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\(3+\frac{6}{8}-\frac{2}{4}\)
\(=\frac{24}{8}+\frac{6}{8}-\frac{2}{4}\)
\(=\frac{15}{4}-\frac{2}{4}\)
\(=\frac{13}{4}\)
~ Ủng hộ nhé ~
\(\left(x^2+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+5=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-5\\x=5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x\in\varnothing\\x=5\end{cases}}}\)
Vậy x=5
TK MK ĐÊ RỒI MK LÀM TIẾP!
\(x-\frac{1}{9}=\frac{8}{3}\)
\(\Leftrightarrow x=\frac{8}{3}+\frac{1}{9}\)
\(\Leftrightarrow x=\frac{24+1}{9}=\frac{25}{9}\)
\(\frac{5}{2}\)x\(\frac{1}{3}\)+\(\frac{1}{4}\)= \(\frac{5}{6}\)+\(\frac{1}{4}\)= \(\frac{10}{12}\)+\(\frac{3}{12}\)= \(\frac{13}{12}\)
\(\frac{5}{2}\)-\(\frac{1}{3}\):\(\frac{1}{4}\)= \(\frac{5}{2}\)-\(\frac{1}{3}\)x\(\frac{4}{1}\)= \(\frac{5}{2}\)-\(\frac{4}{3}\)= \(\frac{15}{6}\)-\(\frac{8}{6}\)=\(\frac{7}{6}\)
Chi tiết lun nha, nhớ k đúng cho mik nha :33
\(\frac{5}{2}\times\frac{1}{3}+\frac{1}{4}\)
\(=\frac{5}{6}+\frac{1}{4}\)
\(=\frac{10}{12}+\frac{3}{12}\)
\(=\frac{13}{12}\)
\(\frac{5}{2}-\frac{1}{3}\div\frac{1}{4}\)
\(=\frac{5}{2}-\frac{1}{3}\times\frac{4}{1}\)
\(=\frac{5}{2}-\frac{4}{3}\)
\(=\frac{15}{6}-\frac{8}{6}\)
\(=\frac{7}{6}\)
Bài 9:
c) Ta có: \(P=\dfrac{a\sqrt{a}-1}{\sqrt{a}-1}\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}-1}\)
\(=a+\sqrt{a}+1\)
d) Ta có: \(Q=\dfrac{a\sqrt{a}+1}{\sqrt{a}+1}\)
\(=\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}+1}\)
\(=a-\sqrt{a}+1\)
\(\frac{x}{12}=\frac{-3}{9}\)
\(\Rightarrow x=\frac{-3}{9}.12=-4\)
Theo đề, ta có:
\(\frac{x}{12}=-\frac{3}{9}\)
\(-\frac{3}{9}=-\frac{1}{3}=-\frac{4}{12}\)
\(\Rightarrow\frac{x}{12}=-\frac{4}{12}\)
\(\Rightarrow x=-4\)
Vậy x = -4