Chứng tỏ mỗi cặp phân thức sau bằng nhau :
a) \(\dfrac{3}{2x-3}\) và \(\dfrac{3x+6}{2x^2+x-6}\)
b) \(\dfrac{2}{x+4}\) và \(\dfrac{2x^2+6x}{x^3+7x^2+12x}\)
Dùng tính chấ cơ bản của phân thức chứng tỏ rằng các cặp phân thức sau bằng nhau :
a) \(\dfrac{x^2+3x+2}{3x+6}\) và \(\dfrac{2x^2+x-1}{6x-3}\)
b) \(\dfrac{15x-10}{3x^2+3x-\left(2x+2\right)}\) và \(\dfrac{5x^2-5x+5}{x^3+1}\)
a ) \(\dfrac{x^2+3x+2}{3x+6}=\dfrac{\left(x+1\right)\left(x+2\right)}{3\left(x+2\right)}=\dfrac{x+1}{3}\) (1)
\(\dfrac{2x^2+x-1}{6x-3}=\dfrac{\left(2x-1\right)\left(x+1\right)}{3\left(2x-1\right)}=\dfrac{x+1}{3}\) (2)
Từ (1) ; (2) \(\Rightarrow\dfrac{x^2+3x+2}{3x+6}=\dfrac{2x^2+x-1}{6x-3}\) (đpcm)
b ) \(\dfrac{15x-10}{3x^2+3x-\left(2x+2\right)}=\dfrac{5\left(3x-2\right)}{\left(3x-2\right)\left(x+1\right)}=\dfrac{5}{x+1}\) (3)
\(\dfrac{5x^2-5x+5}{x^3+1}=\dfrac{5\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5}{x+1}\) (4)
Từ (3) và (4) \(\Rightarrow\dfrac{15x-10}{3x^2+3x-\left(2x+2\right)}=\dfrac{5x^2-5x+5}{x^3+1}\) (đpcm)
a) \(\dfrac{x^2+3x+2}{3x+6}=\dfrac{x^2+x+2x+2}{3\left(x+2\right)}=\dfrac{\left(x^2+x\right)+\left(2x+2\right)}{3\left(x+2\right)}=\dfrac{x\left(x+1\right)+2\left(x+1\right)}{3\left(x+2\right)}=\dfrac{\left(x+1\right)\left(x+2\right)}{3\left(x+2\right)}=\dfrac{x+1}{3}\left(1\right)\) \(\dfrac{2x^2+x-1}{6x-3}=\dfrac{2x^2+2x-x-1}{3\left(2x-1\right)}=\dfrac{2x\left(x+1\right)-\left(x+1\right)}{3\left(2x-1\right)}=\dfrac{\left(2x-1\right)\left(x+1\right)}{3\left(2x-1\right)}=\dfrac{x+1}{3}\left(2\right)\) Từ (1)và (2)=> \(\dfrac{x^2+3x+2}{3x+6}=\dfrac{2x^2+x-1}{6x-3}\) b)\(\dfrac{15x-10}{3x^2+3x-\left(2x+2\right)}=\dfrac{5\left(3x-2\right)}{3x\left(x+1\right)-2\left(x+1\right)}=\dfrac{5\left(3x-2\right)}{\left(3x-2\right)\left(x+1\right)}=\dfrac{5}{x+1}\left(3\right)\) \(\dfrac{5x^2-5x+5}{x^3+1}=\dfrac{5\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5}{x+1}\left(4\right)\) Từ (3) và (4) => \(\dfrac{15x-10}{3x^2+3x-\left(2x+2\right)}=\dfrac{5x^2-5x+5}{x^3+1}\)
1. Chứng tỏ cặp phân thức sau = nhau:
\(\dfrac{x^2-10x+21}{^{ }x^3-7x^2+x-7}\) và \(\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}\)
2. Điền vào chỗ trống những đa thức thích hợp:
a) \(\dfrac{....}{x^2+3x+2}\)=\(\dfrac{3x^2+4x-4}{3x^2+7x+2}\) b) \(\dfrac{5x^2+7x-2}{2x+1}=\dfrac{5x^3-8x^2-23x+6}{...}\)
c) \(\dfrac{...}{6x^2+8x+2}=\dfrac{3x^2+4x-4}{3x^2+7x+2}\)
Câu 1:
\(\dfrac{x^2-10x+21}{x^3-7x^2+x-7}=\dfrac{\left(x-7\right)\left(x-3\right)}{\left(x-7\right)\left(x^2+1\right)}=\dfrac{x-3}{x^2+1}\)
\(\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}=\dfrac{2x^2-6x+5x-15}{\left(2x+5\right)\left(x^2+1\right)}=\dfrac{\left(2x+5\right)\left(x-3\right)}{\left(2x+5\right)\left(x^2+1\right)}=\dfrac{x-3}{x^2+1}\)
Do đó: \(\dfrac{x^2-10x+21}{x^3-7x^2+x-7}=\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}\)
a, \(\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
b, \(\dfrac{x+2}{x-3}-\dfrac{x^2+6}{x^2-3x}\)
c, \(\dfrac{1}{9x-18}+\dfrac{16-7x}{72-18x}+\dfrac{5}{12x-24}\)
a.\(\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}=\dfrac{3x-x+6}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+3\right)}=\dfrac{1}{x}\)
2.Dùng định nghĩa hai phân thức bằng nhau,hãy tìm đa thức A trong đảng thức sau
a,\(\dfrac{A}{3x+1}\)=\(\dfrac{9x^2-6x-1}{3x-1}\) b,\(\dfrac{2x-3}{A}\)=\(\dfrac{6x^2-7x-3}{12x+4}\)
c,\(\dfrac{12x+4}{4x+28}\)=\(\dfrac{A}{2x^2+8x-21}\) d,\(\dfrac{x^2+4x+4}{x^2-4}\)=\(\dfrac{x^2+3x+2}{A}\)
d: \(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{\left(x+1\right)\left(x+2\right)}{A}\)
hay A=x-2
Hãy chứng tỏ các phân thức sau bằng nhau
a/ \(\dfrac{x+3}{2x-5}=\dfrac{x^2+3x}{2x^2-5x}\)
b/ \(\dfrac{3-x}{x+3}=\dfrac{x^2-6x+9}{9-x^{ }}\)
c/ \(\dfrac{x^3+64}{\left(3-x\right)\left(x^2-4x+16\right)}\)\(=\dfrac{x-4}{x-3}\)
d/ \(\dfrac{x^3+6x^2-x-30}{x^3+3x^2-25x-75}=\dfrac{x-2}{x-5}\)
AI GIÚP MK VS Ạ AI NHANH MK SẼ VOTE Ạ
\(a,VP=\dfrac{x\left(x+3\right)}{x\left(2x-5\right)}=\dfrac{x+3}{2x-5}=VT\\ b,VP=\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}=\dfrac{3-x}{x+3}=VT\\ c,VP=\dfrac{\left(x+4\right)\left(x^2-4x+16\right)}{\left(3-x\right)\left(x^2-4x+16\right)}=\dfrac{x+4}{3-x}=VP\left(bạn.sửa.lại.đề.đi\right)\\ d,VT=\dfrac{x^3-2x^2+8x^2-16x+15x-30}{x^3-5x^2+8x^2-40x+15x-75}\\ =\dfrac{\left(x-2\right)\left(x^2+8x+15\right)}{\left(x-5\right)\left(x^2+8x+15\right)}=\dfrac{x-2}{x-5}=VP\)
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) \(\dfrac{5x^2}{x^3+6x^2+12x+8};\dfrac{4x}{x^2+4x+4};\dfrac{3}{2x+4}\)
Dùng định nghĩa hai phân thức bằng nhau, hãy tìm đa thức A trong mỗi đẳng thức sau :
a) \(\dfrac{A}{2x-1}=\dfrac{6x^2+3x}{4x^2-1}\)
b) \(\dfrac{4x^2-3x-7}{A}=\dfrac{4x-7}{2x+3}\)
c) \(\dfrac{4x^2-7x+3}{x^2-1}=\dfrac{A}{x^2+2x+1}\)
d) \(\dfrac{x^2-2x}{2x^2-3x-2}=\dfrac{x^2+2x}{A}\)
Giải phương trình:
a) \(\dfrac{3x-2}{x^2-12x+20}-\dfrac{4x+3}{x^2+6x-16}=\dfrac{7x+11}{x^2-2x-80}\)
b) \(\dfrac{2x-5}{x^2+5x-36}-\dfrac{x-6}{x^2+3x-28}=\dfrac{x+8}{x^2+16x+63}\)
a: \(\Leftrightarrow\dfrac{3x-2}{\left(x-2\right)\left(x-10\right)}-\dfrac{4x+3}{\left(x+8\right)\left(x-2\right)}=\dfrac{8x+11}{\left(x-10\right)\left(x+8\right)}\)
=>(3x-2)(x+8)-(4x+3)(x-10)=(8x+11)(x-2)
=>3x^2+24x-2x-16-4x^2+40x-3x+30=8x^2-16x+11x-22
=>-x^2+59x+14-8x^2+5x+22=0
=>-9x^2+54x+36=0
=>x^2-6x-4=0
=>\(x=3\pm\sqrt{13}\)
b: \(\Leftrightarrow\dfrac{2x-5}{\left(x+9\right)\left(x-4\right)}-\dfrac{x-6}{\left(x+7\right)\left(x-4\right)}=\dfrac{x+8}{\left(x+9\right)\left(x+7\right)}\)
=>(2x-5)(x+7)-(x-6)(x+9)=(x+8)(x-4)
=>2x^2+14x-5x-35-x^2-9x+6x+54=x^2+4x-32
=>x^2+6x+19=x^2+4x-32
=>2x=-51
=>x=-51/2
Đưa các phân thức sau về cùng mẫu
a) \(\dfrac{x}{2x^2+7x-15}\); \(\dfrac{x+2}{x^2+3x-10}\); \(\dfrac{1}{x+5}\)
b) \(\dfrac{1}{-x^2+3x-2}\); \(\dfrac{1}{x^2+5x-6}\); \(\dfrac{1}{-x^2+4x-3}\)
c)\(\dfrac{3}{x^3-1}\); \(\dfrac{2x}{x^2+x+1}\); \(\dfrac{x}{x-1}\)
d)\(\dfrac{x}{x^2-2xy+y^2-x^2}\); \(\dfrac{y}{x^2+2yz-y^2-z^2}\); \(\dfrac{z}{x^2-2xz-y^2+z^2}\)
a: \(\dfrac{x}{2x^2+7x-15}=\dfrac{x}{\left(x+5\right)\left(2x-3\right)}=\dfrac{x^2-2x}{\left(x+5\right)\left(x-2\right)\left(2x-3\right)}\)
\(\dfrac{x+2}{x^2+3x-10}=\dfrac{x+2}{\left(x+5\right)\left(x-2\right)}=\dfrac{\left(x+2\right)\left(2x-3\right)}{\left(2x-3\right)\left(x+5\right)\left(x-2\right)}\)
\(\dfrac{1}{x+5}=\dfrac{\left(2x-3\right)\left(x-2\right)}{\left(2x-3\right)\left(x-2\right)\left(x+5\right)}\)
b: \(\dfrac{1}{-x^2+3x-2}=\dfrac{-1}{\left(x-1\right)\left(x-2\right)}=\dfrac{-\left(x+6\right)\left(x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x+6\right)\left(x-3\right)}\)
\(\dfrac{1}{x^2+5x-6}=\dfrac{1}{\left(x+6\right)\left(x-1\right)}=\dfrac{\left(x-2\right)\left(x-3\right)}{\left(x+6\right)\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)
\(\dfrac{1}{-x^2+4x-3}=\dfrac{-1}{\left(x-1\right)\left(x-3\right)}=\dfrac{-\left(x-2\right)\left(x+6\right)}{\left(x-1\right)\left(x-3\right)\left(x+6\right)\left(x-2\right)}\)
c: \(\dfrac{3}{x^3-1}=\dfrac{3}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\dfrac{2x}{x^2+x+1}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\dfrac{x}{x-1}=\dfrac{x\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)