Bài 1:

\(M=\left|x+13\right|+64\)

Vì \(\left|x+3\right|\ge0\)

=> \(\left|x+3\right|+64\ge64\)

Vậy GTNN của M là 64 khi x=-13

\(A=\left|x+3\right|+\left|x+5\right|=\left|-\left(x+3\right)\right|+\left|x+5\right|\)

Áp dụng bđt \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\) ta có:

\(A\ge\left|-x-3+x+5\right|=2\)

Vaayj GTNN của A là 2 khi \(-3\le x\le5\)

Bài 2:

a) \(\left(x+10\right)^2=0\)

\(\Leftrightarrow x+10=0\Leftrightarrow x=-10\)

b) \(\left(x-\sqrt{121}\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow x-\sqrt{121}=0\) (vì \(x^2+1>0\) )

\(\Leftrightarrow x=11\)

Bài 1:

a)Ta thấy: \(\left|x+13\right|\ge0\)

\(\Rightarrow\left|x+13\right|+64\ge64\)

\(\Rightarrow M\ge64\)

Dấu = khi x=-13

b)\(\left|x+3\right|+\left|x+5\right|=\left|x+3\right|+\left|-x-5\right|\)

Áp dụng Bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(\left|x+3\right|+\left|-x-5\right|\ge\left|x+3+\left(-x\right)-5\right|=2\)

\(\Rightarrow A\ge2\)

Dấu = khi \(\left(x+3\right)\left(x+5\right)\ge0\)\(\Rightarrow3\le x\le5\)

\(\Rightarrow\begin{cases}\left(x+3\right)\left(x+5\right)=0\\3\le x\le5\end{cases}\)\(\Rightarrow\)\(\begin{cases}x=-3\\x=-5\end{cases}\)

Vậy MinA=2 khi \(\begin{cases}x=-3\\x=-5\end{cases}\)

a) (x+10)²=10

=>x²+20x+100=10

=>x²+20x+90=0

Delta=20²-4(1.90)=40

\(\Rightarrow x_{1,2}=\frac{-20\pm\sqrt{40}}{2}\)

TK: Tìm Min (x^4 + 1) (y^4 + 1) với x + y = căn10 ; x , y > 0 - Thanh Truc

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

Áp dụng Bunhiacopski:

\(\left[\left(x-\dfrac{1}{2}\right)+\left(y-\dfrac{1}{2}\right)\right]^2\le\left(1^2+1^2\right)\left[\left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2\right]=2\cdot\dfrac{1}{2}=1\\ \Leftrightarrow A\le1+1=2\)\(A_{max}=2\Leftrightarrow x=y=1\)

\(x^2+y^2\ge0\Rightarrow x+y=x^2+y^2\ge0\)

\(A_{min}=0\) khi \(x=y=0\)

Cách tìm max khác:

Ta có:

$(x-1)^2\geq 0, \forall x\in\mathbb{R}$

$\Rightarrow x^2+1\geq 2x$

Tương tự: $y^2+1\geq 2y$

$\Rightarrow 2(x+y)\leq x^2+y^2+2=x+y+2$

$\Rightarrow x+y\leq 2$ hay $A\leq 2$
Vậy $A_{\max}=2$ khi $x=y=1$