ĐKXĐ:...

Đặt \(\sqrt{x^2-5x+5}=t\)(...)

PT trở thành:

\(3t=t^2+2\Leftrightarrow t^2-3t+2=0\)(*)

PT (*) có các hệ số a=1; b=-3; c=2 

\(\Rightarrow a+b+c=1+\left(-3\right)+2=0\) nên pt có 2 nghiệm

\(\left\{{}\begin{matrix}t_1=1\\t_2=2\end{matrix}\right.\)

\(\Rightarrow...\)

HITANDRUN(NEW), tra ở trong não ấy bạn.

Hướng dẫn:

\(A=\left|2x-2\right|+\left|2x-2023\right|\)

\(=\left|2x-2\right|+\left|2023-2x\right|\)

\(\ge\left|2x-2+2023-2x\right|=2021\)

Vậy GTNN của A là 2021, đạt được khi và chỉ khi \(\left(2x-2\right)\left(2023-2x\right)\ge0\)\(\Leftrightarrow1\le x\le\dfrac{2023}{2}\)

Hướng dẫn:

ĐKXĐ:...

Ta có:

\(\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{\left(x+4\right)-\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}+\dfrac{\left(x+5\right)-\left(x+4\right)}{\left(x+4\right)\left(x+5\right)}+\dfrac{\left(x+6\right)-\left(x+5\right)}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+6}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{3}{\left(x+3\right)\left(x+6\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\left(x+3\right)\left(x+6\right)=54\Leftrightarrow...\)

1. We wanted to watch Pinocchio, so we turned to the Movie channel

3.  I like reading books about history of television because I'm interested in it.

5.The Haunted Theatre is frightening, but children love it

Đặt \(x^2-3=t\)

\(\Rightarrow P=t\left(t+5\right)=t^2+5t\)

\(=t^2+2.t.\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{25}{4}\)

\(=\left(t+\dfrac{5}{2}\right)^2-\dfrac{25}{4}\ge-\dfrac{25}{4}\)

\(\Rightarrow...\)

Đồ thị hàm số trên cắt trục hoành tại điểm có hoành độ là

\(\dfrac{-\left(k^2-2k\right)}{k-2}\)\(\Rightarrow2=\dfrac{-k\left(k-2\right)}{k-2}\Leftrightarrow-k=2\Leftrightarrow k=-2\left(tm\right)\)

Xét \(y=0\)\(\Rightarrow...\)

Xét \(y\ne0\). Ta có:

\(\left\{{}\begin{matrix}x^2+y^2+xy+2x=5y\\\left(x^2+2x\right)\left(x+y-3\right)=-3y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+2x=5y-y^2-xy\left(1\right)\\\left(x^2+2x\right)\left(x+y-3\right)=-3y\left(2\right)\end{matrix}\right.\)

Thay (1) vào (2), ta có:

\(\left(5y-y^2-xy\right)\left(x+y-3\right)=-3y\)

\(-y\left(x+y-5\right)\left(x+y-3\right)=-3y\)

\(\Leftrightarrow\left(x+y-5\right)\left(x+y-3\right)=3\left(\cdot\right)\)

Đặt \(x+y-5=t\), phương trình \(\left(\cdot\right)\) trở thành

\(t\left(t+2\right)=3\)\(\Leftrightarrow t^2+2t+1=4\Leftrightarrow\left(t+1\right)^2=4\)

\(\Leftrightarrow\left[{}\begin{matrix}t+1=2\\t+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y-5=1\\x+y-5=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y=6\\x+y=2\end{matrix}\right.\)\(\Rightarrow...\)