49x² - 4 = 0

<=> (7x)² - 2² = 0

<=> (7x - 2)(7x + 2) = 0

<=> \(\left[{}\begin{matrix}7x-2=0\\7x+2=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{2}{7}\\x=-\dfrac{2}{7}\end{matrix}\right.\)

\(49x^2-4=0\)

\(\Rightarrow49x^2=4\)

\(\Rightarrow x^2=\dfrac{4}{49}\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{7}\\x=-\dfrac{2}{7}\end{matrix}\right.\)

49x²- 4=0

(7x)² -2² = 0

(7x - 2).(7x + 2) = 0

7x-2 = 0 hay 7x + 2 = 0

   x = 2/7 hay       x = -2/7

vậy S= ( 2/7,-2/7)

\(\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\\ \Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\\ \Leftrightarrow\left(x+2\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)

⇔(5x-4-49x)(5x-4+49x)=0

⇔(-44x-4)(54x-4)=0

⇒-44x-4=0 hoặc 54x-4=0

TH1:-44x-4=0                TH2:54x-4=0

     ⇔x=-1/11                   ⇔x=2/27

Vậy xϵ{-1/11;2/27}

(5x-4-7x)(5x-4+7x)=0

(-2x-4)(12x-4)=0

⇒-2x-4=0 hoặc 12x-4=0

    x=-2                  x=1/3

b)(5x – 4)2 – 49x2 = (5x – 4)2 – (7x)2 = (5x – 4 – 7x)(5x – 4 + 7x)

= (12x – 4)(-2x – 4) = -8(3x – 1)(x + 2)

Vậy (3x – 1)(x + 2) = 0 ⇒ 3x - 1 = 0 hoặc x + 2 = 0

⇒ x = 1/3 hoặc x = -2

Vì \(AD\) là tia phân giác của \(BAC\) nên ta có :

\(\dfrac{BD}{DC}=\dfrac{AB}{AC}=\dfrac{6}{8}=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{BD}{3}=\dfrac{DC}{4}\)

Theo dãy tính chất ti số bằng nhau , ta có :

\(\dfrac{BD}{3}=\dfrac{DC}{4}=\dfrac{BC+DC}{3}=\dfrac{10}{7}\)

\(\Rightarrow BD=3.\dfrac{10}{7}=\dfrac{30}{7}\)

\(\Rightarrow DC=4.\dfrac{10}{7}=\dfrac{40}{7}\)

\(\Rightarrow x=\dfrac{30}{7};y=\dfrac{40}{7}\Rightarrow S=4100\)

Vậy S = 4100

Đáp án cần chọn là: C                          THAM KHẢO

a,x(x-2)+x-2=0

⇔ (x-2)(x+1)=0

⇔ x=2;x=-1

b,x³+x²+x+1=0

⇔ x²(x+1)+x+1=0

⇔ (x+1)(x²+1)=0

⇔ x=-1