Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
câu d
\(D=\dfrac{\left(1-x^2\right)}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{\left(1-x^2\right)\left(x^2-x-3\right)+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{x^2-x-3-x^4+x^3-3x^2+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x^4+x^3+x^2-15x}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x\left(x^3-x^2-x+15\right)}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-\left(x^3-x^2-x+15\right)}{\left(x+3\right)}\end{matrix}\right.\)
a: \(4-6x+\dfrac{9}{4}x^2=\left(2-\dfrac{3}{2}x\right)^2\)
c: \(x^6-3x^5+3x^4-x^3=\left(x^2-x\right)^3\)
\(a.\left(2xy-3\right)^2=4x^2y^2-12xy+9\)
\(b.\left(\dfrac{1}{2}x+\dfrac{1}{3}\right)^2=\dfrac{1}{4}x^2+\dfrac{1}{3}x+\dfrac{1}{9}\)
a) \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{9x^2-6x+1}\)
\(=-\dfrac{9x^2+3x+2x-6x^2}{\left(3x-1\right)\left(3x+1\right)}.\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)
\(=-\dfrac{x\left(3x+5\right)}{\left(3x-1\right)^2}.\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)
\(=\dfrac{-1}{2}\)
b) \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\left(\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{3x-9-x^2}{3x\left(x+3\right)}\right)\)
\(=\dfrac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\dfrac{3x\left(x+3\right)}{-x^2+3x-9}\)
\(=\dfrac{x^2-3x+9}{x-3}.\dfrac{3}{-\left(x^2-3x+9\right)}\)
\(=-\dfrac{3}{x-3}\)
a,\(\dfrac{3}{x-3}\) - \(\dfrac{6x}{9-x^2}\) + \(\dfrac{x}{x+3}\) (*)
đkxđ: x khác 3, x khác -3
(*) \(\dfrac{3(x+3)}{\left(x-3\right).\left(x+3\right)}\)- \(\dfrac{6x}{\left(x-3\right).\left(x+3\right)}\) + \(\dfrac{x\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}\)
=>3x+9 -6x + x2+3x
<=>x2 + 3x-6x+3x + 9
<=>x2 +9
<=>(x-3).(x+3)
Lời giải:
1.
\(\frac{a^3-4a^2-a+4}{a^3-7a^2+14a-8}=\frac{a^2(a-4)-(a-4)}{(a^3-8)-(7a^2-14a)}=\frac{(a-4)(a^2-1)}{(a-2)(a^2+2a+4)-7a(a-2)}\)
\(=\frac{(a-4)(a-1)(a+1)}{(a-2)(a^2-5a+4)}=\frac{(a-4)(a-1)(a+1)}{(a-2)(a-1)(a-4)}=\frac{a+1}{a-2}\)
2.
\(\frac{x^2y^2+1+(x^2-y)(1-y)}{x^2y^2+1+(x^2+y)(1+y)}=\frac{x^2y^2+1+x^2-x^2y-y+y^2}{x^2y^2+1+x^2+x^2y+y+y^2}\)
\(=\frac{(x^2y^2-x^2y+x^2)+(y^2-y+1)}{(x^2y^2+x^2y+x^2)+(y^2+y+1)}\)
\(=\frac{x^2(y^2-y+1)+(y^2-y+1)}{x^2(y^2+y+1)+(y^2+y+1)}=\frac{(x^2+1)(y^2-y+1)}{(x^2+1)(y^2+y+1)}=\frac{y^2-y+1}{y^2+y+1}\)
\(a,VP=\dfrac{x^2+4x+3}{x^2+6x+9}=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x+3\right)^2}=\dfrac{x+1}{x+3}=VT\)
Vậy ta có đpcm
b, \(VP=\dfrac{3x\left(x+y\right)^2}{9x^2\left(x+y\right)}=\dfrac{x+y}{3x}=VT\)
Vậy ta có đpcm
a) Ta có: \(\dfrac{x^2+4x+3}{x^2+6x+9}\)
\(=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x+3\right)\left(x+3\right)}\)
\(=\dfrac{x+1}{x+3}\)
b: Ta có: \(\dfrac{3x\left(x+y\right)^2}{9x^2\left(x+y\right)}\)
\(=\dfrac{3x\left(x+y\right)\left(x+y\right)}{3x\cdot3x\cdot\left(x+y\right)}\)
\(=\dfrac{x+y}{3x}\)
1) \(\left(3x-2\right)^2=9x^2-12x+4\)
\(\left(\dfrac{1}{2}x^2+\dfrac{1}{3}\right)^2=\dfrac{1}{4}x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\)
\(\left(a+b\sqrt{3}\right)^2=a^2+2\sqrt{3}ab+3b^2\)
2) \(4a^2+4a+1=\left(2a+1\right)^2\)
\(9x^2-6x+1=\left(3x-1\right)^2\)
\(\dfrac{1}{4}x^2-\dfrac{1}{3}xy+\dfrac{1}{9}y^2=\left(\dfrac{1}{2}x-\dfrac{1}{3}y\right)^2\)