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a) \(\dfrac{x^3}{x+1}+\dfrac{x^2}{x-1}+\dfrac{1}{x+1}+\dfrac{1}{1-x}\)
\(=\dfrac{x^3}{x+1}+\dfrac{x^2}{x-1}+\dfrac{1}{x+1}+\dfrac{-1}{x-1}\)
\(=\dfrac{x^3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{-1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^4-x+x^3+x+x-1-x+1}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^4+x^3}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^3\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{x^3}{x-1}\)
b) \(\dfrac{x^3}{x-1}-\dfrac{x^2}{x+1}-\dfrac{1}{x-1}+\dfrac{1}{x+1}\)
\(=\dfrac{x^3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x^2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{1\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^3\left(x+1\right)-x^2\left(x-1\right)-1\left(x+1\right)+1\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^4+x^3-x^3+x^2-x-1+x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^4+x^2-2}{\left(x-1\right)\left(x+1\right)}\)
c) \(\dfrac{4-2x+x^2}{2+x}-2-x\)
\(=\dfrac{4-2x+x^2}{2+x}-\dfrac{2\left(2+x\right)}{2+x}-\dfrac{x\left(2+x\right)}{2+x}\)
\(=\dfrac{4-2x+x^2-4-2x-2x-x^2}{2+x}\)
\(=\dfrac{-6x}{2+x}\)
Còn lại thì dễ rồi, bạn tự làm nha ^^
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)
\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)
Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)
\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)
Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)
\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)
Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
\(A=\dfrac{x^3}{9y^2}-\dfrac{1}{8}x^2y+\dfrac{2}{15}xy^2\\ B=\dfrac{2a-b}{a+1}-\dfrac{\left(a-1\right)^2}{b-2}\cdot\dfrac{\left(b-2\right)\left(b+2\right)}{\left(a-1\right)\left(a+1\right)}\\ B=\dfrac{2a-b}{a+1}-\dfrac{\left(a-1\right)\left(b+2\right)}{a+1}\\ B=\dfrac{2a-b-\left(a-1\right)\left(b+2\right)}{a+1}\\ B=\dfrac{2a-b-ab-2a+b+2}{a+1}=\dfrac{2-ab}{a+1}\)
a: \(A=\left(5xy-2xy+4xy\right)+3x-2y-y^2\)
\(=7xy+3x-2y-y^2\)
b: \(B=\left(\dfrac{1}{2}ab^2-\dfrac{7}{8}ab^2-\dfrac{1}{2}ab^2\right)+\left(\dfrac{3}{4}a^2b-\dfrac{3}{8}a^2b\right)\)
\(=\dfrac{-7}{8}ab^2+\dfrac{3}{8}a^2b\)
c: \(C=\left(2a^2b+5a^2b\right)+\left(-8b^2-3b^2\right)+\left(5c^2+4c^2\right)\)
\(=7a^2b-11b^2+9c^2\)
a/ (x-1)2-(4x+3)(2-x)=x2-2x+1-(8x-4x2+6-3x)
=x2-2x+1-8x+4x2-6+3x=5x2-7x-6
b/ (15x3y2 - 6x2y3) : 3x2y2 = 5x - 2y
c/ \(\dfrac{x+7}{x-7}-\dfrac{x-7}{x+7}+\dfrac{4x^2}{x^2-49}\)=\(\dfrac{\left(x+7\right)^2-\left(x-7\right)^2+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{x^2+14x+49-\left(x^2-14x+49\right)+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{28x+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x\left(x+7\right)}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x}{x-7}\)
2.
\(P=\left(\dfrac{a+6}{3\left(a+3\right)}-\dfrac{1}{a+3}\right).\dfrac{27a}{a+2}=\left(\dfrac{a+3}{3\left(a+3\right)}\right).\dfrac{27a}{a+2}=\dfrac{27a}{3\left(a+2\right)}=\dfrac{9a}{a+2}\)
ĐKXĐ là :
\(a\ne0;-3;-2\)
Vs a = 1 ta có:
=> P=3
1.
\(M=\left(\dfrac{2a}{2a+b}-\dfrac{4a^2}{\left(2a+b\right)^2}\right):\left(\dfrac{2a}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{1}{2a-b}\right)=\left(\dfrac{4a^2+2ab-4a^2}{\left(2a+b\right)^2}\right).\left(\dfrac{\left(2a+b\right)\left(2a-b\right)}{b}\right)=\dfrac{2a.\left(2a-b\right)}{\left(2a+b\right)}\)
ĐKXĐ: \(a\ne1\)
a. \(\dfrac{3a^2-a+3}{a^3-1}+\dfrac{1-a}{a^2+a+1}+\dfrac{2}{1-a}\)
\(=\dfrac{3a^2-a+3+\left(1-a\right).\left(a-1\right)-2.\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{3a^2-a+3-a^2+2a-1-2a^2-2a-2}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{-a+1}{\left(a-1\right).\left(a^2+a+1\right)}\)
\(=-\dfrac{1}{a^2+a+1}\)
a) Ta có: \(\dfrac{3a^2-a+3}{a^3-1}+\dfrac{1-a}{a^2+a+1}+\dfrac{2}{1-a}\)
\(=\dfrac{3a^2-a+3}{\left(a-1\right)\left(a^2+a+1\right)}-\dfrac{\left(a-1\right)^2}{\left(a-1\right)\left(a^2+a+1\right)}-\dfrac{2\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{3a^2-a+3-\left(a^2-2a+1\right)-2a^2-2a-2}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{a^2-3a+1-a^2+2a-1}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{-a}{\left(a-1\right)\left(a^2+a+1\right)}\)
b) Ta có: \(x-\dfrac{xy}{x+y}-\dfrac{x^3}{x^2y^2}\)
\(=x-\dfrac{xy}{x+y}-\dfrac{x}{y^2}\)
\(=\dfrac{xy^2\cdot\left(x+y\right)}{y^2\cdot\left(x+y\right)}+\dfrac{y^2\cdot xy}{y^2\cdot\left(x+y\right)}-\dfrac{x\cdot\left(x+y\right)}{y^2\cdot\left(x+y\right)}\)
\(=\dfrac{x^2y^2+xy^3+xy^3-x^2-xy}{y^2\cdot\left(x+y\right)}\)
\(=\dfrac{x^2y^2+2xy^3-x^2-xy}{y^2\cdot\left(x+y\right)}\)
e) = \(\dfrac{3}{2\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\)
= \(\dfrac{3x}{2x\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\) = \(\dfrac{3x-x+6}{2x\left(x+3\right)}\)
= \(\dfrac{2x-6}{2x\left(x+3\right)}\)
= \(\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\)
c) = \(\dfrac{2\left(a^3-b^3\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)
= \(\dfrac{-2\left(a+b\right)\left(a^2-2ab+b^2\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)
= \(\dfrac{-2\left(a+b\right)}{1}\) . \(\dfrac{2}{1}\) = -4 (a+b)