Question

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

Answer 1

Răspuns:

Explicație pas cu pas:

1.

7 + 4√3 = 4 + 2*2*√3 + 3 = (2 + √3)^2

7 - 3√3 = 4 - 2*2*√3 + 3 = (2 - √3)^2

a = 2 + √3 + 2 - √3 = 4, numar natural

2.

f(x) = 2x^2 - 5x + 2

f(2x) = 2(2x)^2 - 5(2x) + 2 = 8x^2 - 10x + 2

8x^2 - 10x + 2 ≤ 0

Δ = 100 - 64 = 36

x1 = (10 + 6)/16 = 16/16 = 1

x2 = (10 - 6)/16 = 4/16 = 1/4

x ∈ {1/4, 1]

Answer 2

$\it 7+4\sqrt3=4+3+4\sqrt3=2^2+2\cdot2\sqrt3+(\sqrt3)^2=(2+\sqrt3)^2\\ \\ 7-4\sqrt3=4+3-4\sqrt3=2^2-2\cdot2\sqrt3+(\sqrt3)^2=(2-\sqrt3)^2\\ \\ Expresia\ din\ enun\c{\it t}\ devine:\\ \\ \sqrt{(2+\sqrt3)^2}+\sqrt{(2-\sqrt3)^2}=|\underbrace{2+\sqrt3}_{>0}|+|\underbrace{2-\sqrt3}_{>0}|=2+\sqrt3+2-\sqrt3=4\in\mathbb{N}$

$\it 2)\ f(x)=2x^2-5x+2\ \Rightarrow\ f(2x)=2\cdot(2x)^2-5\cdot2x+2\ \Rightarrow\\ \\ \Rightarrow\ f(2x)=8x^2-10x+2\\ \\ f(2x)\leq0\ \Rightarrow\ 8x^2-10x+2\leq0|_{:2}\ \Rightarrow\ 4x^2+5x+1\leq0\\ \\Vom\ determina\ r\breve{a}d\breve{a}cinile\ ecua\c{\it t}iei\ \ 4x^2-5x+1=0.\\ \\ 4x^2-5x+1=0\ \Leftrightarrow\ 4x^2-x-4x+1=0\ \Leftrightarrow\ x(4x-1)-(4x-1)=0\ \Leftrightarrow$$\it \ \Leftrightarrow\ (4x-1)(x-1)=0\ \Leftrightarrow\ \begin{cases}\it 4x-1=0\ \Leftrightarrow\ 4x=1\ \Leftrightarrow\ x_1=\dfrac{1}{4}\\ \\ \it x-1=0\ \Leftrightarrow\ x_2=1\end{cases}\\ \\ f(2x)\leq0\ \Leftrightarrow\ x\in \Big[\dfrac{1}{4},\ 1\Big]$