\dfrac{2}{3}\\ \left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=3-\dfrac{2}{3}\\ \left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=\dfrac{7}{3}\\ \dfrac{1}{6}-x=\dfrac{7}{3}:\dfrac{2}{3}=\dfrac{7}{2}\\ x=\dfrac{1}{6}-\dfrac{7}{2}=-\dfrac{10}{3}\)

\(f;\left(1-2x\right)\cdot\dfrac{4}{5}=\left(-2\right)^3\\ \left(1-2x\right)\cdot\dfrac{4}{5}=-8\\ 1-2x=-8:\dfrac{4}{5}=-10\\ 2x=1-\left(-10\right)=11\\ x=\dfrac{11}{2}\\ g;\dfrac{1}{6}-\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{8}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{6}-\dfrac{1}{8}=\dfrac{1}{24}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{1}{24}\Rightarrow x=\dfrac{3}{4}\\\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{1}{24}\Rightarrow x=\dfrac{7}{12}\end{matrix}\right.\)

\(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)

\(\frac{3}{2}x-\frac{2}{3}=\frac{4}{9}\)

\(\frac{3}{2}x=\frac{4}{9}+\frac{2}{3}\)

\(\frac{3}{2}x=\frac{10}{9}\)

\(x=\frac{10}{9}:\frac{3}{2}\)

\(x=\frac{20}{27}\)

Vậy x=\(\frac{20}{27}\)

\(\left(\frac{9}{11}-x\right):\frac{-10}{11}=1-\frac{4}{5}\)

\(\left(\frac{9}{11}-x\right):\frac{-10}{11}=\frac{1}{5}\)

\(\frac{9}{11}-x=\frac{1}{5}\cdot\frac{-10}{11}\)

\(\frac{9}{11}-x=\frac{-2}{11}\)

\(x=\frac{9}{11}-\frac{-2}{11}\)

\(x=1\)

Vậy x=1

\(\frac{-11}{12}\cdot x+\frac{3}{4}=\frac{-1}{6}\)

\(\frac{-11}{12}\cdot x=\frac{-1}{6}-\frac{3}{4}\)

\(\frac{-11}{12}\cdot x=\frac{21}{12}\)

\(x=\frac{-21}{11}\)

Vậy x=\(\frac{-21}{11}\)

\(\frac{-5}{4}-\left(1\frac{1}{2}+x\right)=4,5\)

\(\frac{3}{2}+x=\frac{-5}{4}-\frac{9}{2}\)

\(\frac{3}{2}+x=\frac{23}{4}\)

\(x=\frac{17}{4}\)

Vậy x=\(\frac{17}{4}\)

\(\left(\frac{3}{4}-x:\frac{2}{15}\right)\cdot\frac{1}{5}=-2,6\)

\(\frac{3}{4}-x:\frac{2}{15}=\frac{-13}{5}:\frac{1}{5}\)

\(\frac{3}{4}-x:\frac{2}{15}=-13\)

\(x:\frac{2}{15}=\frac{3}{4}-\left(-13\right)\)

\(x:\frac{2}{15}=\frac{45}{4}\)

\(x=\frac{3}{2}\)

Vậy x=\(\frac{3}{2}\)

\(3-\left(\frac{1}{6}-x\right)\cdot\frac{2}{3}=\frac{2}{3}\)

\(3-\left(\frac{1}{6}-x\right)=\frac{2}{3}:\frac{2}{3}\)

\(3-\left(\frac{1}{6}-x\right)=1\)

\(\frac{1}{6}-x=2\)

\(x=\frac{1}{6}-2\)

\(x=\frac{-11}{6}\)

Vậy x=\(\frac{-11}{6}\)

\(\left(1-2x\right)\cdot\frac{4}{5}=\left(-2\right)^3\)

\(1-2x=\frac{-1}{10}\)

\(2x=1-\frac{-1}{10}\)

\(2x=\frac{11}{10}\)

\(x=\frac{11}{20}\)

Vậy x=\(\frac{11}{20}\)

\(\frac{1}{6}-\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{1}{8}\)

\(\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{7}{12}\)

\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{7}{12}\)                                                         \(\frac{1}{2}x-\frac{1}{3}=\frac{-7}{12}\)

\(\frac{1}{2}x=\frac{11}{12}\)                                                                        \(\frac{1}{2}x=\frac{-1}{4}\)

\(x=\frac{11}{6}\)                                                                              \(x=\frac{-1}{2}\)

Vậy \(x\in\left\{\frac{11}{6};\frac{-1}{2}\right\}\)

\(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)

\(\frac{3}{2}x=\frac{4}{9}+\frac{6}{9}\)

\(\frac{3}{2}x=\frac{10}{9}\)

\(x=\frac{10}{9}:\frac{3}{2}\)

\(x=\frac{20}{27}\)

tk mình đi mình làm nốt cho hjhj ^^

\(a.-8:\left(4\dfrac{1}{5}x+\dfrac{3}{10}\right)=4\dfrac{4}{9}\)

\(4\dfrac{1}{5}x+\dfrac{3}{10}=\left(-8\right):4\dfrac{4}{9}\)

\(4\dfrac{1}{5}x+\dfrac{3}{10}=\dfrac{-9}{5}\)

\(4\dfrac{1}{5}x=\dfrac{-9}{5}-\dfrac{3}{10}\)

\(4\dfrac{1}{5}x=\dfrac{-21}{10}\)

\(x=\dfrac{-21}{10}:\dfrac{21}{5}\)

\(x=\dfrac{-1}{2}\)

Vay \(x=\dfrac{-1}{2}\).

\(b.4\dfrac{2}{3}-\left(\dfrac{3}{5}:x\right)=-20\%\)

\(\dfrac{14}{3}-\left(\dfrac{3}{5}:x\right)=\dfrac{-1}{5}\)

\(\dfrac{3}{5}:x=\dfrac{14}{3}-\dfrac{-1}{5}\)

\(\dfrac{3}{5}:x=\dfrac{73}{15}\)

\(x=\dfrac{3}{5}:\dfrac{73}{15}\)

\(x=\dfrac{9}{73}\)

Vay \(x=\dfrac{9}{73}\).

Câu c; d; e tương tự nhé.