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\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+c}=a\left(\frac{a}{b+c}\right)+b\left(\frac{b}{a+c}\right)+c\left(\frac{c}{a+b}\right)\)
\(=a\left(\frac{a+b+c}{b+c}-1\right)+b\left(\frac{a+b+c}{a+c}-1\right)+c\left(\frac{a+b+c}{a+b}-1\right)\)
\(=\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)-a-b-c\)
\(=a+b+c-a-b-c=0\)
\(A=3x^2+\left(x-2\right)^2+1\)
\(A=3x^2+x^2-4x+4+1\)
\(A=4x^2-4x+1+4\)
\(A=\left(2x-1\right)^2+4\ge4\forall x\)
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 \(Min_A=4\Leftrightarrow x=\frac{1}{2}\)
(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 : \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=2^2=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=4-2\cdot\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=4-2\cdot\left(\dfrac{a+b+c}{abc}\right)=4-2\cdot\dfrac{abc}{abc}=4-2\cdot1=2\)
\(\left(x-1\right)^3+\left(2x-1\right)^3=\left(3x-2\right)^3\)
\(\left(3x-2\right)\left[\left(x-1\right)^2-\left(x-1\right)\left(2x-1\right)+\left(2x-1\right)^2-\left(3x-2\right)^2\right]=0\)
\(\left(3x-2\right).\left(-3\right)\left(2x^2-3x+1\right)=0\)
\(\left(3x-2\right)\left(x-1\right)\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{2}\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy ....
\(x^8+98x^4+1\)
\(=\left(x^8+2x^4+1\right)+96x^4\)
\(=\left(x^4+1\right)^2+16\left(x^4+1\right)x^2+64x^4-16\left(x^4+1\right)x^2+32x^4\)
\(=\left(x^4+1+8x^2\right)^2-16x^2\left(x^4+1-2x^2\right)\)
\(=\left(x^4+8x^2+1\right)^2-\left[4x\left(x^2-1\right)\right]^2\)
\(=\left(x^4+8x^2+1-4x^3+4x\right)\left(x^4+8x^2+1+4x^3-4x\right)\)
x8 + 98x4 + 1
= (x8 + 2x4 + 1 ) + 96x4
= (x4 + 1)2 + 16x2(x4 + 1) + 64x4 - 16x2(x4 + 1) + 32x4
= (x4 + 1 + 8x2)2 – 16x2(x4 + 1 – 2x2)
= (x4 + 8x2 + 1)2 - 16x2(x2 – 1)2
= (x4 + 8x2 + 1)2 - (4x3 – 4x )2
= (x4 + 4x3 + 8x2 – 4x + 1)(x4 - 4x3 + 8x2 + 4x + 1)
\(\left(x^2+3x+4\right)^2=\left(x^2+3x+\frac{9}{4}+\frac{7}{4}\right)^2\)
\(=[\left(x+\frac{3}{2}\right)^2+\frac{7}{4}]^2\ge\left(\frac{7}{4}\right)^2=\frac{49}{16}\)
Vay GTNN của A là \(\frac{49}{18}\) đạt được khi x= \(-\frac{3}{2}\)