20x1+20x2+20x3+20x4+20x5+20x6= 20x(1+2+3+4+5+6)= 20 x 21 = 420

Còn câu 2 mik thấy khó quá cho mik xin lỗi nha

=20+40+60+80+100+120

=420

Con thứ nhất: = 20.(1+2+3+4+5+6)

                      =420

Con 2 để mình nghĩ nhé!

đây nè, tick đúng cho mình nha

ĐK:  \(x\ne\left\{1;3;8;20\right\}\)

\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)

\(\Leftrightarrow\)\(\frac{1}{x-3}-\frac{1}{x-1}+\frac{1}{x-8}-\frac{1}{x-3}+\frac{1}{x-20}-\frac{1}{x-8}-\frac{1}{x-20}=-\frac{3}{4}\)

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

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

\(\Leftrightarrow\)\(x=\frac{7}{3}\)(t/m)

Vậy...

Mình thấy bài có vẻ sai ngay từ bước đầu

Bài sai ngay từ đầu là đúng rồi vì phải ghi điều kiện là "\(x\notin\left\{1;3;8;20\right\}\)" chứ không phải "\(x\ne\left\{1;3;8;20\right\}\)".

a) \(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(x-\frac{20}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10.\frac{4}{55}=\frac{3}{11}\)

\(x-\frac{8}{11}=\frac{3}{11}\)

x = 1

b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\) ( nhân cho cả tử và mẫu của các số hạng trên ( ngoại trừ 2/x.(x+1) ) là 2)

\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

\(2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(2.\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\frac{1}{x+1}=\frac{1}{18}\)

=> x + 1 = 18

x = 17

\(a,x-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(\Rightarrow x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(\Rightarrow x-10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)

\(\Rightarrow x-\frac{8}{11}=\frac{3}{11}\)

\(\Rightarrow x=1\)

\(b,\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{18}=\frac{1}{x+1}\)

\(\Rightarrow x+1=18\Leftrightarrow x=17\)

Tự làm dễ mà 

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

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

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

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

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

\(x-3=7\)

\(x=10\)

bn l-ike cho mk trước đã

Ta có : A=20/11×13 + 20/13×15 +20/15×17+...+20/53×55

A = 10 ×( 2/11×13+2/13×15+...12/53×55)

A = 10 ×(1/11-1/13+1/13-1/15+1/15-1/17+...+1/53-1/55)

A = 10 × (1/11-1/55)

A =10 × 4/55

A = 8/11

`#040911`

`a,`

`15 + 25 \div (2x - 1) = 20`

`\Rightarrow 25 \div (2x - 1) = 20 - 15`

`\Rightarrow 25 \div (2x - 1) = 5`

`\Rightarrow 2x - 1 = 25 \div 5`

`\Rightarrow 2x - 1 = 5`

`\Rightarrow 2x = 6`

`\Rightarrow x = 3`

Vây, `x = 3.`

`b,`

\(3^{x-1}+2\cdot3^x=21\)

`\Rightarrow 3^x \div 3 + 2. 3^x = 21`

`\Rightarrow 3^x . \frac{1}{3} + 2. 3^x = 21`

`\Rightarrow 3^x . (\frac{1}{3} + 2) = 21`

`\Rightarrow 3^x . \frac{7}{3} = 21`

`\Rightarrow 3^x = 21 \div \frac{7}{3}`

`\Rightarrow 3^x = 9`

`\Rightarrow 3^x = 3^2`

`\Rightarrow x = 2`

Vậy, `x = 2.`

`c,`

\(2^{x-3}+2^{x+1}=17\)

`\Rightarrow 2^x \div 2^3 + 2^x . 2 = 17`

`\Rightarrow 2^x . \frac{1}{8} + 2^x . 2 = 17`

`\Rightarrow 2^x . (\frac{1}{8} + 2) = 17`

`\Rightarrow 2^x . \frac{17}{8} = 17`

`\Rightarrow 2^x = 17 \div \frac{17}{8}`

`\Rightarrow 2^x = 8`

`\Rightarrow 2^x = 2^3`

`\Rightarrow x = 3`

Vậy, `x = 3`

`d,`

\(5^x-5^{x-1}=20\)

`\Rightarrow 5^x - 5^x \div 5 = 20`

`\Rightarrow 5^x - 5^x . \frac{1}{5} = 20`

`\Rightarrow 5^x . (1 - \frac{1}{5} = 20`

`\Rightarrow 5^x . \frac{4}{5} = 20`

`\Rightarrow 5^x = 20 \div \frac{4}{5}`

`\Rightarrow 5^x = 25`

`\Rightarrow 5^x = 5^2`

`\Rightarrow x = 2`

Vậy, `x = 2.`

\(a.25:\left(2x-1\right)=5\)

\(2x-1=5\Leftrightarrow2x=6\Leftrightarrow x=3\)

\(b.3^x:3+2.3^x=21\)\(\Leftrightarrow3^x.\dfrac{1}{3}+2.3^x=21\)

\(\Leftrightarrow3^x\left(\dfrac{1}{3}+2\right)=21\)

\(\Leftrightarrow3^x.\dfrac{7}{3}=21\)

\(\Leftrightarrow3^x=9\Leftrightarrow x=2\)

\(c.2^x:2^3+2^x.2=17\Leftrightarrow2^x.\dfrac{1}{8}+2^x.2=17\)

\(\Leftrightarrow2^x.\dfrac{17}{8}=17\Leftrightarrow2^x=8\Leftrightarrow x=3\)

\(d.5^x-5^x:5=20\Leftrightarrow5^x-5^x.\dfrac{1}{5}=20\)

\(\Leftrightarrow5^x\left(1-\dfrac{1}{5}\right)=20\Leftrightarrow5^x=20:\dfrac{4}{5}\Leftrightarrow5^x=25\Leftrightarrow x=2\)