\(\dfrac{\text{80x3-125x}}{3\left(x-3\right)-\left(x-3\right)\left(8-4x\right)}\) \(\dfrac{9-\left(x+5^{ }\right)^2}{x^2+4x+4}\)
\(\dfrac{32x-8x^2+2x^3}{x^3+64}\) \(\dfrac{x^2+5x+6}{x^2+4x+4}\)
Rút gọn các phân thức :
a) \(\dfrac{14xy^5\left(2x-3y\right)}{21x^2y\left(2x-3y\right)^2}\)
b) \(\dfrac{8xy\left(3x-1\right)^3}{12x^3\left(1-3x\right)}\)
c) \(\dfrac{20x^2-45}{\left(2x+3\right)^2}\)
d) \(\dfrac{5x^2-10xy}{2\left(2y-x\right)^3}\)
e) \(\dfrac{32x-8x^2+2x^3}{x^3+64}\)
f) \(\dfrac{9-\left(x+5\right)^2}{x^2+4x+4}\)
g) \(\dfrac{80x^3-125x}{3\left(x-3\right)-\left(x-3\right)\left(8-4x\right)}\)
h) \(\dfrac{5x^3+5x}{x^4-1}\)
i) \(\dfrac{x^2+5x+6}{x^2+4x+4}\)
Giải phương trình:
1, \(\left(x^2+x+1\right)\left(x^4+2x^3+7x^2+26x+37\right)=5\left(x+3\right)^3\)
2, \(\left(x+1\right)^3+\left(x+3\right)^3+6\left(x+1\right)\left(x+7\right)\left(x+3\right)=8\left(x+2\right)^3\)
3, \(x^3+\left(x-1\right)^3+3x\left(x-1\right)\left(x^4+x\right)=\left(2x-1\right)^3\)
4, \(\dfrac{\left(x+1\right)^3}{3x+1}+\dfrac{x^3+5x+2}{x^3+2x+1}=x+3\)
5, \(\dfrac{5x^3+x^2+x+1}{4x^2+1}+\dfrac{6\left(4x^2+1\right)}{x^3+x^2+1}=x+7\)
6, \(\left(x^2-4x+1\right)^3+\left(8x-x^2+4\right)^3+\left(x-5\right)^3=125x^3\)
\(\dfrac{1}{x+3}+\dfrac{8-x}{4x^2+8x}\)
\(\dfrac{3-2x}{\left(x-5\right)\left(x+2\right)}+\dfrac{1}{x+5}\)
a) Ta có: \(\dfrac{1}{x+3}+\dfrac{8-x}{4x^2+8x}\)
\(=\dfrac{1}{x+3}+\dfrac{8-x}{4x\left(x+2\right)}\)
\(=\dfrac{4x\left(x+2\right)}{4x\left(x+3\right)\left(x+2\right)}+\dfrac{\left(8-x\right)\left(x+3\right)}{4x\left(x+3\right)\left(x+2\right)}\)
\(=\dfrac{4x^2+8x+8x+24-x^2-3x}{4x\left(x+3\right)\left(x+2\right)}\)
\(=\dfrac{3x^2+13x+24}{4x\left(x+3\right)\left(x+2\right)}\)
b) Ta có: \(\dfrac{3-2x}{\left(x-5\right)\left(x+2\right)}+\dfrac{1}{x+5}\)
\(=\dfrac{\left(3-2x\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)\left(x+2\right)}+\dfrac{\left(x-5\right)\left(x+2\right)}{\left(x+5\right)\left(x-5\right)\left(x+2\right)}\)
\(=\dfrac{3x+15-2x^2-10x+x^2+2x-5x-10}{\left(x+5\right)\left(x-5\right)\left(x+2\right)}\)
\(=\dfrac{-x^2-10x+5}{\left(x+5\right)\left(x-5\right)\left(x+2\right)}\)
Giải các phương trình:
\(1.2x\left(x-3\right)+5\left(x-3\right)\)
\(2.\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
\(3.\dfrac{x}{2x-6}+\dfrac{x}{2x-2}=\dfrac{-2x}{\left(x+1\right)\left(3-x\right)}\)
\(1,\) thiếu đề
\(2,\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
\(\Leftrightarrow\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)}{30}-\dfrac{150}{30}\)
\(\Leftrightarrow5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)
\(\Leftrightarrow25x+10-80x+10=24x+12-150\)
\(\Leftrightarrow-55x+20=24x-138\)
\(\Leftrightarrow24x-138+55x-20=0\)
\(\Leftrightarrow79x-158=0\)
\(\Leftrightarrow x=2\)
\(3,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne3\end{matrix}\right.\\ \dfrac{x}{2x-6}+\dfrac{x}{2x-2}=\dfrac{-2x}{\left(x+1\right)\left(3-x\right)}\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x-1\right)}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x-1\right)}-\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow x\left(\dfrac{1}{2\left(x-3\right)}+\dfrac{1}{2\left(x-1\right)}-\dfrac{2}{\left(x+1\right)\left(x-3\right)}\right)=0\)
\(\Leftrightarrow x\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4\left(x-1\right)}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)
\(\Leftrightarrow x\left(\dfrac{x^2-1}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{x^2-2x-3}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4x-4}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)
\(\Leftrightarrow x.\dfrac{x^2-1+x^2-2x-3-4x+4}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{2x^2-6x}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{2x\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x=0\)
Giải phương trình
a) \(\dfrac{3}{5x-1}\)+ \(\dfrac{2}{3-5x}\)=\(\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
b) \(\dfrac{5-x}{4x^2-8x}\)+\(\dfrac{7}{8x}\)=\(\dfrac{x-1}{2x\left(x-2\right)}\)+\(\dfrac{1}{8x-16}\)
a:Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
=>3x-9-10x+2=-4
=>-7x-7=-4
=>-7x=3
=>x=-3/7
b: =>\(\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)
=>\(2\left(5-x\right)+7\left(x-2\right)=4\left(x-1\right)+x\)
=>10-2x+7x-14=4x-4+x
=>5x-4=5x-4
=>0x=0(luôn đúng)
Vậy: S=R\{0;2}
\(\left(3-x\right)^3=-\dfrac{27}{64};\left(x-5\right)^3=\dfrac{1}{-27};\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8};\left(2x-1\right)^2=\dfrac{1}{4};\left(2-3x\right)^2=\dfrac{9}{4};\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\)
\(\left(3-x\right)^3=-\dfrac{27}{64}\)
\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)
\(=>3-x=\dfrac{-3}{4}\)
\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)
\(x=\dfrac{15}{4}\)
________
\(\left(x-5\right)^3=\dfrac{1}{-27}\)
\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)
\(=>x-5=\dfrac{-1}{3}\)
\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)
\(x=\dfrac{14}{3}\)
_____________
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)
\(\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{3}{2}\right)^3\)
\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}+\dfrac{1}{2}\)
\(x=2\)
________
\(\left(2x-1\right)^2=\dfrac{1}{4}\)
\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\) hoặc \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(=>2x-1=\dfrac{1}{2}\) \(2x-1=\dfrac{-1}{2}\)
\(2x=\dfrac{1}{2}+1=\dfrac{1}{2}+\dfrac{2}{2}\) \(2x=\dfrac{-1}{2}+1=\dfrac{-1}{2}+\dfrac{2}{2}\)
\(2x=\dfrac{3}{2}\) \(2x=\dfrac{1}{2}\)
\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\) \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)
\(x=\dfrac{3}{4}\) \(x=\dfrac{1}{4}\)
____________
\(\left(2-3x\right)^2=\dfrac{9}{4}\)
\(\left(2-3x\right)^2=\left(\dfrac{3}{2}\right)^2\) hoặc \(\left(2-3x\right)^2=\left(\dfrac{-3}{2}\right)^2\)
\(=>2-3x=\dfrac{3}{2}\) \(2-3x=\dfrac{-3}{2}\)
\(3x=2-\dfrac{3}{2}=\dfrac{4}{2}-\dfrac{3}{2}\) \(3x=2-\dfrac{-3}{2}=\dfrac{4}{2}+\dfrac{3}{2}\)
\(3x=\dfrac{1}{2}\) \(3x=\dfrac{7}{2}\)
\(x=\dfrac{1}{2}.\dfrac{1}{3}\) \(x=\dfrac{7}{2}.\dfrac{1}{3}\)
\(x=\dfrac{1}{6}\) \(x=\dfrac{7}{6}\)
______________
\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này
(3-x)3=(-\(\dfrac{3}{4}\))3
3-x=-\(\dfrac{3}{4}\)
x=3-(-\(\dfrac{3}{4}\))
x=\(\dfrac{15}{4}\)
Tính:
a) \(\left(2x^3+x^2-8x+3\right):\left(2x-3\right)\)
b) \(\dfrac{x-5}{x^2-4x+4}:\dfrac{x^2-25}{2x-4}\)
a: \(=\dfrac{2x^3-3x^2+4x^2-6x-2x+3}{2x-3}=x^2+2x-1\)
b: \(=\dfrac{x-5}{\left(x-2\right)^2}\cdot\dfrac{2\left(x-2\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{2}{\left(x-2\right)\left(x+5\right)}\)
Quy đồng mẫu thức:
a) \(\dfrac{x^2-20}{x^2-4}+\dfrac{x-5}{x+2}-\dfrac{3}{2-x}\)
b) \(\dfrac{5}{2x-3}-\dfrac{2}{2x+3}-\dfrac{2x-9}{9-4x^2}\)
c) \(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}\)
\(a,=\dfrac{x^2-20+x^2-7x+10+3x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\\ b,=\dfrac{10x+15-4x+6+2x-9}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{4\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{4}{2x-3}\\ c,=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}\\ =\dfrac{1}{x}-\dfrac{1}{x+4}=\dfrac{x+4-x}{x\left(x+4\right)}=\dfrac{4}{x\left(x+4\right)}\)
a) \(x-\dfrac{\dfrac{x}{2}-\dfrac{3+x}{4}}{2}=\dfrac{2x-\dfrac{10-7x}{3}}{3}-\left(x-1\right)\)
b) \(x^2-6x-2+\dfrac{14}{x^2-6x+7}=0\)
c) \(\dfrac{8x^2}{3\left(1-4x^2\right)}=\dfrac{2x}{6x-3}-\dfrac{1+8x}{4+8x}\)
d) \(\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}=\dfrac{6}{x^2-9}\)
e) \(\left(1-\dfrac{2x-1}{x+1}\right)^3+6\left(1-\dfrac{2x-1}{x+1}\right)^2=\dfrac{12\left(2x-1\right)}{x+1}-20\)
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26