a) x+3/22=-27/121.11/9

⇒ x+3/22=-3/11

⇒ x=-9/22

b) (3/2/20-2/1/20)x+2/7=3/5

⇒ (31/10-41/20)x=11/35

⇒ 21/20x=11/35

⇒x=44/147

c) |3/2x-5|-1/2=-1/4

⇒ |3/2x-5|=1/4

⇒ hoặc 3/2x-5=1/4⇒3/2x=21/4⇒x=7/2

hoặc 3/2x-5=-1/4⇒3/2x=19/4⇒x=19/6

Bài 1:

a) \(\dfrac{9}{20}-\dfrac{8}{15}\times\dfrac{5}{12}\)

\(=\dfrac{9}{20}-\dfrac{2}{9}\)

\(=\dfrac{41}{180}\)

b) \(\dfrac{2}{3}\div\dfrac{4}{5}\div\dfrac{7}{12}\)

\(=\dfrac{2}{3}\times\dfrac{5}{4}\times\dfrac{12}{7}\)

\(=\dfrac{5}{6}\times\dfrac{12}{7}\)

\(=\dfrac{10}{7}\)

c) \(\dfrac{7}{9}\times\dfrac{1}{3}+\dfrac{7}{9}\times\dfrac{2}{3}\)

\(=\dfrac{7}{9}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)

\(=\dfrac{7}{9}\times1\)

\(=\dfrac{7}{9}\)

Bài 2:

a) \(2\times\left(x-1\right)=4026\)

\(\left(x-1\right)=4026\div2\)

\(x-1=2013\)

\(x=2014\)

Vậy: \(x=2014\)

b) \(x\times3,7+6,3\times x=320\)

\(x\times\left(3,7+6,3\right)=320\)

\(x\times10=320\)

\(x=320\div10\)

\(x=32\)

Vậy: \(x=32\)

c) \(0,25\times3< 3< 1,02\)

\(\Leftrightarrow0,75< 3< 1,02\) ( S )

=> \(0,75< 1,02< 3\)

a: Ta có: \(\dfrac{1}{4}:x=3\dfrac{4}{5}:40\dfrac{8}{15}\)

\(\Leftrightarrow x=\dfrac{1}{4}\cdot\dfrac{\dfrac{608}{15}}{3+\dfrac{4}{5}}\)

\(\Leftrightarrow x=\dfrac{152}{15}:\dfrac{19}{5}=\dfrac{8}{3}\)

b: Ta có: \(\left(x+1\right):\dfrac{5}{6}=\dfrac{20}{3}\)

\(\Leftrightarrow x+1=\dfrac{50}{9}\)

hay \(x=\dfrac{41}{9}\)

c: Ta có: \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

\(\Leftrightarrow x^2-1=63\)

\(\Leftrightarrow x^2=64\)

hay \(x\in\left\{8;-8\right\}\)

c. \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\) 

    \(7.9=\left(x-1\right).\left(x+1\right)\) 

    \(63=x^2-1\) 

             \(x^2=63+1\) 

             \(x^2=64\) 

             \(x^2=8^2\)

             \(x=8\)          

a) \(\frac{x^7}{3^5}=9\)

=> \(\frac{x^7}{3^5}=3^2\)

=> x⁷ = 3² . 3⁵ = 3⁷

=> x = 3

b) \(\frac{32}{x^7}=4\)

=> \(\frac{2^5}{x^7}=2^2\)

=> \(2^2\cdot x^7=2^5\)

=> \(x^7=\frac{2^5}{2^2}=2^3\)

=> không tìm được x

2. \(\frac{2^{30}}{3^{20}}-\left(\frac{2}{3}\right)^{20}\cdot2^{10}\)

\(=\frac{2^{30}}{3^{20}}-\frac{2^{20}}{3^{20}}\cdot2^{10}\)

\(=\frac{2^{30}}{3^{20}}-\frac{2^{30}}{3^{20}}=0\)

Câu 6: Khôg có cau nào đúng

Câu 7: C

Câu 8: B

Câu 9: B

Câu 10: D

6Không có đáp án nào đúng x=11/4

7C

8B

9B

10D

`a, x-2=7/15`

`=>x=7/15 +2`

`=>x= 7/15+ 30/15`

`=>x= 37/15`

`b, 9/20 +x=2/5 +3/20`

`=> 9/20 +x=8/20 +3/20`

`=> 9/20 +x=11/20`

`=>x=11/20-9/20`

`=>x= 2/20=1/10`

x = \(\dfrac{7}{15}\) + 2 = \(\dfrac{37}{15}\) 

x = \(\dfrac{2}{5}\) + \(\dfrac{3}{20}\) - \(\dfrac{9}{20}\) = \(\dfrac{1}{10}\)

a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

                 \(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

                \(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

           \(=>2x+\frac{3}{5}=\frac{3}{5}\)

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

                                    \(2x=0\)

                                      \(x=0:2\)

                                      \(x=0\)

b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)

=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.

\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)

a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

\(\left(2x+\frac{3}{5}\right)^2=0+\frac{9}{25}\)

\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)

\(\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)

b) \(\left(3x-1\right)\left(-\frac{1}{2}x+5\right)=0\)

\(\left(3x-1\right)\left(-\frac{x}{2}+5\right)=0\)

\(\left(3x-1\right)\left(5-\frac{x}{2}\right)=0\)

\(\orbr{\begin{cases}3x-1=0\\5-\frac{x}{2}=0\end{cases}}\)

\(\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)

Tìm x , biết : 

|x-2| = -3 

Vô lí vì |x-2| phải lớn hơn hoặc bằng 0

=> không có giá trị x nào thỏa mãn đề bài trên .

|x-7| - 5 + 3  = 1

<=> |x-7| - 5 = -2 

<=> |x-7| = 3

<=> x - 7 = 3 hoặc x - 7 = -3 

<=> x = 10 hoặc x = 4

Vậy x€ { 10 ; 4  }

|x+5| - 5 = 4 - (-3) 

<=> |x+5| - 5 = 7

<=> |x+5| = 7 + 5

<=> |x +5| = 12

<=> x + 5 = 12 hoặc x + 5 =-12 

<=> x = 7 hoặc x = -17

Vậy x € { 7 ; -17 } 

Cảm ơn bạn tasaka mk là mi rất vui được làm quen