Tìm x
\(\left(2x+\frac{1}{3}\right)^3=\frac{-8}{27}\)
Câu 1: Tìm x
1/\(\frac{1+3x}{2}=\frac{7}{3}-\frac{x+1}{6}\)
2,\(\left|\frac{2x+1}{3}\right|=\left|\frac{x-1}{5}\right|\)
3,\(\left(\frac{4}{3}+2x\right)^3=\frac{8}{-27}\)
Tìm x
\(\left(2x-1\right)^3=\frac{8}{27}\)
\(\Rightarrow\left(2x-1\right)^3=\left(\frac{2}{3}\right)^3\)
\(\Rightarrow2x-1=\frac{2}{3}\)
biểu thức trở nên đơn giản r đó bn tự giải nha
tíc mình nha
(2x- 1)3 = \(\frac{8}{27}\)
(2x- 1)3= \(\frac{2^3}{3^3}\)
2x- 1= \(\frac{2}{3}\)
x= (\(\frac{2}{3}\)+1): 2
x= \(\frac{5}{6}\)
\(\left(2x-1\right)^3=\frac{8}{27}\)
\(\left(2x-1\right)^3=\left(\frac{2}{3}\right)^3\)
\(2x-1=\frac{2}{3}\)
\(2x=\frac{2}{3}+1\)
\(2x=\frac{5}{3}\)
\(x=\frac{5}{3}:2\)
\(x=\frac{5}{3}.\frac{1}{2}\)
\(x=\frac{5}{6}\)
k chi mình nhé, xong mình k cho các bạn :))
Bài 8: Tìm số nguyên x biết
a) \(\left(\frac{-12}{27}+\frac{2}{3}\right)+\frac{-2}{9}\le x\le\left(\frac{11}{7}+\frac{2}{5}\right)+\frac{7}{5}+\frac{3}{7}\) b\(\frac{-x}{2}+\frac{2x}{3}+\frac{x+1}{4}+\frac{2x+1}{6}=\frac{8}{3}\)
c)\(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)
Tìm x, biết:
a) \(\left(\frac{-3}{4}\right)^{3x-1}=\frac{-27}{64}\)
b) \(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{265}\)
c) \(\frac{\left(x+3\right)^5}{\left(x+3\right)^2}=\frac{64}{27}\)
d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)
a) \(\left(-\frac{3}{4}\right)^{3x-1}=\frac{-27}{64}\)
\(\Leftrightarrow\left(-\frac{3}{4}\right)^{3x-1}=\left(-\frac{3}{4}\right)^3\)
\(\Leftrightarrow3x-1=3\)
\(\Leftrightarrow3x=4\)
\(\Leftrightarrow x=\frac{4}{3}\)
b) Đề sai ! Sửa :
\(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{625}\)
\(\Leftrightarrow\left(\frac{4}{5}\right)^{2x+5}=\left(\frac{4}{5}\right)^4\)
\(\Leftrightarrow2x+5=4\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\frac{1}{2}\)
c) \(\frac{\left(x+3\right)^5}{\left(x+5\right)^2}=\frac{64}{27}\)
\(\Leftrightarrow\left(x+3\right)^3=\left(\frac{4}{3}\right)^3\)
\(\Leftrightarrow x+3=\frac{4}{3}\)
\(\Leftrightarrow x=-\frac{5}{3}\)
d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)
\(\Leftrightarrow\left(x-\frac{2}{15}\right)^3=\left(\frac{2}{15}\right)^3\)
\(\Leftrightarrow x-\frac{2}{15}=\frac{2}{15}\)
\(\Leftrightarrow x=\frac{4}{15}\)
Tìm x:
a)\(2.\left(3x-\frac{1}{2}\right)-2x=\frac{1}{2}\left(2x-3\right)\)
b)\(\left(2x-\frac{3}{5}\right)^2=\frac{4}{25}\)
c)\(\left(3x-1\right)^3=27\)
d)\(5-\left|x\right|=2\)
e)|2x+1|-3=3
f)|3-2x|=5
\(\left(5-x\right)\left(3x-\frac{1}{4}\right)=0\)
Tìm x biết
\(|x+\frac{1}{3}|+\frac{4}{5}=|\left(-3,2\right)+\frac{2}{5}|+\left(27-\frac{3}{5}\right)\left(27-\frac{3^2}{6}\right)\left(27-\frac{3^3}{7}\right)...\left(27-\frac{3^{2010}}{2014}\right)\)
\(\left|x+\frac{1}{3}\right|+\frac{4}{5}=\left|-3,2+\frac{2}{5}\right|+\left(27-\frac{3}{5}\right)\left(27-\frac{3^2}{6}\right)...\left(27-\frac{3^5}{9}\right)...\left(27-\frac{3^{2010}}{2014}\right)\)
\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}+\left(27-\frac{3^2}{6}\right)\left(27-\frac{3^3}{7}\right)...\left(27-27\right)...\left(27-\frac{3^{2010}}{2014}\right)\)
\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\)
\(\Leftrightarrow\left|x+\frac{1}{3}\right|=2\)
\(\Rightarrow\hept{\begin{cases}x+\frac{1}{3}=2\\x+\frac{1}{3}=-2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=-\frac{7}{3}\end{cases}}}\)
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.
tìm x:
\(|x+\frac{1}{3}|+\frac{4}{5}=|\left(-3,2\right)+\frac{2}{5}|+\left(27-\frac{3}{5}\right)\left(27-\frac{3^2}{7}\right)\left(27-\frac{3^3}{7}\right)...\left(27-\frac{3^{2010}}{2014}\right)\)
tìm x biết :
a) \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
b) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
c) \(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
Bổ sung câu a: \(\Rightarrow\) \(\left[\begin{array}{nghiempt}\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\\\left(x+\frac{1}{5}\right)^2=\left(-\frac{3}{5}\right)^2\end{array}\right.\)\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=\frac{2}{5}\\x=-\frac{4}{5}\end{array}\right.\)
Tìm x:
a)\(\left(2x+3\right)^2=\frac{9}{21}\)
b)\(\left(3x-1\right)^3=\frac{-8}{27}\)
c)\(x^{10}=25.x^8\)
d)\(\left(x^4\right)^2=\frac{x^{12}}{x^{15}}\)
e)\(\left(x+3\right)^3=\left(x+3\right)^5\)
Cầu giúp đỡ~
e)
\(\left(x+3\right)^3=\left(x+3\right)^5\)
\(\Rightarrow\)\(x+3=1;0\)
TH1: TH2
\(x+3=0\) \(x+3=1\)
\(x=-3\) \(x=-2\)
\(x\in\left\{-3;-2\right\}\)
Tìm x:
a, \(\left(\frac{x}{3}-\frac{2}{5}\right):\frac{-2}{5}+\frac{1}{2}x=\frac{-3}{4}\)
b. \(\left(\frac{-1}{8}x-\frac{3}{4}\right)-\frac{-8}{5}=\frac{5}{3}-x\)
c. \(\left(x-\frac{1}{3}\right)^{x+1}=\left(x-\frac{1}{3}\right)^x\)
d. \(\frac{x+2}{3}+\frac{2x-1}{4}\)
e. \(\left(\frac{3}{2}-x\right)^4=64^2\)
f. \(\left(5-\frac{x}{2}\right)^3-\frac{1}{27}=0\)
Nhờ các bạn giúp mik tí nha!