mấy bạn ơi giúp mik gấp,mik đang cần gấp ,cảm ơn trc ạ
Cho phân thức M=(a2+b2+c2)(a+b+c)2+(ab+bc+ca)2 / (a+b+c)2-(ab+bc+ca)
a,Tìm các giá trị của a,b,c để phân thức được xác định(tức để mẫu ≠0)
b,Rút gọn M
quy đồng mẫu thức phân thức
2/x^2-5x+6 và 3/x-3
x^2-4x+4/x^2-2x và x+1/x^2-1
x^3-2^3/x2-4 và 3/x+2
2x/x2+3x+2 và 3x/x2+4x+3
quy đồng mẫu thức phân thức 4/x^2-3x+2 và 1/x^2-x
\(\dfrac{4}{x^2-3x+2}\) và \(\dfrac{1}{x^2-x}\)
\(\dfrac{4}{x^2-3x+2}=\dfrac{4}{\left(x-1\right)\left(x-2\right)}\)
\(\dfrac{1}{x^2-x}=\dfrac{1}{x\left(x-1\right)}\)
`MSC: x(x-1)(x-2)`
\(\dfrac{4}{\left(x-1\right)\left(x-2\right)}=\dfrac{4\cdot x}{x\left(x-1\right)\left(x-2\right)}=\dfrac{4x}{x\left(x-1\right)\left(x-2\right)}\)
\(\dfrac{1}{x\left(x-1\right)}=\dfrac{1\cdot\left(x-2\right)}{x\left(x-1\right)\left(x-2\right)}=\dfrac{x-2}{x\left(x-1\right)\left(x-2\right)}\)
Rút gọn phân thức
\(a,\dfrac{3x^2y+4xy^2}{6x+8y}=\dfrac{xy\left(3x+4y\right)}{2\left(3x+4y\right)}=\dfrac{xy}{2}\\ b,\dfrac{-3x^2-6x}{4-x^2}=\dfrac{-3x\left(x-2\right)}{-\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x+2}\\ c,\dfrac{8x^2y^2\left(x+y\right)}{4xy\left(x^2-y^2\right)}=\dfrac{4xy.\left(x+y\right).2xy}{4xy\left(x+y\right)\left(x-y\right)}=\dfrac{2xy}{x-y}\)
\(d,\dfrac{9x^3-18x}{3.\left(x^4-4\right)}=\dfrac{9x\left(x^2-2\right)}{3.\left(x^2-2\right)\left(x^2+2\right)}=\dfrac{9x}{3\left(x^2+2\right)}=\dfrac{3x}{x^2+2}\\ e,\dfrac{x\left(x+3\right)}{x^2\left(3+x\right)}=\dfrac{1}{x}\\ f,\dfrac{x^2-2x+1}{x^2-3x+2}=\dfrac{\left(x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}=\dfrac{x-1}{x-2}\)
Bài 3"
\(a,3ac^2.2ab=\left(3.2\right)\left(a.a\right).b.c^2=6a^2bc^2\\ a^2b.6c^2=6a^2bc^2\\ Vì:6a^2bc^2=6a^2bc^2\\ \Rightarrow3ac^2.2ab=6c^2.a^2b\\ \Rightarrow\dfrac{3ac^2}{a^2b}=\dfrac{6c^2}{2ab}\\ ---\\ b,\left(2x+y\right).4x^2=2x.4x^2+4x^2.y=8x^3+4x^2y\\ 2.\left(4x^2+2xy\right)=2.4x^2+2.2xy=8x^2+4xy\\ Vì:8x^3+4x^2y\ne8x^2+4xy\\ \Rightarrow\dfrac{2x+y}{4x^2+2xy}\ne\dfrac{2}{4x^2}\)
Bài 4:
\(a,\dfrac{3x^2y}{9xy^2}=\dfrac{3x^2y:3xy}{9xy^2:3xy}=\dfrac{x}{3y}\\ b,\dfrac{x^2+2x}{x^2+4x+4}=\dfrac{x\left(x+2\right)}{\left(x+2\right)^2}=\dfrac{x}{x+2}\\ c,\dfrac{x^2y^2-xy}{xy-1}=\dfrac{xy\left(xy-1\right)}{xy-1}=xy\)
3:
a: \(\dfrac{3ac^2}{a^2b}=\dfrac{a\cdot3c^2}{a\cdot ab}=\dfrac{3c^2}{ab}=\dfrac{3c^2\cdot2}{ab\cdot2}=\dfrac{6c^2}{2ab}\)
b: \(\dfrac{2x+y}{4x^2+2xy}=\dfrac{2x+y}{2x\cdot2x+2x\cdot y}=\dfrac{2x+y}{2x\left(2x+y\right)}=\dfrac{1}{2x}=\dfrac{1\cdot2}{2x\cdot2}=\dfrac{2}{4x}< >\dfrac{2}{4x^2}\)
Bài 1 :
\(\dfrac{3x+2}{x-5}\) xác định \(\Leftrightarrow x-5\ne0\Leftrightarrow x\ne5\)
\(\dfrac{x+15}{x-2y}\) xác định \(\Leftrightarrow x-2y\ne0\Leftrightarrow x\ne2y\)
\(\dfrac{x+3}{x^2+1}\) xác định \(\Leftrightarrow x^2+1\ne0\Leftrightarrow\forall x\in R\)
\(\dfrac{5x-4}{\left(3x-2\right)\left(3x+2\right)}\) xác định \(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)\ne0\Leftrightarrow\left\{{}\begin{matrix}3x-2\ne0\\3x+2\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{2}{3}\\x\ne-\dfrac{2}{3}\end{matrix}\right.\)
1:
a: ĐKXĐ: x-5<>0
=>x<>5
b: ĐKXĐ: x-2y<>0
=>x<>2y
c: ĐKXĐ: x^2+1<>0
=>x^2<>-1(luôn đúng)
d: ĐKXĐ: (3x-2)(3x+2)<>0
=>3x-2<>0 và 3x+2<>0
=>x<>2/3 và x<>-2/3
2:
a: ĐKXĐ: x<>-1
\(A=\dfrac{3x\left(x+1\right)}{\left(x+1\right)^2}=\dfrac{3x}{x+1}\)
Khi x=-2 thì \(A=\dfrac{3\cdot\left(-2\right)}{-2+1}=6\)
b: ĐKXĐ: x<>y; x<>-y
\(B=\dfrac{y\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{y}{x+y}\)
Khi x=4 và y=-2 thì \(B=\dfrac{-2}{-2+4}=-\dfrac{2}{2}=-1\)
Bài 2 :
\(A=\dfrac{3x^2+3x}{x^2+2x+1}\)
\(\Leftrightarrow A=\dfrac{3x\left(x+1\right)}{\left(x+1\right)^2}\left(x\ne-1\right)\)
\(\Leftrightarrow A=\dfrac{3x}{x+1}\)
\(\Leftrightarrow A=\dfrac{3.\left(-2\right)}{-2+1}=6\left(x=-2\right)\)
\(B=\dfrac{xy-y^2}{x^2-y^2}\)
\(\Leftrightarrow B=\dfrac{x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\left(x\ne\pm y\right)\)
\(\Leftrightarrow B=\dfrac{x}{\left(x+y\right)}\)
\(\Leftrightarrow B=\dfrac{4}{4-2}=2\left(x=4;y=-2\right)\)
1) \(5\left(x-y\right)-y\left(x-y\right)\)
\(=\left(x-y\right)\left(5-y\right)\)
2) \(x\left(y+1\right)+8\left(y+1\right)\)
\(=\left(y+1\right)\left(x+8\right)\)
3) \(5\left(x-y\right)-x\left(x-y\right)\)
\(=\left(x-y\right)\left(5-x\right)\)
4) \(z\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(z-5\right)\)
5) \(3x\left(x+5\right)-2\left(5+x\right)\)
\(=3x\left(x+5\right)-2\left(x+5\right)\)
\(=\left(x+5\right)\left(3x-2\right)\)
6) \(x^2\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+4\right)\)
#Urushi
1: =(x-y)(5-y)
2: =(y+1)(x+8)
3: =(x-y)(5-x)
4: =(x+y)(z-5)
5: =(x+5)(3x-2)
6: =(x-1)(x^2+4)