Question

Giải các phương trình sau :

a, ( x ² - x + 1) + ( x ² - 2 x + 3) + ( x ² - 3 x + 5) + ... + ( x ² - 100 x + 199) = 300

b, x² + \(\frac{4x^2}{x^2-4x+4}\)

c, \(\frac{3x}{9x^2+9x+2}+\frac{2x}{9x^2+3x+2}\) = \(\frac{1}{3}\)

Answer 1

a/ Tách 300 thành 100 chữ số 3 rồi chuyển vế dồn từng số 3 vào ( ) có \(\left(x^2-x-2\right)+\left(x^2-2x\right)+\left(x^2-3x+2\right)+...+\left(x^2-100x+196\right)\)

=0 \(\Leftrightarrow\left(x-2\right)\left(x+1\right)+x\left(x-3\right)+\left(x-1\right)\left(x-2\right)+...+\left(x-96\right)\left(x-4\right)+\left(x-97\right)\left(x-3\right)+\left(x-98\right)\left(x-2\right)\)=0\(\Leftrightarrow\left(x-2\right)\left(2x-97\right)+\left(x-3\right)\left(2x-97\right)+...=0\Rightarrow x=2\)

Answer 2

b tường đương \(x^2-4+\frac{4x^2}{x^2-4x+4}-1=0\Leftrightarrow\left(x-2\right)\left(x+2\right)+\frac{3x^2+4x-4}{\left(x-2\right)^2}=0\Leftrightarrow\left(x-2\right)\left(x+2\right)+\frac{\left(x+2\right)\left(3x-2\right)}{\left(x-2\right)^2}=0\Leftrightarrow\left(x-2\right)\left(x+2+\frac{3x-2}{\left(x-2\right)^2}\right)=0\Leftrightarrow x=2\)

Answer 3

c/ \(\Leftrightarrow\frac{3x}{9x^2+9x+2}-\frac{1}{6}+\frac{2x}{9x^2+3x+2}-\frac{1}{6}=0\)

\(\Leftrightarrow\frac{18x-9x^2-9x-2}{6\left(9x^2+9x+2\right)}+\frac{12x-9x^2-3x-2}{6\left(9x^2+3x+2\right)}=0\)

\(\Leftrightarrow\left(-9x^2+9x-2\right)\left(...\right)=0\Rightarrow\left(3x-1\right)\left(-3x+2\right)=0\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\x=\frac{2}{3}\end{matrix}\right.\)

Answer 4

câu b sửa thành x² + \(\frac{4x^2}{x^2-4x+4}=5\)