x = từ 1 đến 10000....0

Bài 1:

Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)

\(\Leftrightarrow2x=\frac{1440}{144}=10\)

\(\Rightarrow x=5\)

Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)

=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)

a)\(5^x.\left(5^3\right)^2=625\)

\(5^x.5^6=5^4\)

\(5^x=5^{-2}\)

\(x=-2\)

b)\(27< 81^3:3^x< 243\)

\(3^3< \left(3^4\right)^3:3^x< 3^5\)

\(3^3< 3^{12}:3^x< 3^5\)

\(3^{12}:3^x=3^4\)

\(3^x=3^3\)

\(x=3\)

c)\(\left(5x+1\right)^2=\frac{36}{49}\) 

\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)

\(5x+1=\frac{6}{7}\)

\(5x=\frac{-1}{7}\)

\(x=\frac{-1}{35}\)

d)\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)

\(\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\)

\(x-\frac{2}{9}=\frac{4}{9}\)

\(x=\frac{6}{9}=\frac{2}{3}\)

\(5^x.\left(5^3\right)^2=625\)

\(\Rightarrow5^x.5^6=5^4\)

\(\Rightarrow5^{x+6}=5^4\Rightarrow x+6=4\Rightarrow x=-2\)

Đề sai rồi bạn : Phải là :

 \(5^x:\left(5^3\right)^2=625\)

\(\Rightarrow5^x:5^6=5^4\)

\(\Rightarrow5^{x-6}=5^4\)

\(\Rightarrow x-6=4\Rightarrow x=10\)

Nhứng nếu đề đúng thì bạn có thể lấy KQ trên

Chỉ biết \(x\) = \(\frac{109}{6075}\) thôi

Dễ thấy (\(\frac{3}{4}\)-81); (\(\frac{3^2}{5}\)-81); (\(\frac{3^3}{6}\)-81);... (\(\frac{3^{2007}}{2010}\)-81) có dạng (\(\frac{3^x}{3+x}\)-81) và x\(\varepsilon\){1;2;3;...2007}.

Nếu x=6 thì \(\frac{3^x}{3+x}\)-81=\(\frac{3^6}{3+6}\)-81=0

=>  (\(\frac{3}{4}\)-81) (\(\frac{3}{4}\)-81)(\(\frac{3^3}{6}\)-81)...​(\(\frac{3^6}{3+6}\)-81)...(\(\frac{3^{2007}}{2010}\)-81)=0

Mà |x-30|-6001=(\(\frac{3}{4}\)-81) (\(\frac{3}{4}\)-81)(\(\frac{3^3}{6}\)-81)...​(\(\frac{3^6}{3+6}\)-81)...(\(\frac{3^{2007}}{2010}\)-81)

=>|x-30|-6001=0

=>|x-30|=6001

=>x-30=6001 hoặc x-30=-6001

=>x=6031 hoặc x=-5971

-------------------The end----------------

\(\text{|x - 30| - 6001 = }\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)...\left(\frac{3^{2007}}{2010}-81\right)\)

\(\Rightarrow\text{ |x - 30| - 6001 = }\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)...\left(\frac{3^6}{9}-3^4\right)...\left(\frac{3^{2007}}{2010}-81\right)\)

\(\Rightarrow\left|x-30\right|- 6001 = \left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)...\left(3^4-3^4\right)...\left(\frac{3^{2007}}{2010}-81\right)\)

\(\Rightarrow|x - 30| - 6001 = \left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)...0...\left(\frac{3^{2007}}{2010}-81\right)\)

\(\Rightarrow\text{|x - 30| - 6001 = }0\)

\(\Rightarrow\left|x-30\right|=6001\) 

\(\Rightarrow x-30=6001\)hoặc \(x-30=-6001\)

\(\Rightarrow x=6031\)hoặc\(x=-5971\)

Vậy: x= 6031 hoặc x= -5971

(Nói thật thì mình mới lớp 7, đây có phải của lớp 8 không?)

Nhân 2 cả 2 vế lên:

\(\left(2x+\frac{2}{1x3}\right)+...+\left(2x+\frac{2}{23x25}\right)=22x+\frac{2}{3}+\frac{2}{9}+\frac{2}{81}+\frac{2}{243}\)2/243

\(24x+\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{23}-\frac{1}{25}\right)=22x+\frac{162+54+6+2}{243}\)

\(24x+\frac{24}{25}=22x+\frac{224}{243}\)

\(2x=\frac{224}{243}-\frac{24}{25}\)

\(2x=-\frac{232}{6025}\)

\(x=\frac{-116}{6075}\)

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

\(12x+\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\right)\right]=11.x+\left(\frac{81}{243}+\frac{27}{243}+\frac{3}{243}+\frac{1}{243}\right)\)

\(12x+\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{25}\right)\right]=11.x+\frac{112}{243}\)

\(12x+\left(\frac{1}{2}.\frac{24}{25}\right)=11.x+\frac{112}{243}\)

\(12x+\frac{12}{25}=11x+\frac{112}{243}\)

\(11x-12x=\frac{112}{243}-\frac{12}{25}\)

\(-1x=-\frac{116}{6075}\)

\(x=-\frac{116}{6075}\div\left(-1\right)\)

\(x=\frac{116}{6075}\)

a,\(8< 2^x\le2^9.2^{-5}\)

\(2^3< 2^x\le2^4\)

\(\Rightarrow x=4\)

b, \(27< 81^3.3^x< 243\)

\(3^3< 3^{12-x}< 3^5\)

\(\Rightarrow3< 12-x< 5\)

12-x=4

x=8

c,\(\left(\frac{2}{5}\right)^x>\left(\frac{2}{5}\right)^3.\left(\frac{2}{5}\right)^2\)

\(\left(\frac{2}{5}\right)^x>\left(\frac{2}{5}\right)^5\)

\(\Rightarrow x>5\)

x=6;7;8........

tìm x, biết:

(5x+1)^2=36/49

\(a)\)\(\left(\frac{3}{5}\right)^{2x+1}=\frac{81}{625}\)

\(\Leftrightarrow\)\(\left(\frac{3}{5}\right)^{2x+1}=\left(\frac{3}{5}\right)^4\)

\(\Leftrightarrow\)\(2x+1=4\)

\(\Leftrightarrow\)\(x=\frac{3}{2}\)

Vậy \(x=\frac{3}{2}\)

\(b)\)\(\left(\frac{2}{3}\right)^x.\left(\frac{2}{3}\right)^3=\frac{32}{243}\)

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

\(\Leftrightarrow\)\(x+3=5\)

\(\Leftrightarrow\)\(x=2\)

Vậy \(x=2\)

\(c)\)\(\left(2x-1\right)^2=\left(2x-1\right)^3\)

\(\Leftrightarrow\)\(\left(2x-1\right)^3-\left(2x-1\right)^2=0\)

\(\Leftrightarrow\)\(\left(2x-1\right)^2\left(2x-1-1\right)=0\)

\(\Leftrightarrow\)\(\left(2x-1\right)^2\left(2x-2\right)=0\)

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

Vậy \(x=\frac{1}{2}\) hoặc \(x=1\)

Chúc bạn học tốt ~ 

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

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

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

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

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 = ......