Question

Cho x thỏa mãn 

\(\sqrt{x^2-6x+36}\)+\(\sqrt{x^2-6x+64}\)=7

TÌm GTBT : A= \(\sqrt{4x^2-24x+256}\)-2.\(\sqrt{x^2-6x+36}\)

Answer 1

đặt \(\sqrt{x^2-6x+36}=\)M;\(\sqrt{x^2-6x+64}=\)N ,hiển nhiên M\(\ne\)N

M+N=7 <=>(M+N)(M-N)=7(M-N) <=>M²-N²=7(M-N) <=>-28=7(M-N) <=>N-M=4

A=2N-2M=2.4=8

Answer 2

Đặt \(\sqrt{x^2-6x+36}=a\ge0\Rightarrow\sqrt{x^2-6x+64}=\sqrt{a^2+28}\)

Vậy ta có phương trình :

\(a+\sqrt{a^2+28}=7\Leftrightarrow\sqrt{a^2+28}=7-a\Leftrightarrow\hept{\begin{cases}a\le7\\a^2+28=a^2-14a+49\end{cases}\Leftrightarrow a=\frac{3}{2}}\)

ta có : \(A=\sqrt{4\left(x^2-6x+36\right)+112}-2\sqrt{x^2-6x+36}=\sqrt{4a^2+112}-2a=8\)

Answer 3

\(\sqrt{x^2-6x+36}-\sqrt{x^2-6x+64}=7\)

<=> \(\sqrt{x^2-6x+64}=\sqrt{x^2-6x+36}-7\)(*)

Ta có :

\(A=\sqrt{4x^2-24x+256}-2\sqrt{x^2-6x+36}\)

\(=\sqrt{4\left(x^2-6x+64\right)}-2\sqrt{x^2-6x+36}\)

\(=2\left(\sqrt{x^2-6x+64}-\sqrt{x^2-6x+36}\right)\). Thay (*), ta có :

\(A=2\left(\sqrt{x^2-6x+36}-7-\sqrt{x^2-6x+36}\right)=2\left(-7\right)=-14\)

Vậy A = -14