a) (x - 45) . 27 = 0

=> x - 45 = 0

x = 45

Vậy x = 45.

b) 23 (42 - x) = 23

42 - x = 23 : 23

42 - x = 1

x = 42 - 1

x = 41

Vậy x = 41.

a) (x - 45) . 27 = 0

x - 45 = 0 : 27

x - 45 = 0

x = 0 + 45

x = 45

b) 23 . (42 - x) = 23

42 - x = 23 : 23

42 - x = 1

- x = 1 + (-42)

-x = -41

x = 41

a) ( x - 45 ) . 27 = 0

Suy ra: x - 45 = 0

x = 0 + 45

x = 45

b) 23 . ( 42 - x) = 23

42 - x = 23 : 23

42 - x = 1

x = 42 - 1

x = 41

\(x\div23+45=8911\div63\)

\(x\div23=96,4444\)

\(x=96,4444\cdot23\)

\(x=2218,2222\)

Câu a: 2\(^3\).\(x\) : |\(\frac{13}{23}\) - 5\(\frac{13}{23}\)| = - 2\(\frac13\)

\(8x\) : |-5| = - \(\frac73\)

8\(x\) : 5 = -\(\frac73\)

8\(x\) = - \(\frac73\times5\)

8\(x\) = - \(\frac{35}{3}\)

\(x\) = - \(\frac{35}{3}\) : 8

\(x\) = - \(\frac{35}{24}\)

Vậy \(x=-\frac{35}{24}\)

Câu b:

||\(x-3\)| - 5| = 7

|\(x-3\)| - 5 = 7 hoặc |\(x-3\)| - 5 = - 7

TH1: |\(x-3\)| - 5 = 7

|\(x-3\)| = 7+ 5

|\(x-3\)| = 12

\(\left[\begin{array}{l}x-3=-12\\ x-3=12\end{array}\right.\)

\(\left[\begin{array}{l}x=-12+3\\ x=12+3\end{array}\right.\)

\(\left[\begin{array}{l}x=-9\\ x=15\end{array}\right.\)

TH2: |\(x-3\)| - 5 = - 7

|\(x-3\)| = - 7 + 5

|\(x\) - 3| = - 2(loại) Vì trị tuyêt đối của một số luôn không âm.

Vậy \(x\in\) {-9; 15}

\(\left(\frac{5}{12}-\frac{5}{7}-\frac{22}{45}+\frac{7}{12}-\frac{23}{45}\right).\left|x\right|-9=-2\)

\(\left(\frac{12}{12}-\frac{5}{7}-\frac{45}{45}\right).\left|x\right|=-2+9\)

\(\left(1-\frac{5}{7}-1\right).\left|x\right|=7\)

\(\frac{-5}{7}.\left|x\right|=7\)

\(\left|x\right|=7:\left(\frac{-5}{7}\right)\)

\(\left|x\right|=\frac{-49}{5}\)

\(\Rightarrow x\in\varnothing\) vì trị tuyệt đối của 1 số luôn dương 

bằng \(\dfrac{-3}{11}\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

\(\Leftrightarrow\left[\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\right]x=\frac{23}{45}\)

\(\Leftrightarrow\left[\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)\right]x=\frac{23}{45}\)

\(\Leftrightarrow\left(\frac{1}{2}.\frac{44}{90}\right)x=\frac{23}{45}\)

\(\Leftrightarrow\frac{11}{45}x=\frac{23}{45}\Rightarrow x=\frac{23}{45}:\frac{11}{45}=\frac{23}{11}\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{48}{45}\)

\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{48}{45}\)

\(\Rightarrow\frac{22}{45}x=\frac{48}{45}\)

\(\Rightarrow x=\frac{24}{11}\)

Vậy...

Có \(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\right)+x=\frac{23}{45}\)

Cho \(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\)

Ta có công thức sau: \(\frac{1}{n\cdot\left(n+1\right)}+\frac{1}{\left(n+1\right)\cdot\left(n+2\right)}=\frac{2}{n\cdot\left(n+1\right)\left(n+1\right)}\)

\(\Rightarrow2A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{8\cdot9\cdot10}\\ =\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\\ =\frac{1}{1\cdot2}-\frac{1}{9\cdot10}=\frac{22}{45}\)

\(\Rightarrow A=\frac{22}{45}:2=\frac{11}{45}\)

Thay vào phép tính trên ta được:

\(\frac{11}{45}\cdot x=\frac{23}{45}\\ x=\frac{23}{45}:\frac{11}{45}\\ x=\frac{23}{11}\)

Vậy \(x=\frac{23}{11}\)

\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\right)x=\frac{23}{45}\)

=> \(\left[\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{8\cdot9\cdot10}\right)\right]x=\frac{23}{45}\)

=>\(\left[\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\right)\right]x=\frac{23}{45}\)

=> \(\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9\cdot10}\right)\right]x=\frac{23}{45}\)

=> \(\left[\frac{1}{2}\cdot\frac{22}{45}\right]x=\frac{23}{45}\)

=> \(\frac{11}{45}x=\frac{23}{45}\)

=> \(x=\frac{23}{45}:\frac{11}{45}=\frac{23}{45}\cdot\frac{45}{11}=\frac{23}{11}\)

Vậy x = 23/11

Ez :))

Theo đầu bài ta có:
\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\right)\cdot x=\frac{23}{45}\)
\(\Rightarrow\frac{\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}}{2}\cdot x=\frac{23}{45}\)
\(\Rightarrow\left(\frac{1}{1\cdot2}-\frac{1}{9\cdot10}\right)\cdot x=\frac{46}{45}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{1}{90}\right)\cdot x=\frac{46}{45}\)
\(\Rightarrow\frac{22}{45}\cdot x=\frac{46}{45}\)
\(\Rightarrow x=\frac{23}{11}\)

                  Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\)

                     \(A=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right)\)

                  \(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)

                 \(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)\)

               \(A=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)

              \(\Rightarrow\frac{11}{45}.x=\frac{23}{45}\)

              \(\Rightarrow x=\frac{23}{45}:\frac{11}{45}=\frac{23}{11}\)

            Ủng hộ mk nha !!! ^_^