1) \(2^3\times x-5^2\times x=2\times\left(5^2+2^2\right)-33\)

\(x\times\left(2^3-5^2\right)=2\times\left(25+4\right)-33\)

\(x\times\left(8-25\right)=2\times29-33\)

\(x\times-17=25\)

\(x=-\dfrac{25}{17}\)

2) \(15\div\left(x+2\right)=\left(3^3+3\right)\div1\)

\(15\div\left(x+2\right)=\left(27+3\right)\div1\)

\(15\div\left(x+2\right)=30\div1\)

\(15\div\left(x+2\right)=30\)

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

\(x=-\dfrac{3}{2}\)

3) \(20\div\left(x+1\right)=\left(5^2+1\right)\div13\)

\(20\div\left(x+1\right)=\left(25+1\right)\div13\)

\(20\div\left(x+1\right)=26\div13\)

\(20\div\left(x+1\right)=2\)

\(x+1=20\div2\)

\(x+1=10\)

\(x=9\)

4) \(320\div\left(x-1\right)=\left(5^3-5^2\right)\div4+15\)

\(320\div\left(x-1\right)=\left(125-25\right)\div4+15\)

\(320\div\left(x-1\right)=100\div4+15\)

\(320\div\left(x-1\right)=25+15\)

\(320\div\left(x-1\right)=40\)

\(x-1=8\)

\(x=9\)

5) \(240\div\left(x-5\right)=2^2\times5^2-20\)

\(240\div\left(x-5\right)=4\times25-20\)

\(240\div\left(x-5\right)=100-20\)

\(240\div\left(x-5\right)=80\)

\(x-5=30\)

\(x=35\)

6) \(70\div\left(x-3\right)=\left(3^4-1\right)\div4-10\)

\(70\div\left(x-3\right)=\left(81-1\right)\div4-10\)

\(70\div\left(x-3\right)=80\div4-10\)

\(70\div\left(x-3\right)=20-10\)

\(70\div\left(x-3\right)=10\)

\(x-3=7\)

\(x=10\)

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 )

=> \(0,75< 1,02< 3\)

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

3: 

\(=\dfrac{1}{7}\cdot\dfrac{3}{5}\cdot\dfrac{5}{6}\cdot\dfrac{5}{8}=\dfrac{1}{7}\cdot\dfrac{1}{2}\cdot\dfrac{5}{8}=\dfrac{5}{112}\)

4:

=>2/3:x=-2-1/3=-7/3

=>x=-2/3:7/3=-2/7

5:

AC=CB=12/2=6cm

IB=6/2=3cm

a)

Dãy trên có số số hạng là:

( 20 - 1 ) : 1 + 1 = 20 ( số hạng )

Tổng của dãy trên là:

( 20 + 1 ) x 20 : 2 = 210 

Đáp số: 210

b)

Dãy trên có số số hạng là:

( 21 - 1 ) : 2 + 1 = 11 ( số hạng )

Tổng của dãy trên là:

( 21 + 1 ) x 11 : 2 = 121

Đáp số: 121

c) ( 2x - 1 ) x 2 = 13

2x - 1 = \(\dfrac{13}{2}\)

2x = \(\dfrac{15}{2}\)
\(x=\dfrac{15}{4}\)

32 x ( x - 10 ) = 32

( x - 10 ) = 1

x = 11

\(A=1+2+3+...+20\)

Số hạng:

\(\left(20-1\right):1+1=20\) (số hạng)

Tổng: \(\left(20+1\right)\cdot20:2=210\)

\(B=1+3+5+...+21\)

Số hạng:

\(\left(21-1\right):2+1=11\) (số hạng) 

Tổng: \(\left(21+1\right)\cdot11:2=121\)

\(\left(2x-1\right)\cdot2=13\)

\(\Rightarrow2x-1=\dfrac{13}{2}\)

\(\Rightarrow2x=\dfrac{15}{2}\)

\(\Rightarrow x=\dfrac{15}{4}\)

\(32\cdot\left(x-10\right)=32\)

\(\Rightarrow x-10=1\)

\(\Rightarrow x=11\)

`a,` Khoảng cách là : `1`

Số số hạng là : \(\dfrac{20-1}{1}+1=20\)

Tổng là : \(\dfrac{20+1\times20}{2}=\)`210`

`b,` Khoảng cách là : `2`

Số số hạng là : \(\dfrac{21-1}{2}+1=11\)

Tổng là : \(\dfrac{21+1\times11}{2}=121\)

\(c,\left(2x-1\right).2=13\\ \Rightarrow2x-1=\dfrac{13}{2}\\ \Rightarrow2x=\dfrac{13}{2}+1\\ \Rightarrow2x=\dfrac{15}{2}\\ \Rightarrow x=\dfrac{15}{2}:2\\ \Rightarrow x=\dfrac{15}{4}\)

\(32.\left(x-10\right)=32\\ \Rightarrow x-10=32:32\\ \Rightarrow x-10=1\\ \Rightarrow x=1+10\\ \Rightarrow x=11\)

Bài 1:

a; (\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\)) x \(\dfrac{3}{4}\) = \(\dfrac{1}{4}\)

     \(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) : \(\dfrac{3}{4}\)

      \(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) x \(\dfrac{4}{3}\)

    \(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) =  \(\dfrac{1}{3}\)

      \(\dfrac{1}{4}x\) = \(\dfrac{1}{3}\) + \(\dfrac{1}{8}\)

       \(\dfrac{1}{4}\) \(x\)=  \(\dfrac{8}{24}\) + \(\dfrac{11}{24}\)

          \(\dfrac{1}{4}x=\dfrac{11}{24}\)

           \(x=\dfrac{11}{24}:\dfrac{1}{4}\)

           \(x=\dfrac{11}{24}\times4\)

           \(x=\dfrac{11}{6}\) 

b; \(\dfrac{12}{5}:x\) = \(\dfrac{14}{3}\) x \(\dfrac{4}{7}\)

     \(\dfrac{12}{5}\) : \(x\) = \(\dfrac{8}{3}\)

            \(x\) = \(\dfrac{12}{5}\) : \(\dfrac{8}{3}\)

            \(x\) = \(\dfrac{12}{5}\) x \(\dfrac{3}{8}\)

             \(x\) = \(\dfrac{9}{10}\)

c; \(x+\dfrac{2}{3}\) = 8:4 - 1

   \(x+\dfrac{2}{3}\) = 2 - 1

   \(x+\dfrac{2}{3}\) = 1

  \(x=1-\dfrac{2}{3}\)

  \(x\) = \(\dfrac{1}{3}\)

a: ĐKXĐ: x>=0

\(5\sqrt{x}-2=13\)

=>\(5\sqrt{x}=15\)

=>\(\sqrt{x}=3\)

=>x=9

b: ĐKXĐ: x>=-1

\(\sqrt{9x+9}+\sqrt{x+1}=20\)

=>\(3\sqrt{x+1}+\sqrt{x+1}=20\)

=>\(4\sqrt{x+1}=20\)

=>x+1=25

=>x=24