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

bằng \(\dfrac{-3}{11}\)

a) \(\left|2x-1\right|=5\)

\(\Rightarrow\left[\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy \(x\left[\begin{matrix}=3\\=-2\end{matrix}\right.\)

b) \(\left(5^x-1\right)3-2=70\)

\(\Rightarrow5^x.3-3=72\)

\(\Rightarrow5^x.3=75\)

\(\Rightarrow5^x=5^2\)

\(\Rightarrow x=2\)

Vậy \(x=2.\)

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

\(\Rightarrow\left(x-1\frac{1}{2}\right)^2=\frac{-1}{2}\)

............. Làm tiếp nhé!

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

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

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

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

\(\Rightarrow\frac{1}{2}.\frac{22}{45}x=\frac{23}{45}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Vậy...

49\(\frac{8}{23}\) - (5\(\frac{7}{32}\) + 14\(\frac{8}{23}\))

= 49\(\frac{8}{23}\) - 5\(\frac{7}{32}\) - 14\(\frac{8}{23}\)

= (49\(\frac{8}{23}\) - 14\(\frac{8}{23}\)) - 5\(\frac{7}{32}\)

= 35 - 5 - \(\frac{7}{32}\)

= 30 - \(\frac{7}{32}\)

= \(\frac{960}{32}\) - \(\frac{7}{32}\)

= 953/32

Câu b:

71\(\frac{38}{45}\) - (43\(\frac{8}{45}\) - 1\(\frac{17}{57}\))

= 71\(\frac{38}{45}\) - 43\(\frac{8}{45}\) - 1\(\frac{17}{57}\)

= 28\(\frac{30}{45}\) - 1\(\frac{17}{57}\)

= 28\(\frac23\) - 1\(\frac{17}{57}\)

= \(\frac{86}{3}\) - \(\frac{74}{57}\)

= \(\frac{1634}{57}\) - \(\frac{74}{57}\)

= \(\frac{1560}{57}\)

= \(\frac{520}{19}\)

Câu a:

(-15 + |\(x\)|) + (25 - |-\(x\)|)

= -15 + |\(x\)| + 25 - |\(x\)|

= (25- 15) + (|\(x\)| - |\(x\)|)

= 10 + 0

= 10

b; \(x-34\) - [(15 + \(x\)) -(23 - \(x\))]

\(x-34\) - 15 - \(x\) + 23 - \(x\)

(\(x-x-x\)) - (34 + 15 - 23)

= (0 - \(x\)) - (49 - 23)

= -\(x\) - 26