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\(\left(x^2+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+5=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-5\\x=5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x\in\varnothing\\x=5\end{cases}}}\)
Vậy x=5
TK MK ĐÊ RỒI MK LÀM TIẾP!
\(x-\frac{1}{9}=\frac{8}{3}\)
\(\Leftrightarrow x=\frac{8}{3}+\frac{1}{9}\)
\(\Leftrightarrow x=\frac{24+1}{9}=\frac{25}{9}\)
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\(=\dfrac{10}{17}-\dfrac{5}{3}+7-\dfrac{8}{13}+\dfrac{11}{25}\)
\(=\dfrac{30}{51}-\dfrac{85}{51}+\dfrac{175}{25}+\dfrac{11}{25}-\dfrac{8}{13}\)
\(=\dfrac{-55}{51}+\dfrac{186}{25}-\dfrac{8}{13}\)
\(=\dfrac{-55\cdot325+186\cdot663-8\cdot1275}{16575}=\dfrac{95243}{16575}\)
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\(\frac{3}{5}+\frac{2}{5}.\frac{-5}{3}-\frac{1}{15}=\frac{3}{5}+\frac{-2}{3}-\frac{1}{15}\)
\(=\frac{9}{15}+\frac{-10}{15}-\frac{1}{15}\)
\(=\frac{9-10-1}{15}=\frac{-2}{15}\)
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\(A=\dfrac{1+2+...+9}{11+12+...+19}=\dfrac{\left(9+1\right)\times9:2}{\left(19+11\right)\times9:2}=\dfrac{45}{135}=\dfrac{1}{3}\)
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\(-\frac{1}{8}+\left(-\frac{5}{3}\right)\)
= \(\frac{-3}{24}+\left(-\frac{40}{24}\right)\)
\(=\frac{-3+\left(-40\right)}{24}=-\frac{43}{24}\)
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a ) 13/20
B)
C..........................................................
minh dang tính
\(\frac{x}{12}=\frac{-3}{9}\)
\(\Rightarrow x=\frac{-3}{9}.12=-4\)
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