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30 tháng 8 2016

\(\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{9}\right)+\left(x+\frac{1}{27}\right)+\left(x+\frac{1}{81}\right)=\frac{51}{81}\)

\(\left(x+x+x+x\right)+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)=\frac{51}{81}\)

\(\left(x+x+x+x\right)+\left(\frac{27}{81}+\frac{9}{81}+\frac{3}{81}+\frac{1}{81}\right)=\frac{51}{81}\)

\(x\times4+\frac{40}{81}=\frac{51}{81}\)

\(x\times4=\frac{51}{81}-\frac{40}{81}\)

\(x\times4=\frac{11}{81}\)

\(\Rightarrow x=\frac{11}{81}\div4=\frac{11}{81}\times\frac{1}{4}\)

\(\Rightarrow x=\frac{11}{324}\)

[ 61 + ( 53 - x ) ] . 17 = 1785

61 + ( 53 - x ) = 1785 : 17

61 + ( 53 - x ) = 105

( 53 - x ) = 105 - 61

53 - x = 44

=> x = 53 - 44

=> x = 9

30 tháng 8 2016

Cảm ơn bạn nhiều nha 

21 tháng 6 2017

\(\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{15}\right)+....+\left(x+\frac{1}{575}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)

\(13x+\left(\frac{1}{1.3}+\frac{1}{3.5}+.....+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)

\(13x+\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)

\(2x+\frac{12}{25}=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)

Đặt \(A=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)

\(3A=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)

\(3A-A=1-\frac{1}{3^5}=\frac{242}{243}=2A\)

=> \(A=\frac{121}{243}\)

=> \(2x+\frac{12}{25}=\frac{121}{243}\)

=> \(2x=\frac{121}{243}-\frac{12}{25}=\frac{109}{6075}\)

=> x = ......

4 tháng 6 2016

Tìm x, biết:

3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x2;5;10;17)

2(x1)(x3) +5(x3)(x8) +12(x8)(x20) 1x20 =34 (x1;3;8;20)

x+110 +2+111 x+112 =x+113 +x+114 

x1030 +x1443 +x595 +x1488 =0

4 tháng 6 2016

Trả lời luôn à bạn

19 tháng 4 2019

Câu a \(\left|2x-\frac{1}{3}\right|+\frac{5}{6}=1\)

19 tháng 4 2019

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

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

Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)

9 tháng 6 2017

\(\frac{1}{x+2}-\frac{1}{x+5}+...+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\frac{1}{x+2}-\frac{1}{x+7}=\frac{x}{\left(x+2\right)\left(x+7\right)}\)

\(\Rightarrow x=1\)

9 tháng 6 2017

a)x=15

24 tháng 6 2020

bạn tự làm đi tính toán thôi mà

30 tháng 7 2020

\(\left(a+\frac{1}{1.3}\right)+\left(a+\frac{1}{3.5}\right)+...+\left(a+\frac{1}{23.25}\right)=11a+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)

\(\Rightarrow12a+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11a+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)\)(1)

Ta có \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\right)=\frac{1}{2}\left(1-\frac{1}{25}\right)=\frac{1}{2}.\frac{24}{25}=\frac{12}{25}\)

Lại có \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}=\frac{3\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)}{2}\)

\(=\frac{1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-\frac{1}{3^4}-\frac{1}{3^5}}{2}=\frac{1-\frac{1}{3^5}}{2}=\frac{1}{2}-\frac{1}{3^5.2}\)

Khi đó (1) <=> \(12a-\frac{12}{25}=11a+\frac{1}{2}-\frac{1}{3^5.2}\)

=> \(a=\frac{12}{25}+\frac{1}{2}-\frac{1}{3^5.2}=\frac{49}{50}-\frac{1}{3^5.2}=\frac{49}{50}-\frac{1}{486}=\frac{23764}{24300}\)

30 tháng 7 2020

Gọi \(A=\left(a+\frac{1}{1.3}\right)+\left(a+\frac{1}{3.5}\right)+\left(a+\frac{1}{5.7}\right)+...+\left(a+\frac{1}{23.25}\right)\)

\(\Rightarrow A=12a+\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{23.25}\right)\)

\(\Rightarrow A=12a+\left[\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{23.25}\right)\right]\)

\(\Rightarrow A=12a+\left[\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\right)\right]\)

\(\Rightarrow A=12a+\left[\frac{1}{2}\left(1-\frac{1}{25}\right)\right]\)

\(\Rightarrow A=12a+\left(\frac{1}{2}.\frac{24}{25}\right)\)

\(\Rightarrow A=12a+\frac{12}{25}\)

Gọi \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

\(\Rightarrow B=\frac{1}{1.3}+\frac{1}{3.3}+\frac{1}{9.3}+\frac{1}{27.3}+\frac{1}{81.3}\)

\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)

\(\Rightarrow3B-B=1-\frac{1}{243}\)

\(\Rightarrow2B=\frac{242}{243}\)

\(\Rightarrow B=\frac{121}{243}\)

\(\Rightarrow A=11a+B\)

\(\Rightarrow12a+\frac{12}{25}=11a+\frac{121}{243}\)

\(\Leftrightarrow12a-11a=\frac{121}{243}-\frac{12}{25}\)

\(\Leftrightarrow a=\frac{109}{6075}\)

22 tháng 7 2016

\(1\)\(70:\frac{4x+720}{x}=\frac{1}{2}\)

\(\Leftrightarrow\frac{4x+720}{x}=70:\frac{1}{2}\)

\(\Leftrightarrow\frac{4x+720}{x}=140\)

\(\Leftrightarrow\left(4x+720\right):x=140\)

\(\Leftrightarrow4x+720=140.x\)

\(\Leftrightarrow4x-140x=-720\)

\(\Leftrightarrow x.\left(-136\right)=-720\)

\(\Leftrightarrow x=-720:\left(-136\right)\)

\(\Leftrightarrow x=\frac{90}{17}\)

\(2\)) Mình đang nghĩ

6 tháng 6 2016

\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)

 TH1:   \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)

TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)

6 tháng 6 2016

\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)

\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)

\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

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

\(\Rightarrow x=\frac{2}{5}\)

\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)

\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)

\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)

\(\Rightarrow3x=\frac{1}{9}\)

\(\Rightarrow x=\frac{1}{27}\)

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

\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)  \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)  \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)