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b: \(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+2\right)\left(x+3\right)+\left(x+1\right)\left(x+3\right)+\left(x+2\right)\left(x+1\right)}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
Đề sai ạ ! Sửa nhé :
\(S=\left(\frac{2+x}{2-x}+\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\frac{x^3-4x}{2x^2-x^3}\)
\(\Leftrightarrow S=\left(\frac{-\left(x+2\right)}{x-2}+\frac{4x^2}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{x+2}\right):\frac{x\left(x^2-4\right)}{x^2\left(2-x\right)}\)
\(\Leftrightarrow S=\left(\frac{-\left(x+2\right)^2+4x^2+\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\right):\frac{\left(x-2\right)\left(x+2\right)}{-x\left(x-2\right)}\)
\(\Leftrightarrow S=\frac{-x^2-4x-4+4x^2+x^2-4x+4}{\left(x+2\right)\left(x-2\right)}.\frac{-x}{\left(x+2\right)}\)
\(\Leftrightarrow S=\frac{-x\left(4x^2-8x\right)}{\left(x+2\right)^2\left(x-2\right)}\)
\(\Leftrightarrow S=\frac{-4x^2\left(x-2\right)}{\left(x+2\right)^2\left(x-2\right)}\)
\(\Leftrightarrow S=\frac{-4x^2}{\left(x+2\right)^2}\)
P/s : nếu làm theo đề của bạn, sẽ ra kq dài... Nên mik tiện sửa, còn nếu đề bạn đúng rồi thì mik sẽ làm lại ạ !
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, Điều kiện xác định: x<>0
b, Điều kiện xác định: x <> -1/3
c, Điều kiện xác định: x<>2
d, Điều kiện xác định: a<>0 và b<>0; b<>2a
A : không rút gọn được
\(B=\frac{4x^2\left(x-2\right)+3\left(x-2\right)}{3x\left(4x^2+3\right)+4x^2+3}=\frac{\left(4x^2+3\right)\left(x-2\right)}{\left(4x^2+3\right)\left(3x+1\right)}=\frac{x-2}{3x+1}\)
\(C=\frac{x^4-1}{x^3+2x^2-x-2}=\frac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x+2\right)\left(x^2-1\right)}=\frac{x^2+1}{x+2}\)
\(D=\frac{a^3+b^3}{a^3+\left(a-b\right)^3}=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(a+a-b\right)\left(a^2-a^2+ab+a^2-2ab+b^2\right)}\)\(=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(2a-b\right)\left(a^2-ab+b^2\right)}=\frac{a+b}{2a-b}\)
\(DKXD:x\ne\pm2;x\ne3;x\ne\frac{3}{2};x\ne0\)
\(A=\left(\frac{2+x}{2-x}+\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-3x}\right)\)
\(=\frac{\left(2+x\right)^2-4x^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{2x^2-3x}{x^2-3x}\)
\(=\frac{4+4x+x^2-4x^2-4+4x-x^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{x\left(2x-3\right)}{x\left(x-3\right)}\)
\(=\frac{8x-4x^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{2x-3}{x-3}\)
\(=\frac{4x\left(2x-3\right)}{\left(2+x\right)\left(x-3\right)}\)
b
Xét hơi bị nhiều TH nhá:(
Để \(A>0\) thì \(\frac{4x\left(2x-3\right)}{\left(2+x\right)\left(x-3\right)}>0\)
TH1:\(4x\left(2x-3\right)>0;\left(2+x\right)\left(x-3\right)>0\)
\(TH2:4x\left(2x-3\right)< 0;\left(2+x\right)\left(x-3\right)< 0\)
Bạn tự xét nốt nhá!
c
\(\left|x-7\right|=4\Rightarrow x-7=4;x-7=-4\)
\(\Rightarrow x=11;x=3\)
Thay vào .....
Lời giải:
a. ĐKXĐ: $x\neq \pm 2; x\neq 0; x\neq 3$
b.
\(S=\left[\frac{-(x+2)^2}{(x-2)(x+2)}+\frac{4x^2}{(x-2)(x+2)}+\frac{(x-2)^2}{(x+2)(x-2)}\right]:\frac{x(x-3)}{x^2(2-x)}\\ =\frac{-(x+2)^2+4x^2+(x-2)^2}{(x-2)(x+2)}:\frac{x-3}{x(2-x)}\\ =\frac{4x^2-8x}{(x-2)(x+2)}.\frac{x(2-x)}{x-3}\\ =\frac{4x(x-2)}{(x-2)(x+2)}.\frac{-x(x-2)}{x-3}=\frac{-4x^2(x-2)}{(x+2)(x-3)}\)
c.
$|x-5|=2\Rightarrow x-5=2$ hoặc $x-5=-2$
$\Rightarrow x=7$ hoặc $x=3$. Mà theo ĐKXĐ thì $x\neq 3$ nên $x=7$
$S=\frac{-4.7^2(7-2)}{(7+2)(7-3)}=\frac{-245}{9}$