\(\left|x+\frac{1}{3}\right|+\frac{4}{5}=\left|-3,2+\frac{2}{5}\right|+\left(27-\frac{3}{5}\right)\left(27-\frac{3^2}{6}\right)...\left(27-\frac{3^5}{9}\right)...\left(27-\frac{3^{2010}}{2014}\right)\)

\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}+\left(27-\frac{3^2}{6}\right)\left(27-\frac{3^3}{7}\right)...\left(27-27\right)...\left(27-\frac{3^{2010}}{2014}\right)\)

\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\)

\(\Leftrightarrow\left|x+\frac{1}{3}\right|=2\)

\(\Rightarrow\hept{\begin{cases}x+\frac{1}{3}=2\\x+\frac{1}{3}=-2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=-\frac{7}{3}\end{cases}}}\)

bạn ơi, có một chỗ chưa chuẩn .bạn kiểm tra lại giú mình. chỗ vế trái bạn thiếu \(\left(27-\frac{3}{5}\right)\). bạn bổ sung vào cho đúng nhé. dù sao vẫn cảm ơn bạn.

lên mạng đi 

\(\frac{x+3}{3}=\frac{27}{x+3}\left(ĐKXĐ:x\ne-3\right)\)

\(\Leftrightarrow\left(x-3\right)^2=81\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=9\\x-3=-9\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=12\\x=-6\end{cases}}\)( Tm ĐKXĐ ) 
Vậy ...

Theo bài ra ta cs (x+3).(x+3)=3.27

=>(x+3)^2=81

=>x+3=9

=>x=6

Vậy...

p/s: Nhớ mãi cái hôm thi vio v19 Gặp câu này hong bt làm :((
lg: Đặt biểu thức= A
$<=> A^3 = 9 + 3\sqrt[3]{9-\frac{x}{27}}+A$
$<=> A(A^2- 3\sqrt[3]{9-\frac{x}{27}}) =9 = 1.9 = -1.-9 = -3.-3 = 3.3= -9.-1=9.1$
.... 

\(\frac{\left(\frac{518}{19}-\frac{342}{13}\right).\left(\frac{177}{236}+\frac{76}{236}-\frac{6}{236}\right)}{\left(\frac{3}{4}+x\right).\frac{27}{33}}=1\)

=>\(\frac{\left(\frac{6734}{247}-\frac{6498}{247}\right).\frac{247}{236}}{\left(\frac{3}{4}+x\right).\frac{27}{33}}=1\)

=>(3/4+x)*27/33=236/247*247/236=1

3/4+x=1:27/33=33/27

x=33/27-3/4=132/108-81/108

x=51/108

Vậy x=51/108

tính tử rồi nhân chéo lên tìm x

uk bạn làm đi

Đặt Q = \(\sqrt[3]{3+\sqrt{\frac{x}{27}}}\)+\(\sqrt[3]{3-\sqrt{\frac{x}{27}}}\)

 \(^{Q^3}\)=  3 + \(\sqrt{\frac{x}{27}}\)+3 - \(\sqrt{\frac{x}{27}}\)+3(\(\sqrt[3]{3+\sqrt{\frac{x}{27}}}\)*\(\sqrt[3]{3-\sqrt{\frac{x}{27}}}\) )(\(\sqrt[3]{3+\sqrt{\frac{x}{27}}}\)+\(\sqrt[3]{3-\sqrt{\frac{x}{27}}}\))

\(Q^3\)= 6 +3 \(\sqrt[3]{\left(3+\sqrt{\frac{x}{27}}\right)\left(3-\sqrt{\frac{x}{27}}\right)}\)\(Q\)

\(Q^3\)= 6+ 3\(\sqrt[3]{\left(3^2-\left(\sqrt{\frac{x}{27}}\right)^2\right)}\)\(Q\)

\(Q^3\)= 6 + 3 \(\sqrt[3]{9-\frac{x}{27}}\)\(Q\)

\(Q^3\)= 6 + 3\(\sqrt[3]{\frac{243-x}{27}}\)\(Q\)

\(Q^3\)= 6 + \(\sqrt[3]{243-x}\)\(Q\)

\(Q\)( \(Q^2\)- \(\sqrt[3]{243-x}\)) =6

\(Q\)=\(\frac{6}{Q^2-\sqrt[3]{243-x}}\)

Vì Q \(\in\)Z nên \(Q^2\)\(\in\)\(Z\), 6\(\in\)\(Z\) nên \(\sqrt[3]{243-x}\)\(\in\)\(Z\); \(Q^2\)- \(\sqrt[3]{243-x}\)\(\in\)\(Ư\left(6\right)\)=\(\left\{+-1;+-2;+-3;+-6\right\}\)

Suy ra 243 -x \(\in\)+ -1; + -8 ;+-27;....

\(Q^2\)-\(\sqrt[3]{243-x}\)= 1 \(\Rightarrow\)\(Q^2\)= 1+\(\sqrt[3]{243-x}\)Vì Q\(\in\)Z nên \(\sqrt[3]{243-x}\)= 8 

Suy ra x=241 hoặc x=245

Vậy......

Không biết  mk lm đúng hay sai mong mấy bn đóng góp ý kiến . Cảm ơn nhiều ạ

bạn làm sai từ cái \(\sqrt[3]{243-x}\in Z\)

a)

\(\begin{array}{l}x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\\x = \frac{{ - 7}}{9}:\frac{{14}}{{27}}\\x = \frac{{ - 7}}{9}.\frac{{27}}{{14}}\\x = \frac{{ - 3}}{2}\end{array}\)                

Vậy \(x = \frac{{ - 3}}{2}\).

b)

\(\begin{array}{l}\left( {\frac{{ - 5}}{9}} \right):x = \frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right):\frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right).\frac{3}{2}\\x = \frac{{ - 5}}{6}\end{array}\)

Vậy \(x = \frac{{ - 5}}{6}\).

c)

\(\begin{array}{l}\frac{2}{5}:x = \frac{1}{{16}}:0,125\\\frac{2}{5}:x = \frac{1}{{16}}:\frac{1}{8}\\\frac{2}{5}:x = \frac{1}{{16}}.8\\\frac{2}{5}:x = \frac{1}{2}\\x = \frac{2}{5}:\frac{1}{2}\\x = \frac{2}{5}.2\\x = \frac{4}{5}\end{array}\)      

Vậy \(x = \frac{4}{5}\)

d)

\(\begin{array}{l} - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\\ - \frac{5}{{12}}x = \frac{4}{6} - \frac{3}{6}\\ - \frac{5}{{12}}x = \frac{1}{6}\\x = \frac{1}{6}:\left( { - \frac{5}{{12}}} \right)\\x = \frac{1}{6}.\frac{{ - 12}}{5}\\x = \frac{{ - 2}}{5}\end{array}\)

Vậy \(x = \frac{{ - 2}}{5}\).

Chú ý: Khi trình bày lời giải bài tìm x, sau khi tính xong, ta phải kết luận.

Ta có:

\(\frac{x+3}{3}=\frac{27}{x-3}\)

\(\Rightarrow\left(x+3\right)\left(x-3\right)=27.3\)

\(\Leftrightarrow x^2-9=27.3\)

\(\Leftrightarrow x^2-9=81\)

\(\Leftrightarrow x^2=81+9\)

\(\Leftrightarrow x^2=90\)

\(\Leftrightarrow x=\sqrt{90}=3\sqrt{10}\)

Cái đoạn(x+3)(x-3)=x²-9 là mình dùng hằng đẳng thức của lớp 8

x+3/3=27/x-3

=> (x+3).(x-3)=27.3

=> x.(x-3)+3.(x-3)=81

=> x²-3x+3x-9=81

=> x²-9=81

=> x²=72

=> x= căn 72

Theo đề ta có : \(\frac{x+3}{3}=\frac{27}{x-3}\)

\(\Rightarrow\left(x+3\right)\left(x-3\right)=27.3\)

\(\Rightarrow x^2-3^2=81\)

\(\Rightarrow x^2=81+9=90\)

\(\Rightarrow x=3\sqrt{10}\) hoặc \(x=-3\sqrt{10}\)

ta có 2/3^x+1 +2/3^x=20/27 suy ra 2/3^x *2/3+2/3^x=20/27

suy ra 2/3^x(2/3+1)=20/27 suy ra 2/3^x*5/3=20/27 suy ra 2/3^x=20/27:5/3=4/9

suy ra2/3^x=2/3^2 suy ra x=2

\(\left(\frac{2}{3}\right)^{x+1}+\left(\frac{2}{3}\right)^x=\frac{20}{27}\)

\(\Leftrightarrow\left(\frac{2}{3}\right)^x\left(\frac{2}{3}+1\right)=\frac{20}{27}\)

\(\Leftrightarrow\left(\frac{2}{3}\right)^x\frac{5}{3}=\frac{20}{27}\)

\(\Leftrightarrow\left(\frac{2}{3}\right)^x=\frac{20}{27}.\frac{3}{5}=\frac{4}{9}\)

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

=> x = 2