Question

[Minh Thư (BKTT)]

Bài 1: \(\left(2x+1\right)^3=9\left(2x+1\right)\)  

Bài 2: Tìm GTNN của \(A=\left(2x-1\right)^2+\left(3-y\right)^2+2017\)

M.n giúp mk nha. Mk cảm ơn m.n nhiều!!!

Answer 1

[Trần Việt Linh]

Bài 1:
\(\left(2x+1\right)^3=9\left(2x+1\right)\)

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

\(\Leftrightarrow\left(2x+1\right)\left[\left(2x+1\right)^2-9\right]=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x+1-3\right)\left(2x+1+3\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x-2\right)\left(2x+4\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x+1=0\\2x-2=0\\2x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=1\\x=-2\end{array}\right.\)

Bài 2:

\(A=\left(2x-1\right)^2+\left(3-y\right)^2+2017\)

Vì: \(\left(2x-1\right)^2+\left(3-y\right)^2\ge0\)

=> \(\left(2x-1\right)^2+\left(3-y\right)^2+2017\ge2017\)

Dấu "=" xảy ra khi \(x=\frac{1}{2};y=3\)

Vậy GTNN của A là 2017 khi \(x=\frac{1}{2};y=3\)

Answer 2

[soyeon_Tiểubàng giải]

Bài 1:

(2x + 1)³ = 9.(2x + 1)

=> (2x + 1)³ - 9.(2x + 1) = 0

=> (2x + 1).[(2x + 1)² - 9] = 0

=> (2x + 1).(2x + 1 - 3).(2x + 1 + 3) = 0

=> (2x + 1).(2x - 2).(2x + 4) = 0

\(\Rightarrow\left[\begin{array}{nghiempt}2x+1=0\\2x-2=0\\2x+4=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x=-1\\2x=2\\2x=-4\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-1}{2}\\x=1\\x=-2\end{array}\right.\)

Vậy \(x\in\left\{\frac{-1}{2};1;-2\right\}\)

Bài 2:

Có: \(\left(2x-1\right)^2\ge0;\left(3-y\right)^2\ge0\forall x;y\)

=> \(A=\left(2x-1\right)^2+\left(3-y\right)^2+2017\ge2017\)

Dấu "=" xảy ra khi và chỉ khi \(\begin{cases}\left(2x-1\right)^2=0\\\left(3-y\right)^2=0\end{cases}\)\(\Rightarrow\begin{cases}2x-1=0\\3-y=0\end{cases}\)\(\Rightarrow\begin{cases}2x=1\\y=3\end{cases}\)\(\Rightarrow\begin{cases}x=\frac{1}{2}\\y=3\end{cases}\)

Vậy GTNN của A là 2017 khi và chỉ khi \(x=\frac{1}{2};y=3\)