b ) \(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}\)
\(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}\)
\(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}\)
\(=\frac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)+2-3x}{2x\left(2x-1\right)}\)
\(=\frac{2x-1-6x^2+3x+6x^2-4x+2-3x}{2x\left(2x-1\right)}\)
\(=\frac{-2x+1}{2x\left(2x-1\right)}\)
\(=-\frac{1}{2x}\)
\(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}\)
\(=\frac{1-3x}{2x}+\frac{3x-2}{2x-1}-\frac{3x-2}{2x\left(2x-1\right)}\)
\(=\frac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)-\left(3x-2\right)}{2x\left(2x-1\right)}\)
\(=\frac{2x-1-6x^2+3x+6x^2-4x-3x+2}{2x\left(2x-1\right)}\)
\(=\frac{-2x+1}{2x\left(2x-1\right)}=\frac{-\left(2x-1\right)}{2x\left(2x-1\right)}=-\frac{1}{2x}\)
điều kiện x khác {0,1/2}
\(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}=\frac{2x-6x^2-1+3x+6x^2-4x-3x+2}{2x\left(2x-1\right)}\)
=\(\frac{-2x+1}{2x\left(2x-1\right)}=-\frac{1}{2x}\)
b ) \(\frac{1-3.x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}\)
\(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}\)
\(=\frac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)+2-3x}{2x\left(2x-1\right)}\)
\(=\frac{2x-1-6x^2+3x+6x^2-4x+2-3x}{2x\left(2x-1\right)}\)
\(=\frac{-2x+1}{2x\left(x-1\right)}\)
\(=-\frac{1}{2x}\)
Tìm x biết :
a) (3x+1)^2=x^2+2x+1
b) Thực hiện phép tính: \(\frac{1-3x}{2x}\)+ (3x-2).\(\left(\frac{1}{2x-1}-\frac{1}{4x^2-2x}\right)\)
em xl vì em ko bít em chỉ mới lớp 5 thôi
Thực hiện phép tính:
a) \(\frac{1}{x}-\frac{1}{x+1}\)
b)\(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}\)
c) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}\)
d) \(\frac{2x-1}{x}-\frac{2x+5}{3x-4x^2}+\frac{2x^2+x+3}{3x-4x^2}\)
e) \(\frac{x}{2x+1}+\frac{1}{4x^2-1}-\frac{x-2}{2x-1}\)
giúp mình với mình cần gấp ! Help me ===)
\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x^2+x}\)
b, \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{y^2-xy-xy+x^2}{\left(xy-x^2\right)\left(y^2-xy\right)}=\frac{x^2+y^2}{xy^3-xyxy-xyxy+x^3y}\)Tu rut gon tiep
c, tt
d, cx r
a) \(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}\)
\(=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)
b) \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)
\(=\frac{y}{xy\left(y-x\right)}-\frac{x}{xy\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)
c) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)
\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)
\(a,\frac{1}{x}-\frac{1}{x+1}\)
\(=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}=\frac{x+1-x}{x\left(x+1\right)}\)
\(=\frac{1}{x\left(x+1\right)}\)
\(b,\frac{1}{xy-x^2}-\frac{1}{y^2-xy}\)
\(=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)
\(=\frac{y}{xy\left(y-x\right)}-\frac{x}{xy\left(y-x\right)}=\frac{x-y}{xy\left(x-y\right)}=\frac{1}{xy}\)
Thực hiện phép cộng các phân thức ko cùng mẫu:
a/\(\frac{x^3+2x}{x^3+1}+\frac{2x}{x^2-x+1}+\frac{1}{x+1}\) b/\(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}\)
Tự trình bày nha
a) MC : x^3 +1
KQ: (x+1)^3 / (x^3 +1)
b) MC: 2x-4x^2
KQ: -1/(2x)
a,\(\frac{3}{x}+\frac{1}{x+3}+\frac{3}{x+6}+\frac{1}{x+7}=\frac{1}{1-x}\)
b, \(\frac{1}{x-5}+\frac{1}{x-2}+\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+3}=\frac{3x-3}{4}\)
c,\(\frac{1}{x-3}+\frac{1}{3x+1}+\frac{10x-13}{4x-6}=\frac{1}{x+1}+\frac{1}{2x-1}+\frac{1}{3x+7}\)
d,\(\frac{x^2+x+1}{2x-1}\left(\frac{3x^2-x+5}{4x-2}-3\right)=8\)
e,\(\frac{2x^2-3}{3x-1}\left(2x-\frac{7+4x}{3x-1}\right)=2\)
f,\(\frac{x\left(3x-1\right)\left(3x^2+1\right)\left(6x^2-3x-1\right)}{\left(x+1\right)^3}=\frac{1}{2}\)
g, \(x\left(x^2+2\right)\left(x^2+2x+8+\frac{12}{x-2}\right)=3\left(x-2\right)\)
Giải các phương trình sau:
a)\(\left|x^2-3x-5\right|+2\left|2x-1\right|=x^2-4\)
b)\(\frac{4}{2x+1}+\frac{3}{2x+2}=\frac{2}{2x+3}+\frac{1}{2x+4}\)
c)\(\frac{2x-5}{2x^2+3x-5}+\frac{3x+1}{1-x}=\frac{x+20}{4x+10}\)
d)\(\frac{1}{x^2+5x+4}+\frac{1}{x^2+11x+28}+\frac{1}{x^2+17x+70}=\frac{3}{4x-2}\)
Bµi 5: Gi¶i PT sau.
\(a,\frac{5x-2}{2-2x}+\frac{2x-1}{2}+\frac{x^2+x-3}{1-x}=1\)
b,\(\frac{6x-1}{2-x}+\frac{9x+4}{x+2}=\frac{3x^2-2x+1}{x^2-4}\)
\(c,\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
d) (x2 + 4x + 8)2 + 3x(x2 + 4x + 8) + 2x2 = 0
e) x4 + 2x3 + 4x2 + 2x + 1 = 0
\(f,\frac{3x-1}{x-1}-\frac{2x+5}{x+3}+\frac{4}{x^2+2x-3}=1\)
a) \(\frac{5x-2}{2-2x}+\frac{2x-1}{2}+\frac{x^2+x-3}{1-x}=1\)
ĐK: x≠1
<=>\(\frac{5x-2}{2\left(1-x\right)}+\frac{2x-1}{2}\frac{x^2+x-3}{1-x}=1\)
<=>\(\frac{5x-2+\left(1-x\right).\left(2x-1\right)+2\left(x^2+x-3\right)}{2\left(1-x\right)}=1\)
<=>\(\frac{5x-2+2x-1-2x^2+x+2x^2+2x-6}{2\left(1-x\right)}=1\)
<=>\(\frac{10x-9}{2\left(1-x\right)}=1\)
<=> 10x-9=2(1-x)
<=>10x-9=2-2x
<=> 10x+2x= 2+9
<=> 12x=11
<=> x= \(\frac{11}{12}\left(tm\right)\)
b) \(\frac{6x-1}{2-x}+\frac{9x+4}{x+2}=\frac{3x^2-2x+1}{x^2-4}\)
ĐK: x≠2, x≠-2
<=>\(\frac{6x-1}{-\left(x-2\right)}+\frac{9x+4}{x+2}-\frac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}=0\)
<=> -(x+2).(6x-1)+(x-2).(9x+4)-(3x2-2x+1)=0
<=> -(6x2-x+12x-2)+9x2+4x-18x-8-3x2+2x-1 = 0
<=> -6x2-11x+2+9x2+4x-18x-8-3x2+2x-1=0
<=> -23x-7=0
<=> -23x=7
<=> x= \(\frac{-7}{23}\left(tm\right)\)
tham khảo câu d trong
https://hoc24.vn/hoi-dap/question/919967.html
c) \(\frac{1}{x-1}\)+\(\frac{2x^2-5}{x^3-1}\)=\(\frac{4}{x^2+x+1}\) (ĐKXĐ:x≠1)
⇔\(\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)+\(\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}\)=\(\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
⇒x2+x+1+2x2-5=4x-4
⇔3x2-3x=0
⇔3x(x-1)=0
⇔x=0 (TMĐK) hoặc x=1 (loại)
Vậy tập nghiệm của phương trình đã cho là:S={0}
giải phương trình
a) \(\frac{4x-8}{2x^2+1}=0\)
b)\(\frac{x^2-x-6}{x-3}=0\)
c)\(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\)
d)\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
a.ĐK: 2x2+1\(\ne0\) \(\forall x\)
Để phương trình bằng 0 thì 4x-8=0 ( Vì 2x2+1 >0 với mọi x)
\(\Leftrightarrow x=2\) (TM)
Vậy ...
b.ĐK: x-3\(\ne0\) \(\Leftrightarrow x\ne3\)
Để phương trình bằng 0 thì x2-x-6=0 (Vì x-3\(\ne0\))
\(\Leftrightarrow\left[{}\begin{matrix}x=2\:\left(TM\right)\\x=-3\:\left(TM\right)\end{matrix}\right.\)
Vậy ...
c. ĐK: x\(\ne\)2
\(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\Leftrightarrow\frac{x+5}{3\left(x-2\right)}-\frac{1}{2}=\frac{2x-3}{2\left(x-2\right)}\)
\(\Leftrightarrow\frac{2\left(x+5\right)-3\left(x-2\right)}{6\left(x-2\right)}=\frac{3\left(2x-3\right)}{6\left(x-2\right)}\)
\(\Leftrightarrow2x+10-3x+6=6x-9\) (x\(\ne\)2)
\(\Leftrightarrow x=\frac{25}{7}\left(TM\right)\)
Vậy ...
d. ĐK: \(x\ne\pm\frac{1}{3}\)
\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
\(\Leftrightarrow\frac{12}{1-9x^2}=\frac{\left(1-3x\right)^2-\left(1+3x\right)^2}{1-9x^2}\)
\(\Leftrightarrow12=1-6x+9x^2-1-6x-9x^2\) (\(x\ne\pm\frac{1}{3}\))
\(\Leftrightarrow x=-2\:\left(TM\right)\)
Vậy...