Quy đồng mẫu thức các phân thức :
a) \(\dfrac{7x-1}{2x^2+6x};\dfrac{5-3x}{x^2-9}\)
b) \(\dfrac{x+1}{x-x^2};\dfrac{x+2}{2-4x+2x^2}\)
c) \(\dfrac{4x^2-3x+5}{x^3-1};\dfrac{2x}{x^2+x+1};\dfrac{6}{x-1}\)
d) \(\dfrac{7}{5x};\dfrac{4}{x-2y};\dfrac{x-y}{8y^2-2x^2}\)
e) \(\dfrac{5x^2}{x^3+6x^2+12x+8};\dfrac{4x}{x^2+4x+4};\dfrac{3}{2x+4}\)
BÀI 6 :rút gọn phân thức
\(\dfrac{x^3+3x^3+3x+1}{x^2+x}\)
b)\(\dfrac{x^3-3x^2+3x-1}{2x-2}\)
c)\(\dfrac{x^2+4x+4}{2x+4}\)
d)\(\dfrac{(x-1)(-x-2)}{x+2}\)
e)\(\dfrac{x^2-y^2}{x+y}\)
f)\(\dfrac{3x^2+4xy^2}{6x+8y}\)
g)\(\dfrac{-3x^2-6x}{4-x^2}\)
BÀI 7 :quy đồng mẫu thức các phân thức
\(\dfrac{2}{5x^3y^2}và \dfrac{3}{4xy}\)
b)\(\dfrac{x}{x^2-2xy+y^2} và \dfrac{x}{x^2-xy}\)
c)\(\dfrac{1}{x+2};\dfrac{2}{2x+4}và \dfrac{3}{3x+6}\)
d)\(\dfrac{1}{x+3};\dfrac{2}{2x-6}và \dfrac{3}{3x-9}\)
6:
a: ĐKXĐ: x<>0
\(\dfrac{x^3+3x^2+3x+1}{x^2+x}\)
\(=\dfrac{\left(x+1\right)^3}{x\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{x}\)
b: ĐKXĐ: x<>1
\(\dfrac{x^3-3x^2+3x-1}{2x-2}\)
\(=\dfrac{\left(x-1\right)^3}{2\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{2}\)
c: ĐKXĐ: x<>-2
\(\dfrac{x^2+4x+4}{2x+4}\)
\(=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}\)
\(=\dfrac{x+2}{2}\)
d: ĐKXĐ: x<>-2
\(\dfrac{\left(x-1\right)\left(-x-2\right)}{x+2}\)
\(=\dfrac{\left(-x+1\right)\left(x+2\right)}{x+2}=-x+1\)
e: ĐKXĐ: x<>-y
\(\dfrac{x^2-y^2}{x+y}=\dfrac{\left(x-y\right)\left(x+y\right)}{x+y}=x-y\)
g: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(\dfrac{-3x^2-6x}{4-x^2}=\dfrac{3x^2+6x}{x^2-4}\)
\(=\dfrac{3x\left(x+2\right)}{\left(x+2\right)\cdot\left(x-2\right)}=\dfrac{3x}{x-2}\)
7:
a: \(\dfrac{2}{5x^3y^2}=\dfrac{2\cdot4}{20x^3y^2}=\dfrac{8}{20x^3y^2}\)
\(\dfrac{3}{4xy}=\dfrac{3\cdot5\cdot x^2y}{20x^3y^2}=\dfrac{15x^2y}{20x^3y^2}\)
b: \(\dfrac{x}{x^2-2xy+y^2}=\dfrac{x}{\left(x-y\right)^2}\)
\(\dfrac{x}{x^2-xy}=\dfrac{x}{x\left(x-y\right)}=\dfrac{1}{x-y}=\dfrac{\left(x-y\right)}{\left(x-y\right)^2}\)
c: \(\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)
\(\dfrac{2}{2x+4}=\dfrac{2}{2\left(x+2\right)}=\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)
\(\dfrac{3}{3x+6}=\dfrac{3}{3\left(x+2\right)}=\dfrac{6}{6\left(x+2\right)}\)
d:
\(\dfrac{2}{2x-6}=\dfrac{2}{2\left(x-3\right)}=\dfrac{1}{x-3};\dfrac{3}{3x-9}=\dfrac{3}{3\left(x-3\right)}=\dfrac{1}{x-3}\)
\(\dfrac{2}{2x-6}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{3}{3x-9}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{1}{x+3}=\dfrac{x-3}{\left(x+3\right)\left(x-3\right)}\)
a) 6-7x+7=-8x+12
b) \(\dfrac{\text{4x-1}}{8}\)\(\)=\(\dfrac{6-x}{2}\)-\(\dfrac{1}{4}\)
c) 2x(x-3)+7x-21+0
d) \(\dfrac{x}{x-3}\)+\(\dfrac{x-2}{x+3}\)\(\dfrac{2\left(x^2+6\right)}{x^2-9}\)
bài 2 giải các phương trình sau
b,\(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\) m,\(\dfrac{3x-1}{x+1}=\dfrac{2x+1}{x-1}\)
d,\(\dfrac{3x-14}{x+5}=\dfrac{2}{3}\) p,\(\dfrac{4x+7}{x-1}=\dfrac{12x+5}{3x+4}\)
f,\(\dfrac{6}{x}-1=\dfrac{2x-3}{3}\) r,\(\dfrac{1}{x+3}+\dfrac{1}{x-1}=\dfrac{10}{\left(x+3\right)\left(x-1\right)}\)
h,\(\dfrac{1}{x-2}+3=\dfrac{x-3}{2-x}\) t,\(\dfrac{3x}{x-2}-\dfrac{x}{x-5}=\dfrac{3x}{\left(x-2\right)\left(5-x\right)}\)
j,\(\dfrac{5}{3x+2}=2x-1\) u,\(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=\dfrac{2\left(x^2+x-1\right)}{x\left(x+1\right)}\)
w,\(\dfrac{5x}{2x+2}+1=-\dfrac{6}{x+1}\) s, \(\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{2x}{\left(x-1\right)\left(x-3\right)}\)
ơ,\(\dfrac{1}{x-1}+\dfrac{2}{x+1}=\dfrac{x}{x^2-1}\) v,\(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)
z,\(\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\) ư,\(\dfrac{x+2}{x-2}-\dfrac{-2}{x^2-2x}=\dfrac{1}{x}\)
o,\(x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}\) ô,\(1-\dfrac{1}{1-x}=\dfrac{x^2}{x^2-1}\) zz,\(\dfrac{12}{8+x^3}=1+\dfrac{1}{x+2}\)
b: =>\(4\left(3-7x\right)=x+1\)
=>12-28x=x+1
=>-29x=-11
=>x=11/29
m:=>(3x-1)(x-1)=(2x+1)(x+1)
=>3x^2-4x+1=2x^2+3x+1
=>x^2-7x=0
=>x=0 hoặcx=7
d: =>9x-42=2x+10
=>7x=52
=>x=52/7
p: \(\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\)
=>12x^2+16x+21x+28=12x^2-12x+5x-5
=>37x+28=7x-5
=>30x=-33
=>x=-11/10
j: =>(2x-1)(3x+2)=5
=>6x^2+4x-3x-2-5=0
=>6x^2-x-7=0
=>6x^2-7x+6x-7=0
=>(6x-7)(x+1)=0
=>x=7/6 hoặc x=-1
a) (X-2)(x+3)-3(4x-2)=(x-4)\(^{^{ }2}\)
b) \(\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\)
c) \(x-\dfrac{2x-2}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)
d) \(\left(2x+5\right)^2=\left(x+2\right)^2\)
e) \(x^2-5+6=0\)
g) \(2x^3+6x^2=x^2+3x\)
h) \(\left(x+\dfrac{1}{2}\right)^2+2\left(x+\dfrac{1}{x}\right)-8=0\)
mọi người giúp e với ạ
\(a,\left(x-2\right)\left(x-3\right)-3\left(4x-2\right)=\left(x-4\right)^2\\ \Leftrightarrow x^2-5x+6-12x+6=x^2-8x+16\\ \Leftrightarrow-9x-4=0\\ \Leftrightarrow x=-\dfrac{4}{9}\)
\(b,\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\\ \Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\\ \Leftrightarrow10x=1\\ \Leftrightarrow x=\dfrac{1}{10}\)
\(c,x-\dfrac{2x-2}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\\ \Leftrightarrow30x-12x+12+5x+40=210+10x-10\\ \Leftrightarrow13x=148\\ \Leftrightarrow x=\dfrac{148}{13}\)
\(d,\left(2x+5\right)^2=\left(x+2\right)^2\\ \Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\)
\(e,x^2-5x+6=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
\(g,2x^3+6x^2=x^2+3x\\ \Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
\(h,\left(x+\dfrac{1}{x}\right)^2+2\left(x+\dfrac{1}{x}\right)-8=0\left(x\ne0\right)\)
Đặt \(x+\dfrac{1}{x}=t\), pt trở thành:
\(t^2+2t-8=0\\ \Leftrightarrow\left(t-2\right)\left(t+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1-2x=0\\x^2+1+4x=0\left(1\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\Delta\left(1\right)=16-4=12>0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\\left[{}\begin{matrix}x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\)
Tick plzz
a: Ta có: \(\left(x-2\right)\left(x+3\right)-3\left(4x-2\right)=\left(x-4\right)^2\)
\(\Leftrightarrow x^2+x-6-12x+6-x^2+8x-16=0\)
\(\Leftrightarrow-3x=16\)
hay \(x=-\dfrac{16}{3}\)
b: Ta có: \(\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\)
\(\Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\)
\(\Leftrightarrow-14x+7+4x-6=0\)
\(\Leftrightarrow10x=1\)
hay \(x=\dfrac{1}{10}\)
c: Ta có: \(x-\dfrac{2x-5}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)
\(\Leftrightarrow30x-12x+30+5x+40=210+10x-10\)
\(\Leftrightarrow23x+70=10x+200\)
\(\Leftrightarrow x=10\)
Giải phương trình sau :
a,\(\dfrac{7-3x}{12}+\dfrac{5x+2}{7}=x+13\)
b,\(\dfrac{3\left(x+3\right)}{4}-\dfrac{1}{2}=\dfrac{5x+9}{7}-\dfrac{7x-9}{4}\)
c,\(\dfrac{2x+1}{3}-\dfrac{5x+2}{7}=x+3\)
d,\(\dfrac{2x-3}{3}-\dfrac{2x+3}{7}=\dfrac{4x+3}{5}-17\)
a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)
\(\Leftrightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow39x-84x=1092-73\)
=>-45x=1019
hay x=-1019/45
b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
=>21x+63-14=20x+36-49x+63
=>21x+49=-29x+99
=>50x=50
hay x=1
c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)
=>14x+7-15x-6-21x-63=0
=>-22x-64=0
hay x=-32/11
d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)
=>70x-105-30x-45=84x+63-1785
=>40x-150-84x+1722=0
=>-44x+1572=0
hay x=393/11
a, msc 12.7=84
Chuyển vế về =0 rồi làm
b,msc 28
c,làm tương tự
a, \(\Rightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow-45x=1019\Leftrightarrow x=-\dfrac{1019}{45}\)
b, \(\Rightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
\(\Leftrightarrow21x+63-14=20x+36-49x+63\)
\(\Leftrightarrow50x=50\Leftrightarrow x=1\)
c, \(\Rightarrow14x+7-15x-6=21x+63\Leftrightarrow-22x=62\Leftrightarrow x=-\dfrac{31}{11}\)
d, \(\Rightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-105.17\)
\(\Leftrightarrow70x-105-30x-45=84x+63-1785\)
\(\Leftrightarrow-44x=-1572\Leftrightarrow x=\dfrac{393}{11}\)
Bài 1 : Tìm x biết
a/ x ( x + 4 ) + x + 4=0
b/ x ( x - 3) + 2x - 6 = 0
Bài 2 : rút gọn biểu thức
a/ \(\dfrac{6x^2y^2}{8xy^5}\) b/ \(\dfrac{3x^2-x}{9x^2-6x+1}\) e/ \(\dfrac{x^2+7x+12}{x^2+5x+6}\)
c/ \(\dfrac{x^2-9}{x^2+6x+9}\) d/ \(\dfrac{x^2+2x+1}{3x+3}\)
Bài 3 : thực hiện phép tính ( các mẫu thức đều không buông )
a/ \(\dfrac{15}{2x+6}+\dfrac{5x}{2x+6}\) b/ \(\dfrac{y}{2x^2-xy}+\dfrac{4x}{y^2-2xy}\) c/ \(\dfrac{x-1}{2x^2-2}-\dfrac{x+3}{4x+4}\)
d/ \(\dfrac{4y^2}{11x^4}.\left(-\dfrac{3x^2}{8y}\right)\) e/ \(\dfrac{5x+10}{4x-8}.\dfrac{4-2x}{x+2}\)
Bài 4 : Rút gọn và tính các giá trị của biểu thức
a/ \(\dfrac{3x^2-x}{9x^2-6x+1}\) tại x = \(\dfrac{1}{3}\) b/\(\dfrac{x^2-2xy+y^2-9}{x^2-xy+3x}\) Tại x = 2016 ; y = 3
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
Bài 2:
\(a,\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)
\(b,\dfrac{3x^2-x}{9x^2-6x+1}=\dfrac{x\left(3x-1\right)}{\left(3x-1\right)^2}=\dfrac{x}{3x-1}\)
\(c,\dfrac{x^2-9}{x^2+6x+9}=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x+3\right)^2}=\dfrac{x-3}{x+3}\)
\(d,\dfrac{x^2+2x+1}{3x+3}=\dfrac{\left(x+1\right)^2}{3\left(x+1\right)}=\dfrac{x+1}{3}\)
\(e,\dfrac{x^2+7x+12}{x^2+5x+6}=\dfrac{x^2+2x+6x+12}{x^2+2x+3x+6}=\dfrac{x\left(x+2\right)+6\left(x+2\right)}{x\left(x+2\right)+3\left(x+2\right)}=\dfrac{\left(x+2\right)\left(x+6\right)}{\left(x+2\right)\left(x+3\right)}=\dfrac{x+6}{x+3}\)
Quy đồng các phân thức sau:
9) \(\dfrac{2}{x^2-2x};\dfrac{x}{3x-6}\)
10) \(\dfrac{x}{x-5};x+1\)
11) \(\dfrac{x}{x^2+x+5};-3\)
12)\(\dfrac{x}{2x-8};\dfrac{x+1}{4x-x^2}\)
\(9,\dfrac{2}{x^2-2x}=\dfrac{6}{3x\left(x-2\right)};\dfrac{x}{3x-6}=\dfrac{x^2}{3x\left(x-2\right)}\\ 10,\dfrac{x}{x-5}=\dfrac{x}{x-5};x+1=\dfrac{\left(x+1\right)\left(x-5\right)}{x-5}\\ 11,-3=\dfrac{-3\left(x^2+x+5\right)}{x^2+x+5}\\ 12,\dfrac{x}{2x-8}=\dfrac{x^2}{2x\left(x-4\right)};\dfrac{x+1}{4x-x^2}=\dfrac{-2\left(x+1\right)}{2x\left(x-4\right)}\)
rút gọn phân thức
a.\(\dfrac{5x-15}{4x+4}:\dfrac{x^2-9}{x^2+2x+1}\)
b.\(\dfrac{6x+48}{7x-7}:\dfrac{x^2-64}{x^2-2x+1}\)
c.\(\dfrac{4x-24}{5x+5}:\dfrac{x^2-36}{x^2+2x+1}\)
d.\(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
a.\(\dfrac{5\left(x-3\right)}{4\left(x+1\right)}\) : \(\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x+1\right)^2}\)
= \(\dfrac{5\left(x-3\right)}{4\left(x+1\right)}\). \(\dfrac{\left(x+1\right)^2}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{5\left(x+1\right)}{4\left(x+3\right)}\)
b. \(\dfrac{6\left(x+8\right)}{7\left(x-1\right)}\). \(\dfrac{\left(x-1\right)^2}{\left(x-8\right)\left(x+8\right)}\)
= \(\dfrac{6\left(x-1\right)}{7\left(x-8\right)}\)
c.Tương tự hai câu trên nka!!
d. (\(\dfrac{1}{x\left(x+1\right)}\)-\(\dfrac{2-x}{x+1}\)).(\(\dfrac{x}{x-1}\))
=( \(\dfrac{1}{x\left(x+1\right)}\)-\(\dfrac{2x-x^2}{x\left(x+1\right)}\)). ....
= \(\dfrac{\left(1-x\right)^2}{x\left(x+1\right)}\). ...
= \(\dfrac{x-1}{x+1}\)
c: \(=\dfrac{4\left(x-6\right)}{5\left(x+1\right)}\cdot\dfrac{\left(x+1\right)^2}{\left(x-6\right)\left(x+6\right)}=\dfrac{4\left(x+1\right)}{5\left(x+6\right)}\)
a: \(=\dfrac{5\left(x-3\right)}{4\left(x+1\right)}\cdot\dfrac{\left(x+1\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{5\left(x+1\right)}{4\left(x+3\right)}\)
b: \(=\dfrac{6\left(x+8\right)}{7\left(x-1\right)}\cdot\dfrac{\left(x-1\right)^2}{\left(x-8\right)\left(x+8\right)}=\dfrac{6\left(x-1\right)}{7\left(x-8\right)}\)
