Ta có: VT =82−32−41−2=1−2​82​−32​−4​

=82−2.42−41−2=82−42−41−2=1−2​82​−2.42​−4​=1−2​82​−42​−4​

=42−41−2=−4(1−2)1−2=−4==1−2​42​−4​=1−2​−4(1−2​)​=−4= V P

Vậy 82−32−41−2=−41−2​82​−32​−4​=−4

b) ĐKXĐ: {�≥0�+2≠0�−2≠0�−4≠0⇔{�≥0�≠2�≠4⇔{�≥0�≠4⎩⎨⎧​x≥0x​+2=0x​−2=0x−4=0​⇔⎩⎨⎧​x≥0x​=2x=4​⇔{x≥0x=4​.

Vậy ĐKXĐ của $P$ là $x\ge 0$, $x\ne 4$.

Với $x\ge 0$, $x\ne 4$ ta có:

�=(2�+2−1�−2+7�−4).(�−1)P=(x​+22​−x​−21​+x−47​).(x​−1)

=(2�+2−1�−2+7(�−2)(�+2)).(�−1)=(x​+22​−x​−21​+(x​−2)(x​+2)7​).(x​−1)

=(2(�−2)−(�+2)+7(�−2)(�+2)).(�−1)=((x​−2)(x​+2)2(x​−2)−(x​+2)+7​).(x​−1)

=2�−4−�−2+7(�−2)(�+2).(�−1)=(x​−2)(x​+2)2x​−4−x​−2+7​.(x​−1)

=�+1(�−2)(�+2).(�−1)=(x​−2)(x​+2)x​+1​.(x​−1)

=�−1�−4=x−4x−1​.

Vậy �=�−1�−4P=x−4x−1​ với $x\ge 0$, $x\ne 4$.

ĐKXĐ: \(x\ne0\)

- Với \(x< 0\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2+1}+3>0\\\dfrac{1}{x}-3< 0\Rightarrow\left(\dfrac{1}{x}-3\right)\left(\sqrt{9x^2-6x+2}+3\right)< 0\end{matrix}\right.\)

\(\Rightarrow\) Phương trình vô nghiệm

- Với \(x\ge\dfrac{1}{3}\) tương tự ta có \(\dfrac{1}{x}-3\le0\Rightarrow\left\{{}\begin{matrix}VT>0\\VT\le0\end{matrix}\right.\) nên pt vô nghiệm

- Với \(0< x< \dfrac{1}{3}\)

\(\Rightarrow x\sqrt{x^2+1}+3x=\left(1-3x\right)\left(\sqrt{\left(1-3x\right)^2+1}+3\right)\)

Đặt \(1-3x=y>0\)

\(\Rightarrow x\sqrt{x^2+1}+3x=y\left(\sqrt{y^2+1}+3\right)\)

\(\Leftrightarrow x\sqrt{x^2+1}-y\sqrt{y^2+1}+3\left(x-y\right)=0\)

\(\Leftrightarrow\dfrac{x^2\left(x^2+1\right)-y^2\left(y^2+1\right)}{x\sqrt{x^2+1}+y\sqrt{y^2+1}}+3\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(\dfrac{\left(x+y\right)\left(x^2+y^2\right)+x+y}{x\sqrt{x^2+1}+y\sqrt{y^2+1}}+3\right)=0\) (1)

Do \(\dfrac{\left(x+y\right)\left(x^2+y^2\right)+x+y}{x\sqrt{x^2+1}+y\sqrt{y^2+1}}+3>0;\forall x;y>0\)

\(\left(1\right)\Leftrightarrow x-y=0\Leftrightarrow x-\left(1-3x\right)=0\)

\(\Rightarrow x=\dfrac{1}{4}\)

b.

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-3\\x_1x_2=-1\end{matrix}\right.\)

\(T=\dfrac{3\left|x_1-x_2\right|}{x_1^2x_2+x_1x_2^2}=\dfrac{3\sqrt{\left(x_1-x_2\right)^2}}{x_1x_2\left(x_1+x_2\right)}=\dfrac{3\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}}{x_1x_2\left(x_1+x_2\right)}\)

\(=\dfrac{3\sqrt{\left(-3\right)^2-4.\left(-1\right)}}{-1.\left(-3\right)}=\sqrt{13}\)

\(P=x^2-x\left(15-x\right)+\left(15-x\right)^2=3x^2-45x+225\)

\(P=3x\left(x-9\right)+225\)

Do \(0\le x\le6\Rightarrow x-9< 0\Rightarrow3x\left(x-9\right)\le0\)

\(\Rightarrow P\le225\)

\(P_{max}=225\) khi \(\left(x;y\right)=\left(0;15\right)\)

\(P=3x^2-45x+162+63=3\left(9-x\right)\left(6-x\right)+63\)

Do \(x\le6\Rightarrow\left\{{}\begin{matrix}9-x>0\\6-x\ge0\end{matrix}\right.\) \(\Rightarrow3\left(9-x\right)\left(6-x\right)\ge0\)

\(\Rightarrow P\ge63\)

\(P_{min}=63\) khi \(\left(x;y\right)=\left(6;9\right)\)

Bài 1: