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Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
\(1,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2x+6}-\frac{x-6}{x\left(2x-6\right)}=\frac{3x-x+6}{x\left(2x-6\right)}=\frac{2x+6}{x\left(2x-6\right)}\)
\(2,\frac{1}{1-x}+\frac{2x}{x^2-1}=\frac{-1\left(x+1\right)+2x}{x^2-1}=\frac{x-1}{x^2-1}=\frac{1}{x+1}\)
\(3,\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)
\(4,\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\frac{2\left(2-x\right)}{x+2}=\frac{-5}{2}\)
\(5,\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\frac{3x}{2\left(1-2x\right)}=\frac{3\left(1+2x\right)}{2x\left(x+4\right)}\)
\(6,\frac{12x}{5y^3}.\frac{15y^4}{8x^3}=\frac{9y}{2x^2}\)
1,\(\frac{3}{2x+6}-\frac{x-6}{x\left(2x+6\right)}\)
=\(\frac{3x}{x\left(2x+6\right)}+\frac{x-6}{x\left(2x+6\right)}\)
=\(\frac{3x+x-6}{x\left(2x+6\right)}\)=\(\frac{4x-6}{x\left(2x+6\right)}=\frac{2\left(2x-3\right)}{x\left(2x+6\right)}\)
1) \(\frac{8xy\left(3x-1\right)^3}{12x^3\left(1-3x\right)}=-\frac{8xy\left(3x-1\right)^3}{12x^3\left(3x-1\right)}=-\frac{2y\left(3x-1\right)^2}{3x^2}\)
2) \(\frac{5x^3+5x}{x^4-1}=\frac{5x\left(x^2+1\right)}{\left(x^2+1\right)\left(x^2-1\right)}=\frac{5x}{x^2-1}\)
3) \(\frac{9-\left(x+5\right)^2}{x^2+4x+4}=\frac{\left(3-x-5\right)\left(3+x+5\right)}{\left(x+2\right)^2}=\frac{-\left(x+2\right)\left(x+8\right)}{\left(x+2\right)^2}=-\frac{x+8}{x+2}\)
3) \(\frac{32x-8x^2+2x^3}{x^3+64}=\frac{2x\left(16-4x+x^2\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\frac{2x}{x+4}\)
$\dfrac{4}{-25x^2+20x-3}=\dfrac{3}{5x-1}-\dfrac{2}{5x-3}$
Điều kiện: $5x-1\ne0,\ 5x-3\ne0$
$\Leftrightarrow x\ne\dfrac{1}{5},\ x\ne\dfrac{3}{5}$
Ta có:
$-25x^2+20x-3=-(5x-1)(5x-3)$
$\Rightarrow \dfrac{-4}{(5x-1)(5x-3)}=\dfrac{3(5x-3)-2(5x-1)}{(5x-1)(5x-3)}$
$\Leftrightarrow \dfrac{-4}{(5x-1)(5x-3)}=\dfrac{5x-7}{(5x-1)(5x-3)}$
Vì $x\ne\dfrac{1}{5},\dfrac{3}{5}$ nên:
$-4=5x-7$
$\Leftrightarrow 5x=3$
$\Leftrightarrow x=\dfrac{3}{5}$
Nhưng $x=\dfrac{3}{5}$ không thỏa mãn điều kiện.
Vậy: phương trình vô nghiệm.
b)$\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}-\dfrac{2}{x^2-4x+3}=0$
Điều kiện:
$x^2-3x+2=(x-1)(x-2)\ne0$
$x^2-5x+6=(x-2)(x-3)\ne0$
$x^2-4x+3=(x-1)(x-3)\ne0$
$\Rightarrow x\ne1,\ x\ne2,\ x\ne3$
Ta có:
$\dfrac{1}{(x-1)(x-2)}+\dfrac{1}{(x-2)(x-3)}-\dfrac{2}{(x-1)(x-3)}=0$
Quy đồng:
$\dfrac{x-3+x-1-2(x-2)}{(x-1)(x-2)(x-3)}=0$
$\Leftrightarrow \dfrac{x-3+x-1-2x+4}{(x-1)(x-2)(x-3)}=0$
$\Leftrightarrow \dfrac{0}{(x-1)(x-2)(x-3)}=0$
Phương trình đúng với mọi $x$ thỏa mãn điều kiện.
Vậy: $x\in\mathbb{R}\setminus{1;2;3}$.
c)$\dfrac{x-1}{2x^2-4x}-\dfrac{7}{8x}=\dfrac{5-x}{4x^2-8x}-\dfrac{1}{8x-16}$
Điều kiện: $x\ne0,\ x\ne2$
Ta có:
$\dfrac{x-1}{2x(x-2)}-\dfrac{7}{8x}=\dfrac{5-x}{4x(x-2)}-\dfrac{1}{8(x-2)}$
Quy đồng mẫu $8x(x-2)$:
$\dfrac{4(x-1)-7(x-2)}{8x(x-2)}=\dfrac{2(5-x)-x}{8x(x-2)}$
$\Leftrightarrow \dfrac{4x-4-7x+14}{8x(x-2)}=\dfrac{10-2x-x}{8x(x-2)}$
$\Leftrightarrow \dfrac{-3x+10}{8x(x-2)}=\dfrac{-3x+10}{8x(x-2)}$
Phương trình đúng với mọi $x$ thỏa mãn điều kiện.
Vậy: $x\in\mathbb{R}\setminus{0;2}$.
d)$\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}$
Điều kiện:
$x^2+9x+20=(x+4)(x+5)\ne0$
$x^2+11x+30=(x+5)(x+6)\ne0$
$x^2+13x+42=(x+6)(x+7)\ne0$
$\Rightarrow x\ne-4,-5,-6,-7$
Ta có:
$\dfrac{1}{(x+4)(x+5)}+\dfrac{1}{(x+5)(x+6)}+\dfrac{1}{(x+6)(x+7)}=\dfrac{1}{18}$
Quy đồng:
$\dfrac{(x+6)(x+7)+(x+4)(x+7)+(x+4)(x+5)}{(x+4)(x+5)(x+6)(x+7)}=\dfrac{1}{18}$
$\Leftrightarrow 18[(x+6)(x+7)+(x+4)(x+7)+(x+4)(x+5)]=(x+4)(x+5)(x+6)(x+7)$
Khai triển và rút gọn:
$\Leftrightarrow x^2+11x-26=0$
$\Leftrightarrow (x-2)(x+13)=0$
$\Leftrightarrow x=2$ hoặc $x=-13$
Cả hai giá trị đều thỏa mãn điều kiện.
Vậy: $x\in{-13;2}$.
a) \(\frac{3x-2}{2xy}-\frac{7x-4}{2xy}\)
\(=\frac{3x-2}{2xy}+\frac{-\left(7x-4\right)}{2xy}\)
\(=\frac{3x-2-7x+4}{2xy}\)
\(=\frac{-4x+2}{2xy}\)
\(=\frac{2.\left(-2x+1\right)}{2xy}.\)
\(=\frac{-2x+1}{xy}.\)
b) \(\frac{6}{x^2+4x}+\frac{3}{2x+8}\)
Ta có:
\(x^2+4x=x.\left(x+4\right)\)
\(2x+8=2.\left(x+4\right)\)
\(MTC:2x.\left(x+4\right)\)
\(\frac{6}{x^2+4x}+\frac{3}{2x+8}\)
\(=\frac{6}{x.\left(x+4\right)}+\frac{3}{2.\left(x+4\right)}\)
\(=\frac{6.2}{2x.\left(x+4\right)}+\frac{3x}{2x.\left(x+4\right)}\)
\(=\frac{12}{2x.\left(x+4\right)}+\frac{3x}{2x.\left(x+4\right)}\)
\(=\frac{12+3x}{2x.\left(x+4\right)}\)
\(=\frac{3.\left(4+x\right)}{2x.\left(x+4\right)}\)
\(=\frac{3.\left(x+4\right)}{2x.\left(x+4\right)}\)
\(=\frac{3}{2x}.\)
Chúc bạn học tốt!
a) \(\frac{x+5}{4}\)-\(\frac{2x-5}{3}\)=\(\frac{6x-1}{3}\)+\(\frac{2x-3}{12}\)
⇔\(\frac{3\left(x+5\right)}{12}\)-\(\frac{4\left(2x-5\right)}{12}\)=\(\frac{4\left(6x-1\right)}{12}\)+\(\frac{2x-3}{12}\)
⇒ 3x+15-8x+20=24x-4+2x-3
⇔3x+15-8x+20-24x+4-2x+3=0
⇔-31x+42=0
⇔x=\(\frac{42}{31}\)
Vậy tập nghiệm của phương trình đã cho là:S={\(\frac{42}{31}\)}
Bài làm
a) \(\frac{4x-5}{8xy}+\frac{5-y}{8xy}=\frac{4x-5+5-y}{8xy}=\frac{4x-y}{8xy}\)
b) \(\frac{4x^2}{x-2}+\frac{3}{x-2}+\frac{19}{2-x}=\frac{4x^2}{x-2}+\frac{3}{x-2}-\frac{19}{x-2}=\frac{4x^2+3-19}{x-2}=\frac{4x^2-16}{x-2}=\frac{2\left(x-2\right)\left(2x+4\right)}{x-2}=2\left(2x+4\right)\)
c) \(\frac{2x^3+5}{x^2-x+1}-\frac{x^3+4}{x^2-x+1}=\frac{2x^3+5-x^3-4}{x^2-x+1}=\frac{2x^2-x^3+1}{x^2-x+1}\)
d) \(\frac{6}{5x-20}-\frac{x-5}{x^2-8x+16}=\frac{6}{5\left(x-4\right)}-\frac{x-5}{\left(x-4\right)^2}=\frac{6\left(x-4\right)}{5\left(x-4\right)^2}-\frac{\left(x-5\right)5}{5\left(x-4\right)^2}=\frac{6x-4-5x+25}{5\left(x-4\right)^2}=\frac{x+21}{5\left(x-4\right)^2}\)
# Học tốt #