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Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
a) Để \(\frac{2x+3}{4x-5}=0\)
=> 2x + 3 = 0
x = -3/2
b) Để \(\frac{\left(x-1\right)\left(x+2\right)}{x^2-4x+3}=\frac{\left(x-1\right).\left(x+2\right)}{\left(x-3\right).\left(x-1\right)}=\frac{x+2}{x-3}=0\)
=> x + 2 = 0=> x = -2
c) để \(\frac{x^2-1}{x^2-2x+1}=\frac{\left(x-1\right).\left(x+2\right)}{\left(x-1\right)^2}=\frac{x+2}{x-1}=0\)
=> x + 2 = 0 => x = -2
d) để \(\frac{x^2-4}{x^2+3x-10}=\frac{\left(x+2\right).\left(x-2\right)}{\left(x-2\right).\left(x+5\right)}=\frac{x+2}{x+5}=0\)
=> ...
e) để \(\frac{x^3-16x}{x^3-3x^2-4x}=\frac{x.\left(x-4\right).\left(x+4\right)}{x.\left(x-4\right).\left(x+1\right)}=\frac{x+4}{x+1}=0\)
=> ....
1. Ta có:
\(\frac{1}{x}+\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+2013\right)\left(x+2014\right)}\)
\(=\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+2013}-\frac{1}{x+2014}\)
\(=\frac{2}{x}-\frac{1}{x+2014}\)
\(=\frac{2\left(x+2014\right)}{x\left(x+2014\right)}-\frac{x}{x\left(x+2014\right)}\)
\(=\frac{2x+4028-x}{x\left(x+2014\right)}=\frac{x+4028}{x\left(x+2014\right)}\)
2a) ĐKXĐ: x \(\ne\)1 và x \(\ne\)-1
b) Ta có: A = \(\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-3\)
A = \(\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x+1\right)^2}{x+1}-3\)
A = \(x-1+x+1-3\)
A = \(2x-3\)
c) Với x = 3 => A = 2.3 - 3 = 3
c) Ta có: A = -2
=> 2x - 3 = -2
=> 2x = -2 + 3 = 1
=> x= 1/2
\(a)\frac{2x-1}{5x-10}\) \(\text{Đ}K:x\ne2\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\frac{1}{2}(TM)\)
\(b)\frac{x^2-x}{2x}\) \(\text{Đ}K:x\ne0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x.(x-1)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0(lo\text{ại})\\x=1(TM)\end{cases}}\)
\(c)\frac{2x+3}{4x-5}\) \(\text{Đ}K:x\ne\frac{5}{4}\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow x=\frac{-3}{2}(TM)\)
\(d)\frac{(x-1).(x+2)}{(x-3).(x-1)}\) \(\text{Đ}K:\hept{\begin{cases}x\ne3\\x\ne1\end{cases}}\)
\(\Leftrightarrow(x-1).(x+2)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1(l\text{oại})\\x=-2(TM)\end{cases}}\)
gửi cho 4 câu trc
Ta có: \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}\)
\(=\frac{x^2y+xy^2+xy^2+y^3}{2x^2+2xy-xy-y^2}\)
\(=\frac{xy\left(x+y\right)+y^2\left(x+y\right)}{2x\left(x+y\right)-y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)\left(xy+y^2\right)}{\left(2x-y\right)\left(x+y\right)}=\frac{xy+y^2}{2x-y}\left(đpcm\right)\)
Ta có: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)
\(=\frac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)
\(=\frac{x\left(x+y\right)+2y\left(x+y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}\)
\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)
Bài 1:
a)
ĐKXĐ: \(x^2+y^2\neq 0\Leftrightarrow x,y\) không cùng đồng thời bằng $0$
Tức là: \(\left[\begin{matrix} x=0; y\neq 0\\ y=0; x\neq 0\\ x\neq 0; y\neq 0\end{matrix}\right.\)
b)
ĐKXĐ: \(x^2-2x+1\neq 0\Leftrightarrow (x-1)^2\neq 0\Leftrightarrow x\neq 1\)
c)
ĐKXĐ: \((x+3)^2+(y-2)^2\neq 0\Leftrightarrow x+3,y-2\) không cùng đồng thời bằng $0$
Tức là \(\left[\begin{matrix} x=-3, y\neq 2\\ x\neq -3; y=2\\ x\neq -3; y\neq 2\end{matrix}\right.\)
d)
ĐKXĐ: \(x^2+6x+10\neq 0\Leftrightarrow (x+3)^2+1\neq 0\Leftrightarrow (x+3)^2\neq -1\)
\(\Leftrightarrow x\in\mathbb{R}\)
Lời giải:
a)
ĐKXĐ: \(x^2+3x-10\neq 0\Leftrightarrow (x-2)(x+5)\neq 0\Leftrightarrow x\neq 2; x\neq -5\)
Để giá trị phân thức bằng $0$ thì: \(x^2-4=0\Leftrightarrow (x-2)(x+2)=0\Rightarrow \left[\begin{matrix} x=2\\ x=-2\end{matrix}\right.\)
Kết hợp với ĐKXĐ suy ra $x=-2$
b)
ĐKXĐ: \(x^3-3x^2-4x\neq 0\Leftrightarrow x(x^2-3x-4)\neq 0\)
\(\Leftrightarrow x(x+1)(x-4)\neq 0\Leftrightarrow x\neq 0; x\neq -1; x\neq 4\)
Để giá trị của phân thức bằng $0$ thì $x^3-16x=0$
$\Leftrightarrow x(x^2-16)=0\Leftrightarrow x(x-4)(x+4)=0$
\(\Leftrightarrow \left[\begin{matrix} x=0\\ x=\pm 4\end{matrix}\right.\)
Kết hợp với ĐKXĐ suy ra $x=-4$
c)
ĐKXĐ: \(x^3+2x-3\neq 0\Leftrightarrow (x-1)(x^2+x+3)\neq 0\Leftrightarrow x\neq 1\)
Để giá trị phân thức bằng $0$ thì:
$x^3+x^2-x-1=0\Leftrightarrow (x-1)(x+1)^2=0\Leftrightarrow x=1$ hoặc $x=-1$
Kết hợp với ĐKXĐ suy ra $x=-1$