\(\frac{27}{4}=\frac{-x}{3}=\frac{3}{y^2}=\frac{\left(x+3\right)^3}{-4}=\)||t| - 2|/ 8. Tìm x; y ;t
\(\frac{27}{4}=\frac{-x}{3}=\frac{3}{y^2}=\frac{\left(z+3\right)^3}{-4}=\frac{\left|\left|t\right|-2\right|}{8}\)
\(x=-20,25\)
\(y=\frac{2}{3}\)
\(z=-6\)
\(t=-56;56\)
Tìm các số nguyên x,y,z,t biết:
$\frac{27}{4}$274 =$\frac{-x}{3}$−x3 =$\frac{\left(z+3\right)^3}{-4}$(z+3)3−4 =$\frac{\left|t-2\right|}{8}$//t/−2/8
chú ý / là giá trị tuyệt đối
\(\frac{27}{4}\)= \(\frac{-x}{3}\)=\(\frac{3}{y^2}\)=\(\frac{\left(z+3\right)^3}{-4}\)=\(\frac{\left|\left|t\right|-2\right|}{8}\)
Tìm x,y,z,t
Tìm x,y,z,t thuộc Z biết: \(\frac{27}{4}=\frac{-x}{3}=\frac{3}{y^2}=\frac{\left(z+3\right)^2}{-4}=\left|t-2\right|\)
t thuộc N
Tìm các số nguyên x, y, z, t biết: \(\frac{27}{4}\)=\(\frac{-x}{3}\)=\(\frac{3}{y^2}\)=\(\frac{\left(z+3\right)^3}{-4}\)=\(\frac{\left|t\right|-2}{8}\)
Ta có :
\(\frac{-x}{3}=\frac{27}{4}\) \(\Rightarrow\) \(x=\frac{-81}{4}\)
\(\frac{3}{y^2}=\frac{27}{4}\) \(\Rightarrow\) \(y=\sqrt{\frac{4}{9}}=\frac{2}{3}\)
\(\frac{\left(z+3\right)^3}{-4}=\frac{27}{4}\) \(\Rightarrow\) \(z=-3\)
\(\frac{\left|t\right|-2}{8}=\frac{27}{4}\) \(\Rightarrow\) \(\orbr{\begin{cases}t=56\\t=-56\end{cases}}\)
Vậy ...
\(\frac{27}{4}=\frac{-x}{3}\Rightarrow x=-\frac{81}{4}\notinℤ\)
\(y^2=\frac{4}{9}=\left(\frac{2}{3}\right)^2\Rightarrow y=\pm\frac{2}{3}\notinℤ\)
\(\frac{27}{4}=\frac{\left(z+3\right)^{^3}}{-4}\Rightarrow\left(z+3\right)^3=-27=\left(-3\right)^3\Rightarrow z+3=-3\Rightarrow Z=-6\)
\(+)|t|-2=-54\Rightarrow|t|=-52\)(vô lí)
\(+)|t|-2=54\Rightarrow|t|=56\Rightarrow t=\pm56\)
Tim cac so x,y,z,t biet:
\(\frac{27}{4}=\frac{-x}{3}=\frac{3}{y^2}=\frac{\left(z+3\right)^3}{-4}=\frac{\left|t\right|-2}{8}\)
Trinh bai cach lanm ra he
Tìm các số nguyên x,y,z,t biết:
$\frac{27}{4}$274 =$\frac{-x}{3}$−x3 =$\frac{\left(z+3\right)^3}{-4}$(z+3)3−4 =$\frac{\left|t-2\right|}{8}$//t/−2/8
ai nhanh mk tick nha
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:
\(|x+\frac{1}{3}|+\frac{4}{5}=|\left(-3,2\right)+\frac{2}{5}|+\left(27-\frac{3}{5}\right)\left(27-\frac{3^2}{7}\right)\left(27-\frac{3^3}{7}\right)...\left(27-\frac{3^{2010}}{2014}\right)\)