1. Đáp án nào đúng:
a) \(\dfrac{3x}{5y}=\dfrac{3x\left(x-2\right)}{5y\left(x-2\right)}\)
b) \(\dfrac{3x}{5y}=\dfrac{2x\left(x-2\right)}{3y\left(x+2\right)}\)
c) \(\dfrac{3x}{5y}=\dfrac{9x}{15y}\)
d) \(\dfrac{3x}{5y}=\dfrac{3x.x}{5y.x}\)
2. Tìm đa thức M trong đẳng thức \(\dfrac{8\left(x-y\right)}{4\left(x^2-y^2\right)}\)= \(\dfrac{ }{x+y}\)
3. Rút gọn phân thức \(\dfrac{6x^2y^3}{8x^3y^3}=\)
4. Rút gọn phân thức \(\dfrac{20xy\left(x+y\right)}{5xy\left(x-y\right)}=\)
5. Rút gọn phân thức \(\dfrac{6x-12}{\left(x+3\right)\left(x-3\right)}=\)
6. Rút gọn phân thức \(\dfrac{4\left(x-1\right)-2\left(1-x\right)}{6\left(x-1\right)}=\)
giúp mình nhé mng mình đang gấp ạ
1A,B,D
2 M=2
3 \(=\dfrac{3}{4x}\)
4 \(=\dfrac{4\left(x+y\right)}{x-y}=\dfrac{4x+4y}{x-y}\)
5 K rút gọn đc
6 \(=\dfrac{4\left(x-1\right)+2\left(x-1\right)}{6\left(x-1\right)}=\dfrac{6\left(x-1\right)}{6\left(x-1\right)}=1\)
\(b,\dfrac{x}{2}-\left(\dfrac{3x}{5}-\dfrac{13}{5}\right)=-\left(\dfrac{7}{5}+\dfrac{7}{10}.x\right)\)
\(c,\dfrac{2x-3}{3}+\dfrac{-3}{2}=\dfrac{5-3x}{6}-\dfrac{1}{3}\)
\(d,\dfrac{2}{3x}-\dfrac{3}{12}=\dfrac{4}{5}-\left(\dfrac{7}{x}-2\right)\)
\(e,2\left(x-1\right)=\left(x-1\right)^2\)
\(b,\Rightarrow\dfrac{x}{2}-\dfrac{3x}{5}-\dfrac{13}{5}=-\dfrac{7}{5}-\dfrac{7x}{10}\\ \Rightarrow\dfrac{1}{2}x-\dfrac{3}{5}x+\dfrac{7}{10}x=\dfrac{6}{5}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{6}{5}\Rightarrow x=2\\ c,\Rightarrow\dfrac{2x-3}{3}-\dfrac{5-3x}{6}=-\dfrac{1}{3}+\dfrac{3}{2}=\dfrac{7}{6}\\ \Rightarrow\dfrac{4x-6-5+3x}{6}=\dfrac{7}{6}\\ \Rightarrow7x-11=7\Rightarrow x=\dfrac{18}{7}\\ d,\Rightarrow\dfrac{2}{3x}+\dfrac{7}{x}=\dfrac{4}{5}+2+\dfrac{3}{12}=\dfrac{61}{20}\\ \Rightarrow\dfrac{23}{3x}=\dfrac{61}{20}\\ \Rightarrow183x=460\\ \Rightarrow x=\dfrac{460}{183}\\ e,\Rightarrow2\left(x-1\right)-\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left(2-x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
e: Ta có: \(\left(x-1\right)^2=2\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
BÀI 1 . chứng tỏ rằng các đơn thức sau là đơn thức đồng dạng :
a) \(\dfrac{2}{3}x^5y^2\) d)\(x^3y^2.\left(x.y^2\right)\)
b)\(-3x^3y.\dfrac{1}{5}x^2y\)
c)\(\dfrac{2}{5}x^3.\dfrac{1}{2}\left(x.y\right)^2\) e) \(\dfrac{1}{2}.\left(x.y\right)^2.\dfrac{2}{5}\left(x.y\right)^2\)
Lời giải:
\(x^3y^2(xy^2)=x^3.x.y^2.y^2=x^4y^4\)
\(-3x^3y.\frac{1}{5}x^2y=\frac{-3}{5}x^3.x^2.y.y=\frac{-3}{5}x^5y^2\)
\(\frac{2}{5}x^3\frac{1}{2}(xy)^2=\frac{1}{5}x^3.x^2.y^2=\frac{1}{5}x^5y^2\)
\(\frac{1}{2}(xy)^2\frac{2}{5}(xy)^2=\frac{1}{5}x^2.x^2.y^2.y^2=\frac{1}{5}x^4y^4\)
Vậy các đơn thức phần a,b,c đồng dạng với nhau; đơn thức d và e đồng dạng với nhau.
quy đồng mẫu thức của các phân tử
a, \(\dfrac{2}{5x^3y};\dfrac{5}{xy^2}\)
b, \(\dfrac{3x}{x-5};\dfrac{-2}{3\left(x-5\right)}\)
c, \(\dfrac{1}{\left(x+3\right)\left(x-3\right)};\dfrac{5x}{x^2-9}\)
Phân tích đa thức \(18x^3-\dfrac{8}{25}x\) thành nhân tử
a. \(\dfrac{2}{25}x\left(9x^2-4\right)=\dfrac{2}{25}x\left(3x-2\right)\left(3x+2\right)\)
b. \(2x\left(9x^2-\dfrac{4}{25}\right)=2x\left(3x-\dfrac{2}{5}\right)\left(3x+\dfrac{2}{5}\right)\)
Cách phân tích nào đúng, a hay b. Giải thích vì sao?
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
\(a,VP=\dfrac{x\left(x+3\right)}{x\left(2x-5\right)}=\dfrac{x+3}{2x-5}=VT\\ b,VP=\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}=\dfrac{3-x}{x+3}=VT\\ c,VP=\dfrac{\left(x+4\right)\left(x^2-4x+16\right)}{\left(3-x\right)\left(x^2-4x+16\right)}=\dfrac{x+4}{3-x}=VP\left(bạn.sửa.lại.đề.đi\right)\\ d,VT=\dfrac{x^3-2x^2+8x^2-16x+15x-30}{x^3-5x^2+8x^2-40x+15x-75}\\ =\dfrac{\left(x-2\right)\left(x^2+8x+15\right)}{\left(x-5\right)\left(x^2+8x+15\right)}=\dfrac{x-2}{x-5}=VP\)
Bài 4: Chứng minh rằng các đẳng thức sau bằng nhau
a)\(\dfrac{3x\left(x+5\right)}{2\left(x+5\right)}\)=\(\dfrac{6x^2+30x}{4}\)
b)\(\dfrac{x+2}{x-1}\)=\(\dfrac{\left(x+2\right)\left(x+1\right)}{x^2-1}\)
a/ ĐK: $x\ne -5$
$\dfrac{6x^2+30x}{4}=\dfrac{6x(x+5)}{4}=\dfrac{3x(x+5)}{2}$
Đề này sai
b/ ĐK: $x\ne \pm 1$
$\dfrac{(x+2)(x+1)}{x^2-1}\\=\dfrac{(x+2)(x+1)}{(x-1)(x+1)}\\=\dfrac{x+2}{x-1}$
$\to$ ĐPCM
a, Xét \(VT=\dfrac{3x\left(x+5\right)}{2\left(x+5\right)}=\dfrac{3x}{2}\)
\(VP=\dfrac{6x^2+30x}{4}=\dfrac{6x\left(x+5\right)}{4}=\dfrac{3x\left(x+5\right)}{2}\)
Vậy \(VT\ne VP\)hay đpcm ko xảy ra
b, \(VP=\dfrac{\left(x+2\right)\left(x+1\right)}{x^2-1}=\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+2}{x-1}=VT\)
Vậy ta có đpcm
Dùng định nghĩa hai phân thức bằng nhau chứng minh các đẳng thức sau :
a) \(\dfrac{x^2y^3}{5}=\dfrac{7x^3y^4}{35xy}\)
b) \(\dfrac{x^2\left(x+2\right)}{x\left(x+2\right)^2}=\dfrac{x}{x+2}\)
c) \(\dfrac{3-x}{3+x}=\dfrac{x^2-6x+9}{9-x^2}\)
d) \(\dfrac{x^3-4x}{10-5x}=\dfrac{-x^2-2x}{5}\)
a. \(x^2y^3.35xy=5.7x^3y^4\)
\(\Leftrightarrow35x^3y^4=35x^3y^4\Rightarrowđpcm\)
\(b.x^2\left(x+2\right).\left(x+2\right)=x\left(x+2\right)^2.x\)
\(\Leftrightarrow x^2\left(x+2\right)^2=x^2\left(x+2\right)^2\Rightarrowđpcm\)
\(c.\left(3-x\right)\left(9-x^2\right)=\left(3+x\right)\left(x^2-6x+9\right)\)
\(\Leftrightarrow\left(3-x\right)\left(3-x\right)\left(3+x\right)=\left(3+x\right)\left(3-x\right)^2\)
\(\Leftrightarrow\left(3-x\right)^2\left(3+x\right)=\left(3-x\right)^2\left(3+x\right)\)
\(\Rightarrowđpcm\)
\(d.5\left(x^3-4x\right)=\left(10-5x\right)\left(-x^2-2x\right)\)
\(\Leftrightarrow5x^3-20x=5x^3-20x\Rightarrowđpcm\)
BÀI 1 : RÚT GỌN CÁC BIỂU THỨC SAU .
a, \(\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}\)
b, \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{9x^2-6x+1}\)
c, \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
d, \(\dfrac{1-x^2}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)
câu d
\(D=\dfrac{\left(1-x^2\right)}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{\left(1-x^2\right)\left(x^2-x-3\right)+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{x^2-x-3-x^4+x^3-3x^2+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x^4+x^3+x^2-15x}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x\left(x^3-x^2-x+15\right)}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-\left(x^3-x^2-x+15\right)}{\left(x+3\right)}\end{matrix}\right.\)
