góc A1 và góc A2 là 2 góc bù nhau => góc A2 = 180*- A1 = 110*

góc A1 và góc A3 là 2 góc đối đỉnh => góc A3 = 70*, góc A2 và góc A4 là 2 góc đối đỉnh => góc A4=110*

Câu 2: góc B1 và góc B2 là 2 góc bù nhau => B1+B2 =180*. Lại có B1-B2 = 60*

=> B1+B2+B1-B2 = 240*

<=> 2B1= 240*

=> B1 = 120*=>B1 =B3=120*(2 góc đối đỉnh)

=> B2 =60*=> B2 =B4 =60*(2 góc đối đỉnh)

Ta có :

\(\left(\frac{3}{8}-\frac{1}{5}\right)+\left(\frac{5}{8}-x\right)=\frac{1}{5}\)

\(\Rightarrow\frac{3}{8}-\frac{1}{5}+\frac{5}{8}-x=\frac{1}{5}\)

\(\Rightarrow\left(\frac{3}{8}+\frac{5}{8}\right)-\frac{1}{5}-x=\frac{1}{5}\)

\(\Rightarrow1-\frac{1}{5}-x=\frac{1}{5}\)

\(\Rightarrow\frac{4}{5}-x=\frac{1}{5}\)

\(\Rightarrow x=\frac{4}{5}-\frac{1}{5}=\frac{3}{5}\)

| 2x - 1 | + 1 = 4

=> | 2x - 1 | = 3

\(\Rightarrow\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=4\\2x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

8 - | 1 - 3x | = 3

=> | 1 - 3x | = 5

\(\Rightarrow\orbr{\begin{cases}1-3x=5\\1-3x=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=1-5=-4\\3x=1-\left(-5\right)=6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-4}{3}\\x=2\end{cases}}\)

\(a)\left(\frac{3}{8}-\frac{1}{5}\right)+\left(\frac{5}{8}-x\right)=\frac{1}{5}.\)

\(\frac{3}{8}-\frac{1}{5}+\frac{5}{8}-x=\frac{1}{5}\)

\(\left(\frac{3}{8}+\frac{5}{8}\right)+\left(-\frac{1}{5}-x\right)=\frac{1}{5}\)

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

\(-\frac{1}{5}-x=\frac{1}{5}-1\)

\(-\frac{1}{5}-x=-\frac{4}{5}\)

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

\(x=\frac{3}{5}\)

Ta có :

( x - 1 )^x+3 = ( x - 1 )^x+4

=> ( x - 1 )^x+4 - ( x - 1 )^x+3 = 0

=> ( x- 1 )^x+3 . ( x - 1 - 1 ) = 0

\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{x+3}=0\\x-1-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

Theo đề ra, ta có: \(\widehat{xAB}=\widehat{yBA}=120^o\) mà hai góc này ở vị trí so le trong => Ax//By

a) ( x - 1 )³ = 27 

=> ( x - 1 )³ = 3³

=> x - 1 = 3

=> x = 4

b) x² + x = 0

=> x . ( x + 1 ) = 0

\(\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

c) ( 2x + 1 )² = 25 

=> ( 2x + 1 )² = 5² = ( -5 )²

\(\Rightarrow\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=4\\2x=-6\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

d) ( 2x - 3 )² = 36

=> ( 2x - 3 )² = 6² = ( -6 )²

\(\Rightarrow\orbr{\begin{cases}2x-3=6\\2x-3=-6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=9\\2x=-3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{9}{2}\\x=\frac{-3}{2}\end{cases}}\)

e) 5^x+2 = 625

=> 5^x+2 = 5⁴

=> x + 2 = 4

=> x = 2

f) ( x - 1 )^x+2 = ( x - 1 )^x+4

=> ( x - 1 )^x+4 - ( x - 1 )^x+2 = 0

=> ( x - 1 )^x+2 . [ ( x - 1 )² - 1 ] = 0

\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^2-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^2=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x-1\in\left\{-1;1\right\}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x\in\left\{0;2\right\}\end{cases}}\)

g) ( 2x - 1 )³ = 8

=> ( 2x - 1 )³ = 2³

=> 2x - 1 = 2

=> 2x = 3

=> x = \(\frac{3}{2}\)

h) \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.....\frac{31}{64}=2^x\)

\(\Rightarrow\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}.\frac{4}{2.5}.....\frac{31}{2.32}=2^x\)

\(\Rightarrow\frac{1.2.3.4.....31}{\left(2.2\right).\left(2.3\right).\left(2.4\right).....\left(2.32\right)}=2^x\)

\(\Rightarrow\frac{2.3.4.....31}{2^{31}.32.\left(2.3.4.....31\right)}=2^x\)

\(\Rightarrow\frac{1}{2^{31}.2^5}=2^x\)\(\Rightarrow\frac{1}{2^{36}}=2^x\)

\(\Rightarrow2^x=2^{-36}\Rightarrow x=-36\)

a)x=4

b)x=0;x=1

c)x=2

d)x=9/2

g)x=-1/2

những cái khác tui lười tính vì nó rắc rối ơn xíu

\(\left(-0,6x-\frac{1}{2}\right)\times\frac{3}{4}-\left(-1\right)=\frac{1}{3}\)

\(\left(-0,6x-\frac{1}{2}\right)\times\frac{3}{4}=\frac{1}{3}+\left(-1\right)\)

\(\left(-0,6x-\frac{1}{2}\right)\times\frac{3}{4}=-\frac{2}{3}\)

\(-0,6x-\frac{1}{2}=-\frac{2}{3}\div\frac{3}{4}\)

\(-0,6x-\frac{1}{2}=-\frac{8}{9}\)

\(-0,6x=-\frac{8}{9}+\frac{1}{2}\)

\(-0,6x=-\frac{7}{18}\)

\(x=-\frac{7}{18}\div\left(-0,6\right)\)

\(x=\frac{35}{54}\)

\(\frac{5}{3\times6}+\frac{5}{6\times9}+\frac{5}{9\times12}+....+\frac{5}{99\times102}.\)

Đặt :

 \(A=\frac{5}{3\times6}+\frac{5}{6\times9}+\frac{5}{9\times12}+....+\frac{5}{99\times102}.\)

\(\frac{3}{5}\times A=\frac{3}{5}\times\left(\frac{5}{3\times6}+\frac{5}{6\times9}+...+\frac{5}{99\times102}\right)\)

\(\frac{3}{5}A=\frac{3}{3\times6}+\frac{3}{6\times9}+...+\frac{3}{96\times99}\)

\(\frac{3}{5}A=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+....+\frac{1}{96}-\frac{1}{99}\)

\(\frac{3}{5}A=\frac{1}{3}-\frac{1}{99}\)

\(\frac{3}{5}A=\frac{32}{99}\)

\(A=\frac{32}{99}\div\frac{3}{5}=\frac{160}{297}\)

\(A=\frac{5}{3\times6}+\frac{5}{6\times9}+.....+\frac{5}{99\times102}\)

\(A=\frac{5}{3}\left[\left(\frac{1}{3}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{9}\right)+.....+\left(\frac{1}{99}-\frac{1}{102}\right)\right]\)

\(A=\frac{5}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{102}\right)\)

\(A=\frac{5}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)

\(A=\frac{5}{3}.\frac{11}{34}\)

\(A=\frac{55}{102}\)