Question

ajutați ma va rog cu ex din sub al doilea am teza mâine va rogggg

Answer 1

$\bf 3. \ \it a=\sqrt{20};\ \ b=3\sqrt2=\sqrt{3^2\cdot2}=\sqrt{9\cdot2}=\sqrt{18}\\ \\ \sqrt{18}<\sqrt{20} \Rightarrow b<a.\\ \\ \bf 4.\ \ \it dreptunghi\\ \\ \bf 5.\ \ \it \mathcal{A}{_{romb}}=\dfrac{d_1\cdot d_2}{2}=\dfrac{8\cdot6^{(2}}{2}=8\cdot3=24cm^2\\ \\ \bf 6.\ \ \it de\ 45^o$

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$\it \dfrac{x+7}{3}+ \dfrac{x+4}{2}= \dfrac{5x-2}{4}- \dfrac{2x+5}{3}\Big|_{\cdot12} \Rightarrow 4x+28+6x+24=15x-6-8x-20 \Rightarrow \\ \\ \\ \Rightarrow 10x+52=7x-26 \Rightarrow 10x-7x=-26-52 \Rightarrow 3x=-78|_{:3} \Rightarrow x=-26$