Violympic toán 8

\(x^2+y^2=\left(x+y\right)^2-2xy=m^2-2n\\ x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=m^3-3mn\\ \Rightarrow x^5+y^5=\left(x^3+y^3\right)\left(x^2+y^2\right)-x^2y^2\left(x+y\right)=\left(m^3-3mn\right)\left(m^2-2n\right)-n^2m\\ \Rightarrow x^7+y^7=\left(x^2+y^2\right)\left(x^5+y^5\right)-x^2y^2\left(x^3+y^3\right)=.....\)

tick mik nha

Bài 2: 

a: Ta có: \(A=3x^2+x+2\)

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

\(=3\left(x^2+2\cdot x\cdot\dfrac{1}{6}+\dfrac{1}{36}+\dfrac{23}{36}\right)\)

\(=3\left(x+\dfrac{1}{6}\right)^2+\dfrac{23}{12}\ge\dfrac{23}{12}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{1}{6}\)

Đề sai.

Theo định lý Talet: $\frac{LM}{LN}=\frac{AL}{LD}=1$

Tuy nhiên, không có cơ sở để khẳng định $\frac{KM}{KN}=1$ (xem hình vẽ)

a: Ta có: \(A=16x^2+8x+5\)

\(=16x^2+8x+1+4\)

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

Dấu '=' xảy ra khi \(x=-\dfrac{1}{4}\)

b: Ta có: \(B=2x^2-5x\)

\(=2\left(x^2-\dfrac{5}{2}x\right)\)

\(=2\left(x^2-2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}-\dfrac{25}{16}\right)\)

\(=2\left(x-\dfrac{5}{4}\right)^2-\dfrac{25}{8}\ge-\dfrac{25}{8}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{4}\)

Bài 3: 

a: Ta có: \(A=16x^2+8x+5\)

\(=16x^2+8x+1+4\)

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

Dấu '=' xảy ra khi \(x=-\dfrac{1}{4}\)

b: Ta có: \(B=2x^2-5x\)

\(=2\left(x^2-\dfrac{5}{2}x\right)\)

\(=2\left(x^2-2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}-\dfrac{25}{16}\right)\)

\(=2\left(x-\dfrac{5}{4}\right)^2-\dfrac{25}{8}\ge-\dfrac{25}{8}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{4}\)

a: Ta có: \(\left(x+3\right)^2-\left(x-2\right)^3=\left(x+5\right)\left(x^2-5x+25\right)-108\)

\(\Leftrightarrow x^2+6x+9-x^3+6x^2-12x+8-x^3-125=-108\)

\(\Leftrightarrow7x^2-6x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{6}{7}\end{matrix}\right.\)

Bài 8:

Ta có: \(A=-x^2+2x+4\)

\(=-\left(x^2-2x-4\right)\)

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

\(=-\left(x-1\right)^2+5\le5\forall x\)

Dấu '=' xảy ra khi x=1

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