Hay chứng minh:
\(\text{a}\)) \(\frac{x^5-1}{x-1}=x^4+x^3+x^2+x+1\)
b)\(\frac{2x^2+xy-y^2}{2x^2-3xy+y^2}\frac{x+y}{x-y}\)
8,Thực hiện phép tính
a,\(\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}\)
b,\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
c,\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)
d,\(\frac{1}{x-y}+\frac{3xy}{y^3-x^3}+\frac{x-y}{x^2+xy+y^2}\)
e,\(\frac{2x+y}{2x^2-xy}+\frac{16x}{y^2-4x^2}+\frac{2x-y}{2x^2+xy}\)
f,\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
Giúp mình với ạ!!! ai trả lời nhanh mình tick luôn nhé
a, \(\frac{2x^2-x}{x^2+x+1}+\frac{x^3-2x^2}{x^2+x+1}+\frac{x-1}{x^2+x+1}\)
b, \(\frac{2x+y}{x\left(y^2-x\right)}-\frac{2x-y}{x\left(y^2-x\right)}\)
c, \(\frac{4}{x+2}+\frac{3}{x-2}+\frac{-5-2}{x^2-4}\)
d, \(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)
e, \(\frac{1}{x-y}+\frac{3xy}{y^3-x^3}+\frac{x-y}{x^2+xy+y^2}\)
f, \(\frac{3}{x^2+2xy+y^2}+\frac{4}{2xy-x^2-y^2}+\frac{5}{x^2-y^2}\)
Bài 2: Rút gọn phân thức
\(A=\frac{10x^2-7+5x-2xy}{1-2x^2+x}\)
Bài 3: Chứng minh rằng
a) \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}=\frac{xy+y^2}{2x-y}\)
b) \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}=\frac{1}{x-y}\)
Bài 4: Quy đồng mẫu thức các phân thức sau
a) \(\frac{5x}{\left(x+3\right)^3}\&\frac{x-4}{3x\left(x+2\right)^2}\)
b) \(\frac{x+1}{x-x^2}\&\frac{x+2}{2x^2+2-4x}\)
Ta có: \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}\)
\(=\frac{x^2y+xy^2+xy^2+y^3}{2x^2+2xy-xy-y^2}\)
\(=\frac{xy\left(x+y\right)+y^2\left(x+y\right)}{2x\left(x+y\right)-y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)\left(xy+y^2\right)}{\left(2x-y\right)\left(x+y\right)}=\frac{xy+y^2}{2x-y}\left(đpcm\right)\)
Ta có: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)
\(=\frac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)
\(=\frac{x\left(x+y\right)+2y\left(x+y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}\)
\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)
cho x,y,z là các số dương thỏa \(x^2+y^2+z^2=3\)
chứng minh:\(\frac{x^2}{y+2\text{z}}+\frac{y^2}{z+2x}+\frac{z^2}{x+2y}+\frac{1}{1+\sqrt{3+2\left(xy+yz+x\text{z}\right)}}\ge\frac{5}{4}\)
+\(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3\)
+\(3+2\left(xy+yz+zx\right)=x^2+y^2+z^2+2\left(xy+yz+zx\right)=\left(x+y+z\right)^2\le9\)
\(\Rightarrow B=\frac{1}{1+\sqrt{3+2\left(xy+yz+zx\right)}}\ge\frac{1}{1+3}=\frac{1}{4}\)
+\(A=\frac{x^2}{y+2z}+\frac{y^2}{z+2x}+\frac{z^2}{x+2y}=\frac{x^4}{x^2y+2zx^2}+\frac{y^4}{y^2z+2xy^2}+\frac{z^4}{z^2x+2yz^2}\)
\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2y+y^2z+z^2x+2\left(xy^2+yz^2+zx^2\right)}\)
Áp dụng bđt Bunhiacopxki
\(x^2y+y^2z+z^2x=x.xy+y.yz+z.zx\le\sqrt{x^2+y^2+z^2}.\sqrt{x^2y^2+y^2z^2+z^2x^2}\)
\(\le\sqrt{x^2+y^2+z^2}.\sqrt{\frac{\left(x^2+y^2+z^2\right)^2}{3}}=3\)
(áp dụng \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\))
Tương tự: \(xy^2+yz^2+zx^2\le3\)
\(\Rightarrow B\ge\frac{3^2}{3+2.3}=1\)
\(VT=A+B\ge1+\frac{1}{4}=\frac{5}{4}=VP\)
Bài 1 :Thực hiện phép tính
a, \(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)
b\(\frac{1}{x-y}+\frac{3xy}{y^3-x^3}+\frac{x-y}{x^2+xy+y^2}\)
c, \(\frac{xy}{2x-y}-\frac{x^2-1}{y-2x}\)
d,\(\frac{2\left(x+y\right)\left(x-y\right)}{x}-\frac{-2y^2}{x}\)
Bài 2: Thực hiện phép tính
a,\(\frac{4x+1}{2}-\frac{3x+2}{3}\)
b,\(\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\)
c,\(\frac{x+3}{x^2+1}-\frac{1}{x^2+2}\)
e,\(\frac{3}{2x^2+2x}+\frac{2x-1}{x^2-1}-\frac{2}{x}\)
d,\(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\)
f,\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
Bài 1:
a) Ta có: \(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)
\(=\frac{2x}{x\left(x+2y\right)}+\frac{y}{y\left(x-2y\right)}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2}{x+2y}+\frac{y}{x-2y}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2\left(x-2y\right)}{\left(x+2y\right)\left(x-2y\right)}+\frac{y\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2x-4y+xy+2y^2+4}{\left(x-2y\right)\cdot\left(x+2y\right)}\)
b) Ta có: \(\frac{1}{x-y}+\frac{3xy}{y^3-x^3}+\frac{x-y}{x^2+xy+y^2}\)
\(=\frac{x^2+xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}-\frac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\frac{\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\frac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\frac{2x^2-4xy+2y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\frac{2\left(x^2-2xy+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\frac{2\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\frac{2x-2y}{x^2+xy+y^2}\)
c) Ta có: \(\frac{xy}{2x-y}-\frac{x^2-1}{y-2x}\)
\(=\frac{xy}{2x-y}+\frac{x^2-1}{2x-y}\)
\(=\frac{x^2+xy-1}{2x-y}\)
d) Ta có: \(\frac{2\left(x+y\right)\left(x-y\right)}{x}-\frac{-2y^2}{x}\)
\(=\frac{2\left(x^2-y^2\right)+2y^2}{x}\)
\(=\frac{2x^2-2y^2+2y^2}{x}\)
\(=\frac{2x^2}{x}=2x\)
Bài 2:
a) Ta có: \(\frac{4x+1}{2}-\frac{3x+2}{3}\)
\(=\frac{3\left(4x+1\right)}{6}-\frac{2\left(3x+2\right)}{6}\)
\(=\frac{12x+3-6x-4}{6}\)
\(=\frac{6x-1}{6}\)
b) Ta có: \(\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\)
\(=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x^2}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)
\(=\frac{x^2-9-x^2+9}{x\left(x-3\right)}=\frac{0}{x\left(x-3\right)}=0\)
c) Ta có: \(\frac{x+3}{x^2+1}-\frac{1}{x^2+2}\)
\(=\frac{\left(x+3\right)\left(x^2+2\right)}{\left(x^2+1\right)\left(x^2+2\right)}-\frac{x^2+1}{\left(x^2+2\right)\left(x^2+1\right)}\)
\(=\frac{x^3+2x+3x^2+6-x^2-1}{\left(x^2+1\right)\left(x^2+2\right)}\)
\(=\frac{x^3+2x^2+2x+5}{\left(x^2+1\right)\left(x^2+2\right)}\)
e) Ta có: \(\frac{3}{2x^2+2x}+\frac{2x-1}{x^2-1}-\frac{2}{x}\)
\(=\frac{3}{2x\left(x+1\right)}+\frac{2x-1}{\left(x+1\right)\left(x-1\right)}-\frac{2}{x}\)
\(=\frac{3\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}+\frac{2x\left(2x-1\right)}{2x\left(x+1\right)\left(x-1\right)}-\frac{2\cdot2\cdot\left(x+1\right)\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}\)
\(=\frac{3x-3+4x^2-2x-4\left(x^2-1\right)}{2x\left(x+1\right)\left(x-1\right)}\)
\(=\frac{4x^2+x-3-4x^2+4}{2x\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x+1}{2x\left(x+1\right)\left(x-1\right)}=\frac{1}{2x\left(x-1\right)}\)
d) Ta có: \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\)
\(=\frac{3x+2}{\left(3x-2\right)\left(3x+2\right)}-\frac{4\left(3x-2\right)}{\left(3x+2\right)\left(3x-2\right)}-\frac{-10x+8}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\frac{3x+2-12x+8+10x-8}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\frac{x+2}{\left(3x-2\right)\left(3x+2\right)}\)
f) Ta có: \(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x-y\right)}\)
\(=\frac{3x\cdot2\cdot\left(x-y\right)}{10\left(x+y\right)\left(x-y\right)}-\frac{x\cdot\left(x+y\right)}{10\left(x-y\right)\left(x+y\right)}\)
\(=\frac{6x^2-6xy-x^2-xy}{10\left(x-y\right)\left(x+y\right)}\)
\(=\frac{5x^2-7xy}{10\left(x-y\right)\left(x+y\right)}\)
Rút gọn:
a) \(\frac{x^3+2x^2+1}{4x^2-4}.\frac{x+2}{x^2+1}.\frac{2x^2-2}{x^3+2x^2+1}\)
b)\(\frac{x^4-y^4}{x^2+y^2-2xy}.\frac{x-y}{xy+x^2}\)
c)\(\frac{x^2-9}{x+5}.\frac{2x}{x+3}+\frac{x^2-9}{x+5}.\frac{5-x}{x+3}\)
chứng minh đẳng thức
\(\left(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{x^2-xy-2y^2}\right):\frac{x^4+4x^2y^2+y^4-4}{x^2+y+xy+x}:\frac{1}{2x^2+y+2}=\frac{x+1}{2y-x}\)
CMR
a) \(\frac{2x^2+3xy+y^2}{2x^3+x^2y-2xy^2-y^3}\)=\(\frac{1}{x-y}\)
b) \(\frac{x^2y-2xy^2+y^3}{2x^2-xy-y^2}\)=\(\frac{y-\left(x-y\right)}{2x+y}\)
c) \(\frac{4x^2-4xy+y^2}{y^3-6y^2x+12yx^2-8x}=\frac{-1}{2x-y}\)
\(\frac{^{x^2}+3xy+y^2}{x^3+2x^2y-xy^2-2y^3}=\frac{1}{x-y}\)
chứng minh đẳng thức trên :