Question

1 si 2
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Dau coroana si 28 de puncte.
Materie clasa a 10-a.
Multumesc anticipat.​

Answer 1

Răspuns:

1...17

2..ceva mai laboriosa , vezi atasament ..nu am explicat in amanunt calculul algebric, adica *1= (-1) *(-1) sper sa  fie...implicit

Explicație pas cu pas:

(√3+√7)²=3+7+2√14=10+2√14

[10+2√14]=10+[2√14]=10+[√56]

7=[7]=[√49]<[√56]<[64]=[8]=8

deci

10+[√56]=10+7=17

pt 2, vezi atasamente

Answer 2

1)

$\it (\sqrt3+\sqrt7)^2=10+2\sqrt{21}=10+\sqrt{84}\\ \\ 9=\sqrt{81}<\sqrt{84}<\sqrt{100}=10 \Rightarrow 9<\sqrt{84}<10|_{+10} \Rightarrow \\ \\ \Rightarrow19<10+\sqrt{84}<20 \Rightarrow [10+\sqrt{84}]=9 \Rightarrow [(\sqrt3+\sqrt7)^2]=19$

$\it 2)\ \dfrac{2x-1}{1-x}\geq \dfrac{3x+2}{1-2x} |_{\cdot{(-1)}} \Rightarrow \dfrac{2x-1}{x-1}\leq \dfrac{3x+2}{2x-1} \Rightarrow\\ \\ \\ \Rightarrow \dfrac{2x-1}{x-1}- \dfrac{3x+2}{2x-1}\leq0 \Rightarrow \dfrac{(2x-1)(2x-1)-(x-1)(3x+2)}{(x-1)(2x-1)}\leq0\Rightarrow\\ \\ \\ \Rightarrow \dfrac{4x^2-4x+1-3x^2-2x+3x+2}{2x^2-3x+1}\leq0 \Rightarrow \dfrac{x^2-3x+3}{2x^2-3x+1}\leq0\ \ \ \ \ (*)$

$\it x^2-3x+3 = \Big(x-\dfrac{3}{2}\Big)^2+\dfrac{3}{4}>0 \stackrel{(*)}{\Longrightarrow} 2x^2-3x+1<0 \Rightarrow\\ \\ \\ \Rightarrow 2x^2-x-2x+1<0 \Rightarrow x(2x-1)-(2x-1)<0 \Rightarrow\\ \\ \Rightarrow (2x-1)(x-1)<0 \Rightarrow x\in\Big(\dfrac{1}{2},\ 1\Big)$