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

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

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

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

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

Đặt \(A=\sqrt{x^2-6x+36}+\sqrt{x^2-6x+64}=18\)

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

\(\Rightarrow A.B=\left(x^2-6x+64\right)-\left(x^2-6x+36\right)=28\)

mà \(A=18\Rightarrow B=\frac{28}{18}=\frac{14}{9}\)

\(P=\sqrt{\left(x-3\right)^2+4^2}+\sqrt{\left(y-3\right)^2+4^2}+\sqrt{\left(z-3\right)^2+4^2}\)

\(P\ge\sqrt{\left(x-3+y-3+z-3\right)^2+\left(4+4+4\right)^2}=6\sqrt{5}\)

\(P_{min}=6\sqrt{5}\) khi \(x=y=z=1\)

Mặt khác với mọi \(x\in\left[0;3\right]\) ta có:

\(\sqrt{x^2-6x+25}\le\dfrac{15-x}{3}\)

Thật vậy, BĐT tương đương: \(9\left(x^2-6x+25\right)\le\left(15-x\right)^2\)

\(\Leftrightarrow8x\left(3-x\right)\ge0\) luôn đúng

Tương tự: ...

\(\Rightarrow P\le\dfrac{45-\left(x+y+z\right)}{3}=14\)

\(P_{max}=14\) khi \(\left(x;y;z\right)=\left(0;0;3\right)\) và hoán vị

Đặt \(\left\{{}\begin{matrix}\sqrt{2x+3}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\)

\(\Rightarrow b\left(b^2+1\right)-3a^2=\left(a^2+1\right)a-3b^2\)

\(\Rightarrow a^3-b^3+3a^2-3b^2+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a-b\right)\left(3a+3b\right)+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+3a+3b+1\right)=0\)

\(\Leftrightarrow a=b\Rightarrow\sqrt{2x+3}=\sqrt{y}\)

\(\Rightarrow y=2x+3\)

\(\Rightarrow M=x\left(2x+3\right)+3\left(2x+3\right)-4x^2-3\) tới đây chắc chỉ cần bấm máy

Bài 1: 

\(\Leftrightarrow\left(x^2-6x-7\right)^2-\left(3x^2-12x-9\right)^2=0\)

\(\Leftrightarrow\left(3x^2-12x-9-x^2+6x+7\right)\left(3x^2-12x-9+x^2-6x-7\right)=0\)

\(\Leftrightarrow\left(2x^2-6x-2\right)\left(4x^2-18x-16\right)=0\)

\(\Leftrightarrow\left(x^2-3x-1\right)\left(2x^2-9x-8\right)=0\)

hay \(x\in\left\{\dfrac{3+\sqrt{13}}{2};\dfrac{3-\sqrt{13}}{2};\dfrac{9+\sqrt{145}}{4};\dfrac{9-\sqrt{145}}{4}\right\}\)

Chọn A