\(\left(\frac{-3}{4}\right)^{3x-1}=\frac{256}{81}\)

\(\Rightarrow\left(\frac{-3}{4}\right)^{3x}:\left(\frac{-3}{4}\right)=\frac{256}{81}\)

\(\Rightarrow\left(\frac{-3}{4}\right)^{3x}=\frac{256}{81}.\left(\frac{-3}{4}\right)\)

\(\Rightarrow\left(\frac{-3}{4}\right)^{3x}=\frac{64}{27}\)

\(\Rightarrow\left(-\frac{3}{4}\right)^{3x}=\left(-\frac{4}{3}\right)^3\)

\(\Rightarrow1=\left(-\frac{4}{3}\right)^3.\left(-\frac{4}{3}\right)^{3x}\)

\(\Rightarrow1=\left(-\frac{4}{3}\right)^{-x}\)

\(\Rightarrow1=\left(-\frac{3x}{4}\right)\)

\(\Rightarrow1=-3x:4\)

\(\Rightarrow-3x=4\)

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

Ta có:

\(\frac{256}{81}=\frac{4^4}{"-3"^4}=\frac{1}{\frac{"-3"^4}{4}}=\frac{1}{"\frac{-3}{4}"^4}="\frac{-3}{4}"^4="\frac{-3}{4}"^{3x-1}\Rightarrow3x-1=-4\Rightarrow3x=-4+1\)

\(=-3\)

\(\Rightarrow x=-3:1=-1\)

tớ làm bừa đấy sai thì thôi nha

a) x=1

b) x=1

c) x= -(245/81)

d) x= 1/27

e) x=3

g) x=4

cần gấp

a)(x-2)^4=4^4

=(x-2)=4 sr x=4+2

sr x=6

xin lỗi nha mình chỉ biết làm bài a thôi mong cậu thông cảm và kb với mình nhé

Mình biết bài nào làm bài đó thôi nhé

a) (x-2)⁴ = 256

=> x-2 = 4

x = 4+2

x = 6

b) \(\frac{x^4}{256}=81\)

\(\Rightarrow\frac{x^4}{4^4}=3^4\)

Từ đây có thể làm theo 2 cách khác nhau

C1 : \(\frac{x^4}{4^4}=81\)

\(\Rightarrow\left(\frac{x}{4}\right)^4=81\)

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

x = 3.4

x = 12

C2 : \(\frac{x^4}{4^4}=3^4\)

=> x⁴ = 3⁴.4⁴

x⁴ = (3.4)⁴

x⁴ = 12⁴

<=> x = 4

c) \(125.\left(x+\frac{4}{5}\right)^3=729\)

\(\left(x+\frac{4}{5}\right)^3=729:125\)

\(\left(x+\frac{4}{5}\right)^3=5,832\)

\(\Rightarrow x+\frac{4}{5}=1,8=\frac{9}{5}\)

\(x=\frac{9}{5}-\frac{4}{5}\)

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

sai đề                                     

Tìm x như thường -_-

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?)