A B C D E F H I K Q R

a) Xét \(\Delta EAB\)và \(\Delta FAC\)có :

\(\widehat{BEA}=\widehat{CFA}\left(=90^0\right)\)

\(\widehat{A}\)chung

\(\Rightarrow\Delta EAB\approx\Delta FAC\)(g.g)

\(\Rightarrow\frac{EA}{FA}=\frac{BA}{CA}\)(2 cặp cạnh tương ứng tỉ lệ)\(\Rightarrow\frac{EA}{BA}=\frac{FA}{CA}\)(tính chất của tỉ lệ thức)

Xét \(\Delta AEF\)và \(\Delta ABC\)có:

\(\widehat{A}\)chung.

\(\frac{EA}{BA}=\frac{FA}{CA}\)(chứng minh trên)

\(\Rightarrow\Delta AEF\approx\Delta ABC\left(c.g.c\right)\)(điều phải chứng minh)

\(\frac{5x+2}{x-2}=-3\)

ĐKXĐ : x khác 2

=> 5x + 2 = -3( x - 2 )

<=> 5x + 3x = 6 - 2

<=> 8x = 4

<=> x = 1/2 ( tm )

Vậy S = { 1/2 }

Trả lời:

\(\frac{5x+2}{x-2}=-3\left(đkxđ:x\ne2\right)\)

\(\Leftrightarrow5x+2=-3\left(x-2\right)\)

\(\Leftrightarrow5x+2=-3x+6\)

\(\Leftrightarrow5x+3x=6-2\)

\(\Leftrightarrow8x=4\)

\(\Leftrightarrow x=\frac{1}{2}\left(tm\right)\)

Vậy \(S=\left\{\frac{1}{2}\right\}\)

Xét ~~~~\(\left(a-\frac{1}{b}\right)\left(b-\frac{1}{c}\right)\left(c-\frac{1}{a}\right)\ge\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)\left(c-\frac{1}{c}\right)\)\(\Leftrightarrow\frac{\left(ab-1\right)\left(bc-1\right)\left(ca-1\right)}{abc}\ge\frac{\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)}{abc}\)\(\Leftrightarrow\left(ab-1\right)\left(bc-1\right)\left(ca-1\right)\ge\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\)(Do a,b,c không nhỏ hơn 1 nên abc > 0)\(\Leftrightarrow a^2b^2c^2-\left(abc^2+ab^2c+a^2bc\right)+\left(ab+bc+ca\right)-1\ge a^2b^2c^2-\left(a^2b^2+b^2c^2+c^2a^2\right)+\left(a^2+b^2+c^2\right)-1\)\(\Leftrightarrow-\left(abc^2+ab^2c+a^2bc\right)+\left(ab+bc+ca\right)\ge-\left(a^2b^2+b^2c^2+c^2a^2\right)+\left(a^2+b^2+c^2\right)\)\(\Leftrightarrow2\left(a^2b^2+b^2c^2+c^2a^2\right)-2\left(abc^2+ab^2c+a^2bc\right)\ge2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)\)\(\Leftrightarrow\left(bc-ca\right)^2+\left(ab-bc\right)^2+\left(ca-ab\right)^2\ge\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)\(\Leftrightarrow c^2\left(a-b\right)^2+b^2\left(a-c\right)^2+a^2\left(b-c\right)^2\ge\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)\(\Leftrightarrow\left(c^2-1\right)\left(a-b\right)^2+\left(b^2-1\right)\left(a-c\right)^2+\left(a^2-1\right)\left(b-c\right)^2\ge0\)(Đúng do a,b,c không nhỏ hơn 1)

Đẳng thức xảy ra khi a = b = c hoặc (a,b,c) = (1,1,k) (k bất kì) và các hoán vị

4ccbuiyhn4c8ysd

\(\frac{x}{x^2-x+1}=\frac{2}{3}\)

\(\Rightarrow3x=2\left(x^2-x+1\right)\)

\(\Leftrightarrow2x^2-2x+2-3x=0\)

\(\Leftrightarrow2x^2-5x+2=0\)

\(\Leftrightarrow2x^2-4x-x+2=0\)

\(\Leftrightarrow2x\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\2x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}\)

Với x = 2 => Q = 4/21

Với x = 1/2 => Q = 4/21 :))

"Trần Nhật Quỳnh" có cách này ngắn gọn hơn nữa.

Ta có: 

\(\frac{x}{x^2-x+1}=\frac{2}{3}\) \(\Rightarrow\frac{x^2-x+1}{x}=\frac{3}{2}\)\(\Rightarrow x-1+\frac{1}{x}=\frac{3}{2}\)

\(\Rightarrow x+\frac{1}{x}=\frac{5}{2}\)

Lại có:

\(Q=\frac{x^2}{x^4+x^2+1}\)

\(\frac{1}{Q}=\frac{x^4+x^2+1}{x^2}\)

\(\frac{1}{Q}=x^2+1+\frac{1}{x^2}\)

\(\frac{1}{Q}=\left(x^2+2x^2.\frac{1}{x^2}+\frac{1}{x^2}\right)-2x^2.\frac{1}{x^2}\)

\(\frac{1}{Q}=\left(x+\frac{1}{x}\right)^2-2\)

Vì \(x+\frac{1}{x}=\frac{5}{2}\)nên

\(\frac{1}{Q}=\left(\frac{5}{2}\right)^2-2\)

\(\frac{1}{Q}=\frac{25}{4}-2\)

\(\frac{1}{Q}=\frac{21}{4}\)

\(\Rightarrow Q=\frac{4}{21}\)

Vậy \(Q=\frac{4}{21}\)