Bài 1.
a) \(-\dfrac{5}{4}-0,75=-\dfrac{5}{4}-\dfrac{3}{4}=\dfrac{-5-3}{4}=\dfrac{-8}{4}=-2\)
b) \(\sqrt{49}+\left(-5\right)^3:\sqrt{25}\)
\(=\sqrt{7^2}+\left(-125\right):\sqrt{5^2}\)
\(=7-125:5\)
\(=7-25\)
\(=-18\)
Bài 2.
a) \(x:\left(-9\right)=\left(-40\right):45\)
\(\Rightarrow\dfrac{x}{-9}=\dfrac{-40}{45}\)
\(\Rightarrow\dfrac{x}{-9}=\dfrac{8}{-9}\)
\(\Rightarrow x=8\)
b) \(\dfrac{1}{4}x^2-\dfrac{2}{3}=8\dfrac{1}{3}\)
\(\Rightarrow\dfrac{x^2}{4}-\dfrac{2}{3}=\dfrac{25}{3}\)
\(\Rightarrow\dfrac{x^2}{4}=\dfrac{25}{3}+\dfrac{2}{3}\)
\(\Rightarrow\dfrac{x^2}{4}=9\)
\(\Rightarrow x^2=9\cdot4\)
\(\Rightarrow x^2=36\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
\(\text{#}Toru\)
Bài 1:
a) \(-\dfrac{5}{4}-0,75\)
\(=-1,25-0,75\)
\(=-2\)
b) \(\sqrt{49}+\left(-5\right)^3:\sqrt{25}\)
\(=7+-125:5\)
\(=7+-25\)
\(=-18\)
Bài 2:
a) \(x:\left(-9\right)=\left(-40\right):45\)
⇔\(-\dfrac{x}{9}=-\dfrac{40}{45}\)
⇔\(-\dfrac{x}{9}=-\dfrac{8}{9}\)
⇒ \(x=8\)
Vậy \(x=8\)
b) \(\dfrac{1}{4}x^2-\dfrac{2}{3}=8\dfrac{1}{3}\)
\(\dfrac{1}{4}x^2=8\dfrac{1}{3}+\dfrac{2}{3}=9\)
TH1:
\(x^2=9:\dfrac{1}{4}=36=6^2\)
⇒ \(x=6\)
TH2:
\(x^2=9:\dfrac{1}{4}=-36=-6^2\)
⇒ \(x=-6\)
Vậy \(x=\left\{\pm6\right\}\)
