\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+c}=a\left(\frac{a}{b+c}\right)+b\left(\frac{b}{a+c}\right)+c\left(\frac{c}{a+b}\right)\)

\(=a\left(\frac{a+b+c}{b+c}-1\right)+b\left(\frac{a+b+c}{a+c}-1\right)+c\left(\frac{a+b+c}{a+b}-1\right)\)

\(=\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)-a-b-c\)

\(=a+b+c-a-b-c=0\)

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

\(A=3x^2+x^2-4x+4+1\)

\(A=4x^2-4x+1+4\)

\(A=\left(2x-1\right)^2+4\ge4\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\)

\(A=3x^2+\left(x-2\right)^2+1=4x^2-4x+5=\left(2x-1\right)^2+4\)

Vì \(\left(2x-1\right)^2\ge0\Rightarrow A\ge4\)

Dấu ''='' xảy ra \(\Leftrightarrow x=\frac{1}{2}\)

Vậy \(Min_A=4\Leftrightarrow x=\frac{1}{2}\)

\(\text{ C = 3 - | x + 2 |}\)

               \(\left|x+2\right|\ge0\)

\(\Rightarrow3-\left|x+2\right|\ge3-0\)

\(\Rightarrow3-\left|x+2\right|\ge3\)

\(\Rightarrow C\ge3\)

\(\Rightarrow C=3\Leftrightarrow\left|x+2\right|=0\)

                    \(\Rightarrow x+2=0\)

                     \(\Rightarrow x=0-2\)

                     \(\Rightarrow x=-2\)

Vậy \(\text{Max C = 3 }\Leftrightarrow x=-2\)

\(!x+2!\ge0\Leftrightarrow3-!x+2!\le3\)

"=" xảy ra khi x=-2

\(!3x-15!\ge0\)

\(!3x-15!+8\ge8\)

dấu = xảy ra khi x=5

(x²-5x+6)\(\sqrt{1-x}\)=0

(x-3)(x-2)\(\sqrt{1-x}\)=0

đk: 1-x ≥ 0⇔ x ≤ 1

\(\left[{}\begin{matrix}x-3=0\\x-2=0\\\sqrt{1-x}=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=3\\x=2\\1-x=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=3\\x=2\\x=1\end{matrix}\right.\)

nhận cả 3 nghiệm này vì điều kiện là của phương trình căn thì đã thỏa rồi nhé bạn

Ta có : \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=2^2=4\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=4-2\cdot\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=4-2\cdot\left(\dfrac{a+b+c}{abc}\right)=4-2\cdot\dfrac{abc}{abc}=4-2\cdot1=2\)