Thực hiện các phép tính sau:
a) \(\dfrac{{x + 2}}{{x - 1}} - \dfrac{{x - 3}}{{x - 1}} - \dfrac{{x - 4}}{{1 - x}}\)
b) \(\dfrac{1}{{x + 5}} - \dfrac{1}{{x - 5}} + \dfrac{{2x}}{{{x^2} - 25}}\)
c) \(x + \dfrac{{2{y^2}}}{{x + y}} - y\)
Thực hiện các phép tính sau:
a) \(\dfrac{{x + 2}}{{x - 1}} - \dfrac{{x - 3}}{{x - 1}} - \dfrac{{x - 4}}{{1 - x}}\)
b) \(\dfrac{1}{{x + 5}} - \dfrac{1}{{x - 5}} + \dfrac{{2x}}{{{x^2} - 25}}\)
c) \(x + \dfrac{{2{y^2}}}{{x + y}} - y\)
\(a,\dfrac{x+2}{x-1}-\dfrac{x-3}{x-1}-\dfrac{x-4}{1-x}\\ =\dfrac{x+2}{x-1}-\dfrac{x-3}{x-1}+\dfrac{x-4}{x-1}\\ =\dfrac{x+2-x+3+x-4}{x-1}\\ =\dfrac{x+1}{x-1}\)
\(b,\dfrac{1}{x+5}-\dfrac{1}{x-5}+\dfrac{2x}{x^2-25}\\ =\dfrac{1}{x+5}-\dfrac{1}{x-5}+\dfrac{2x}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{x-5-x-5+2x}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{2x-10}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{2\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{2}{x+5}\)
\(c,x+\dfrac{2y^2}{x+y}-y\\ =\dfrac{x\left(x+y\right)+2y^2-y\left(x+y\right)}{x+y}\\ =\dfrac{x^2+xy+2y^2-xy-y^2}{x+y}\\ =\dfrac{x^2+y^2}{x+y}\)
Thực hiện các phép tính sau:
A = (x⁴-x²+ 2x -1) : (x²+ x -1) - (x²-x)
B = \(\dfrac{x+1}{x-2}+\dfrac{4}{x+2}+\dfrac{2-7x}{x^2-4}\)
1) \(A=\left[x^4-\left(x-1\right)^2\right]:\left(x^2+x-1\right)-x^2+x=\left[\left(x^2-x+1\right)\left(x^2+x-1\right)\right]:\left(x^2+x-1\right)-x^2+x=x^2-x+1-x^2+x=1\)
2) \(B=\dfrac{\left(x+1\right)\left(x+2\right)+4\left(x-2\right)+2-7x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-4}{x^2-4}=1\)
a: \(A=\dfrac{x^4+x^3-x^2-x^3-x^2+x+x^2+x-1}{x^2+x-1}-x^2+x\)
\(=x^2-x+1-x^2+x=1\)
b: \(B=\dfrac{x^2+3x+2+4x-8+2-7x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-8}{\left(x-2\right)\left(x+2\right)}\)
thực hiện phép tính
\(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)
\(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)
\(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)
a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)
\(=\dfrac{3x+2-3x+2-3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)
\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-5}{2x\left(x-5\right)}\)
c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)
\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)
Thực hiện phép tính:
a) \(\dfrac{x}{2x-y}-\dfrac{2x-y}{4x-2y}\)
b)\(\dfrac{3x+1}{x^2-1}-\dfrac{x}{2x-2}\)
c) \(\dfrac{x-2}{x^2-4}-\dfrac{-8-x}{3x^2+6x}\)
d) \(\dfrac{2}{2x-3}-\dfrac{x}{2x+3}-\dfrac{2x+1}{9-4x^2}\)
a: \(=\dfrac{2x-2x+y}{2\left(2x-y\right)}=\dfrac{y}{2\left(2x-y\right)}\)
b: \(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{6x+2-x^2-x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2+5x+2}{2\left(x-1\right)\left(x+1\right)}\)
c: \(=\dfrac{1}{x+2}+\dfrac{x+8}{3x\left(x+2\right)}\)
\(=\dfrac{3x+x+8}{3x\left(x+2\right)}=\dfrac{4x+8}{3x\left(x+2\right)}=\dfrac{4}{3x}\)
d: \(=\dfrac{4x+6-2x^2+3x+2x+1}{\left(2x-3\right)\left(2x+3\right)}\)
\(=\dfrac{-2x^2+9x+7}{\left(2x-3\right)\left(2x+3\right)}\)
Thực hiện phép tính
a) \(^{\dfrac{x^2+2}{x^3-1}}\) +\(\dfrac{2}{x^2+x+1}\) +\(\dfrac{1}{1-x}\)
b) \(\dfrac{1}{x+2}\) +\(\dfrac{3}{x^2-4}\) +\(\dfrac{x-14}{\left(x^2+4x+4\right)\left(x-2\right)}\)
c)\(\dfrac{1}{x-y}\) -\(\dfrac{3xy}{x^3-y^3}\) +\(\dfrac{x-y}{x^2+xy+y^2}\)
d) \(\dfrac{1}{a-b}\) +\(\dfrac{1}{a+b}\) +\(\dfrac{2a}{a^2+b^2}\) +\(\dfrac{4a^3}{a^4+b^4}\)
e) \(\dfrac{1}{a^2-a}\) + \(\dfrac{1}{a^2-3a+2}\) +\(\dfrac{1}{a^2-5a+6}\) +\(\dfrac{1}{a^2-7a+12}\)
a) \(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}=\dfrac{x^2+2+2\left(x-1\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
b) \(=\dfrac{1}{x+2}+\dfrac{3}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x+2\right)\left(x-2\right)+3\left(x+2\right)+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\)
c) \(=\dfrac{x^2+xy+y^2-3xy+\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{x^2-2xy+y^2+x^2-2xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x^2-2xy+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)
Thực hiện phép tính :
a. \(\dfrac{4}{\sqrt{x}+1}+\dfrac{2}{1-\sqrt{x}}-\dfrac{\sqrt{x}-5}{x-1}\)
b. \(\left(\dfrac{x+1}{x+2}+\dfrac{x+2}{x+3}\right):\dfrac{x+3}{x+1}\)
\(a,=\dfrac{4\sqrt{x}-4-2\sqrt{x}-2-\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\left(x\ge0;x\ne1\right)\\ =\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}+1}\\ b,=\dfrac{x^2+4x+3+x^2+4x+4}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{x+1}{x+3}\left(x\ne-1;x\ne-2;x\ne-3\right)\\ =\dfrac{\left(2x^2+8x+7\right)\left(x+1\right)}{\left(x+2\right)\left(x+3\right)^2}\)
Thực hiện phép tính :
a. \(\dfrac{4}{\sqrt{x}+1}+\dfrac{2}{1-\sqrt{x}}-\dfrac{\sqrt{x}-5}{x-1}\)
b. \(\left(\dfrac{x+1}{x+2}+\dfrac{x+2}{x+3}\right):\dfrac{x+3}{x+1}\)
\(a,\dfrac{4}{\sqrt{x}+1}+\dfrac{2}{1-\sqrt{x}}-\dfrac{\sqrt{x}-5}{x-1}\)
\(=\dfrac{4\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}-4-2\sqrt{x}-2-\sqrt{x}+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}+1}\)
\(b,\left(\dfrac{x+1}{x+2}+\dfrac{x+2}{x+3}\right):\dfrac{x+3}{x+1}\)
\(=\left(\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)}+\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x+3\right)}\right).\dfrac{x+1}{x+3}\)
\(=\left(\dfrac{x^2+4x+3}{\left(x+2\right)\left(x+3\right)}+\dfrac{x^2+4x+4}{\left(x+2\right)\left(x+3\right)}\right).\dfrac{x+1}{x+3}\)
\(=\dfrac{x^2+4x+3+x^2+4x+4}{\left(x+2\right)\left(x+3\right)}.\dfrac{x+1}{x+3}\)
\(=\dfrac{2x^2+8x+7}{\left(x+2\right)\left(x+3\right)}.\dfrac{x+1}{x+3}\)
\(=\dfrac{\left(2x^2+8x+7\right)\left(x+1\right)}{\left(x+2\right)\left(x+3\right)^2}\)
\(=\dfrac{\left(2x^2+8x+7\right).x+2x^2+8x+7}{\left(x+2\right)\left(x+3\right)^2}\)
\(=\dfrac{2x^3+8x^2+7x+2x^2+8x+7}{\left(x+2\right)\left(x+3\right)^2}\)
\(=\dfrac{2x^3+10x^2+15x+7}{\left(x+2\right)\left(x+3\right)^2}\)
Thực hiện các phép tính sau:
a) \(\dfrac{{8y}}{{3{x^2}}} \cdot \dfrac{{9{x^2}}}{{4{y^2}}}\)
b) \(\dfrac{{3x + {x^2}}}{{{x^2} + x + 1}} \cdot \dfrac{{3{x^3} - 3}}{{x + 3}}\)
c) \(\dfrac{{2{x^2} + 4}}{{x - 3}} \cdot \dfrac{{3x + 1}}{{x - 1}}:\dfrac{{{x^2} + 2}}{{6 - 2x}}\)
d) \(\dfrac{{2{x^2}}}{{3{y^3}}}:\left( { - \dfrac{{4{x^3}}}{{21{y^2}}}} \right)\)
e) \(\dfrac{{2x + 10}}{{{x^3} - 64}}:\dfrac{{{{\left( {x + 5} \right)}^2}}}{{2x - 8}}\)
f) \(\dfrac{1}{{x + y}}\left( {\dfrac{{x + y}}{{xy}} - x - y} \right) - \dfrac{1}{{{x^2}}}:\dfrac{y}{x}\)
\(a,\dfrac{8y}{3x^2}.\dfrac{9x^2}{4y^2}=\dfrac{72x^2y}{12x^2y^2}=\dfrac{6}{y}\\b,\dfrac{3x+x^2}{x^2+x+1}.\dfrac{3x^3-3}{x+3}=\dfrac{x\left(x+3\right)3\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x+3\right)}=3x\left(x-1\right)=3x^2-3x \)
\(c,\dfrac{2x^2+4}{x-3}.\dfrac{3x+1}{x-1}.\dfrac{6-2x}{x^2+2}=\dfrac{2\left(x^2+2\right)\left(3x+1\right)2\left(3-x\right)}{\left(x-3\right)\left(x-1\right)\left(x^2+2\right)}=\dfrac{-4\left(3x+1\right)}{x-1}=\dfrac{-12x-4}{x-1}\)
\(d,\dfrac{2x^2}{3y^3}:\left(-\dfrac{4x^3}{21y^2}\right)=\dfrac{-2x^2.21y^2}{3y^3.4x^3}=\dfrac{-42x^2y^2}{12x^3y^3}=\dfrac{-7}{2xy}\)
\(e,\dfrac{2x+10}{x^3-64}:\dfrac{\left(x+5\right)^2}{2x-8}=\dfrac{2\left(x+5\right)}{\left(x-4\right)\left(x^2+4x+16\right)}.\dfrac{2\left(x-4\right)}{\left(x+5\right)^2}=\dfrac{4}{\left(x+5\right)\left(x^2+4x+16\right)}=\dfrac{4}{x^3+9x^2+16x+80}\)
\(f,\dfrac{1}{x+y}\left(\dfrac{x+y}{xy}-x-y\right)-\dfrac{1}{x^2}:\dfrac{y}{x}=\dfrac{1}{x+y}\left(\dfrac{\left(x+y\right)\left(1-xy\right)}{xy}\right)-\dfrac{x}{x^2y}=\dfrac{1-xy}{xy}-\dfrac{x}{x^2y}=\dfrac{-x^2y}{x^2y}=-1\)
1.Dùng tính chất của phân thức giải thích tại sao các cặp phân thức bằng nhau
\(\dfrac{3-x}{2-x}=\dfrac{x-3}{2-x}\) ; \(\dfrac{x}{x-1}=\dfrac{x^2-x}{x^2-2x-1}\)
2.Thực hiện phép tính
a.\(\dfrac{1+x}{x+1}-\dfrac{x-1}{x^2+x}\) b.\(\dfrac{2x}{x+23}.\dfrac{3x}{x-1}+\dfrac{2x}{x+23}.\dfrac{23-2x}{x-1}\)
3.Cho hình bình hành ABCD có góc A= 60 độ và BC = 2AB.Gọi M và N lần lượt là trung điểm của BC và DA.Trên tia đối tia BA lấy E sao cho BE =BA.C/m
a.AMDN là hbh b.ABMN là hình thoi c.AM=NE
Bài 2:
a: ĐKXĐ: \(x\notin\left\{0;-1\right\}\)
\(\dfrac{1+x}{x+1}-\dfrac{x-1}{x^2+x}\)
\(=\dfrac{x\left(x+1\right)-x+1}{x\left(x+1\right)}\)
\(=\dfrac{x^2+x-x+1}{x^2+x}=\dfrac{x^2+1}{x^2+x}\)
b: ĐKXĐ: \(x\notin\left\{-23;1\right\}\)
\(\dfrac{2x}{x+23}\cdot\dfrac{3x}{x-1}+\dfrac{2x}{x+23}\cdot\dfrac{23-2x}{x-1}\)
\(=\dfrac{2x}{x+23}\cdot\left(\dfrac{3x}{x-1}+\dfrac{23-2x}{x-1}\right)\)
\(=\dfrac{2x}{x+23}\cdot\dfrac{3x+23-2x}{x-1}\)
\(=\dfrac{2x}{x+23}\cdot\dfrac{x+23}{x-1}=\dfrac{2x}{x-1}\)
Bài 3:
a: Sửa đề: AMCN
Ta có: ABCD là hình bình hành
=>BC=AD(1)
Ta có: M là trung điểm của BC
=>\(BM=MC=\dfrac{BC}{2}\left(2\right)\)
Ta có: N là trung điểm của AD
=>\(NA=ND=\dfrac{AD}{2}\left(3\right)\)
Từ (1),(2),(3) suy ra BM=MC=NA=ND
Xét tứ giác AMCN có
MC//AN
MC=AN
Do đó: AMCN là hình bình hành
b: Xét tứ giác ABMN có
BM//AN
BM=AN
Do đó: ABMN là hình bình hành
Hình bình hành ABMN có \(AB=BM\left(=\dfrac{BC}{2}\right)\)
nên ABMN là hình thoi
c: Ta có: BM//AD
=>\(\widehat{EBM}=\widehat{EAD}\)(hai góc đồng vị)
=>\(\widehat{EBM}=60^0\)
Xét ΔBEM có BE=BM(=BA) và \(\widehat{EBM}=60^0\)
nên ΔBEM đều
=>\(\widehat{BEM}=60^0\)
Xét hình thang ANME có \(\widehat{MEA}=\widehat{EAN}=60^0\)
nên ANME là hình thang cân
=>AM=NE
Thực hiện phép tính
\(\dfrac{3x^2+5x+14}{x^2+1}+\dfrac{x-1}{x^2-x+1}-\dfrac{4}{x+1}\)
\(=\dfrac{3x^2+5x+14+x^2-1-4x^2+4x-4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{9x+9}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{9}{x^2-x+1}\)