Violympic toán 8

1,Ta có: \(A=a^3+b^3+ab\)

\(=\left(a+b\right)\left(a^2-ab+b^2\right)+ab\)

\(=a^2-ab+b^2+ab\)

\(=a^2+b^2\)

\(=\left(a+b\right)^2-2ab\)

\(=1-2ab\)

Vì \(a+b=1\Rightarrow a=1-b\)

Khi đó \(A=1-2\left(1-b\right)b\)

\(=1-2b-2b^2\)

\(=2\left(b^2-b+\dfrac{1}{4}\right)+\dfrac{1}{2}\)

\(=2\left(b-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\)

Vì \(2\left(b-\dfrac{1}{2}\right)^2\ge0\Rightarrow A=2\left(b-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(\left(b-\dfrac{1}{2}\right)^2=0\Leftrightarrow b=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{2}\)

Vậy \(MinA=\dfrac{1}{2}\Leftrightarrow a=b=\dfrac{1}{2}\)

2, \(B=\dfrac{2}{6x-5-9x^2}=\dfrac{-2}{9x^2-6x+5}=\dfrac{-2}{\left(3x-1\right)^2+4}\)

Vì \(\left(3x-1\right)^2\ge0\Rightarrow\left(3x-1\right)^2+4\ge4\)

\(\Rightarrow\dfrac{1}{\left(3x-1\right)^2+4}\le\dfrac{1}{4}\)

\(\Rightarrow B=\dfrac{-2}{\left(3x-1\right)^2+4}\ge\dfrac{-2}{4}=\dfrac{-1}{2}\)

Dấu "=" xảy ra khi \(3x-1=0\Leftrightarrow x=\dfrac{1}{3}\)

Vậy \(MinB=\dfrac{-1}{2}\Leftrightarrow x=\dfrac{1}{3}\)

Cách khác :

Bài 1. Ta có : \(a^3+b^3+ab=\left(a+b\right)\left(a^2-ab+b^2\right)+ab=a^2+b^2\)

Áp dụng BĐT Bunhiacopxki , ta có :

\(\left(a^2+b^2\right)\left(1^2+1^2\right)\) ≥ \(\left(a+b\right)^2\)

⇔ \(a^2+b^2\) ≥ \(\dfrac{\left(a+b\right)^2}{2}=\dfrac{1}{2}\)

⇔ GTNN của \(a^2+b^2\) là \(\dfrac{1}{2}\) . Đẳng thức xảy ra khi : \(x=y=\dfrac{1}{2}\)

Bài 2. \(B=\dfrac{2}{6x-5-9x^2}=\dfrac{-2}{9x^2-6x+5}\)

\(B=\dfrac{-4}{2\left(9x^2-6x+5\right)}=\dfrac{-9x^2+6x-5+9x^2-6x+1}{2\left(9x^2-6x+5\right)}\)

\(B=\dfrac{-1}{2}+\dfrac{\left(3x-1\right)^2}{2\left(3x-1\right)^2+8}\)

Do : \(\dfrac{\left(3x-1\right)^2}{2\left(3x-1\right)^2+8}\) ≥ 0 ∀x

⇒ \(\dfrac{-1}{2}+\dfrac{\left(3x-1\right)^2}{2\left(3x-1\right)^2+8}\) ≥ \(\dfrac{-1}{2}\)

⇒ \(B_{Min}=\dfrac{-1}{2}\) ⇔ \(x=\dfrac{1}{3}\)

.

Giải:

a) \(2\sqrt{5}+\sqrt{\left(1-\sqrt{5}\right)^2}\)

\(=2\sqrt{5}+1-\sqrt{5}\)

\(=\sqrt{5}+1\)

Vậy ...

b) \(2\sqrt{5}+\sqrt{6-2\sqrt{5}}\)

\(=2\sqrt{5}+\sqrt{5-2.\sqrt{5}.1+1}\)

\(=2\sqrt{5}+\sqrt{\left(5-1\right)^2}\)

\(=2\sqrt{5}+5-1\)

\(=2\sqrt{5}+4\)

Vậy ...

a: Xét ΔABD và ΔACB có

góc ABD=góc ACB

góc BAD chung

Do đo: ΔABD đồng dạng với ΔACB

b: Ta có: ΔABD đồng dạng với ΔACB

nên AD/AB=AB/AC
=>AD/2=2/4=1/2

=>AD=1cm

=>DC=3cm

Giải:

\(x^8-x^5-\dfrac{1}{x}+\dfrac{1}{x^4}\ge0\)

\(\Leftrightarrow x^4\left(x^8-x^5-\dfrac{1}{x}+\dfrac{1}{x^4}\right)\ge0\)

\(\Leftrightarrow x^{12}-x^9-x^3+1\ge0\)

\(\Leftrightarrow x^9\left(x^3-1\right)-\left(x^3-1\right)\ge0\)

\(\Leftrightarrow\left(x^3-1\right)\left(x^9-1\right)\ge0\)

\(\Leftrightarrow\left(x^3-1\right)\left(x^3-1\right)\left(x^6+x^3+1\right)\ge0\)

\(\Leftrightarrow\left(x^3-1\right)^2\left(x^6+x^3+1\right)\ge0\) (luôn đúng)

Vậy ...

\(\dfrac{9}{\sqrt{x-19}}+\dfrac{16}{\sqrt{y-5}}+\dfrac{25}{\sqrt{z-91}}=24-\sqrt{x-19}-\sqrt{y-5}-\sqrt{z-91}\\ \Leftrightarrow\left(\dfrac{9}{\sqrt{x-19}}+\sqrt{x-19}\right)+\left(\dfrac{16}{\sqrt{y-5}}+\sqrt{y-5}\right)+\left(\dfrac{25}{\sqrt{z-91}}+\sqrt{z-91}\right)=24\)

Áp dụng BDT: Cô-si:

\(\Rightarrow\left(\dfrac{9}{\sqrt{x-19}}+\sqrt{x-19}\right)+\left(\dfrac{16}{\sqrt{y-5}}+\sqrt{y-5}\right)+\left(\dfrac{25}{\sqrt{z-91}}+\sqrt{z-91}\right)\ge2\sqrt{\dfrac{9}{\sqrt{x-19}}\cdot\sqrt{x-19}}+2\sqrt{\dfrac{16}{\sqrt{y-5}}\cdot\sqrt{y-5}}+2\sqrt{\dfrac{25}{\sqrt{z-91}}\cdot\sqrt{z-91}}\\ =2\cdot3+2\cdot4+2\cdot5=24\)Dấu "=" xảy ra khi:\(\left\{{}\begin{matrix}\dfrac{9}{\sqrt{x-19}}=\sqrt{x-19}\\\dfrac{16}{\sqrt{y-5}}=\sqrt{y-5}\\\dfrac{25}{\sqrt{z-91}}=\sqrt{z-91}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-19=9\\y-5=16\\z-91=25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=28\\y=21\\z=116\end{matrix}\right.\)

Vậy các số \(\left\{x;y;z\right\}=\left\{28;21;116\right\}\)

a) \(VT=\left(a-1\right)\left(a-3\right)\left(a-4\right)\left(a-6\right)+10=\left(a^2-7a+6\right)\left(a^2-7a+12\right)+10\)

Đặt : \(a^2-7a+9=t\) , ta có :

\(VT=\left(t-3\right)\left(t+3\right)+10=t^2-9+10=t^2+1>0\)

⇒ đpcm

b) \(\left(ab+bc+ca\right)^2\) ≥ \(3abc\left(a+b+c\right)\)

⇔ \(\left(ab+bc+ca\right)^2-3ab.ac-3ab.bc-3ac.bc\) ≥ 0

Đặt : x = ab ; y = bc ; z = ac . Ta có :

\(\left(x+y+z\right)^2-3xz-3xy-3yz\) ≥ 0

⇔ \(x^2+y^2+z^2-xy-yz-xz\) ≥ 0

⇔ \(2\left(x^2+y^2+z^2-xy-yz-xz\right)\) ≥ 0

⇔ \(x^2-2xy+y^2+y^2-2yz+z^2+x^2-2xz+z^2\) ≥ 0

⇔ \(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\) ≥ 0 ( Luôn đúng )

⇒ đpcm

Ta có: \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b\right)=a^2-ab+b^2=a^2+2ab+b^2-3ab=\left(a+b\right)^2-3ab=1-3ab\)

áp dụng bđt: (a + b)² ≥ 4ab có:

=> \(\dfrac{\left(a+b\right)^2}{4}\ge ab\) <=> \(ab\le\dfrac{1}{4}\)

=> \(ab\le\dfrac{1}{4}\Leftrightarrow-3ab\ge-\dfrac{3}{4}\Rightarrow1-3ab\ge1-\dfrac{3}{4}=\dfrac{1}{4}\)

hay \(a^3+b^3\ge\dfrac{1}{4}\left(đpcm\right)\)

dấu ''='' xảy ra khi a = b = \(\dfrac{1}{2}\)