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5 tháng 1 2022

\(\dfrac{1}{x^2+x}=\dfrac{x-1}{x\left(x-1\right)\left(x+1\right)};\dfrac{x^2-4}{x^2-1}=\dfrac{x\left(x^2-4\right)}{x\left(x-1\right)\left(x+1\right)}\\ \dfrac{1}{y-1}-\dfrac{1}{y}=\dfrac{y-y+1}{y\left(y-1\right)}=\dfrac{1}{y\left(y-1\right)}\)

NV
23 tháng 12 2022

a.

\(\dfrac{x^3}{x-1}-\dfrac{x^2}{x+1}-\dfrac{1}{x-1}+\dfrac{1}{x+1}=\dfrac{x^3-1}{x-1}-\dfrac{x^2-1}{x+1}\)

\(=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x-1}-\dfrac{\left(x-1\right)\left(x+1\right)}{x+1}\)

\(=x^2+x+1-\left(x-1\right)=x^2+2\)

b.

\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)

\(=\dfrac{\left(x+y\right)^2}{2\left(x-y\right)\left(x+y\right)}-\dfrac{\left(x-y\right)^2}{2\left(x-y\right)\left(x+y\right)}+\dfrac{4y^2}{2\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{\left(x+y\right)^2-\left(x-y\right)^2+4y^2}{2\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{4xy+4y^2}{2\left(x-y\right)\left(x+y\right)}=\dfrac{4y\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{2y}{x-y}\)

c.

\(\dfrac{x+5}{2x-4}.\dfrac{4-2x}{x+2}=\dfrac{x+5}{2x-4}.\dfrac{-\left(2x-4\right)}{x+2}=-\dfrac{x+5}{x+2}\)

d.

\(\dfrac{8}{x^2+2x-3}+\dfrac{2}{x+3}+\dfrac{1}{x-1}=\dfrac{8}{\left(x-1\right)\left(x+3\right)}+\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}+\dfrac{x+3}{\left(x-1\right)\left(x+3\right)}\)

\(=\dfrac{8+2\left(x-1\right)+x+3}{\left(x-1\right)\left(x+3\right)}=\dfrac{3x+9}{\left(x-1\right)\left(x+3\right)}\)

\(=\dfrac{3\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}=\dfrac{3}{x-1}\)

a: \(=\dfrac{3b+4a}{6ab}\)

b: \(=\dfrac{x^2-2x+1-x^2-2x-1}{x^2-1}=\dfrac{-4x}{x^2-1}\)

c: \(=\dfrac{xz+yz-xy-xz}{xyz}=\dfrac{yz-xy}{xyz}=\dfrac{z-x}{xz}\)

d: \(=\dfrac{2x+6-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x+3}\)

e: \(=\dfrac{x-2+2}{\left(x-2\right)^2}=\dfrac{x}{\left(x-2\right)^2}\)

15 tháng 12 2023

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

9 tháng 11 2021

\(\dfrac{x+y}{2-x}=\dfrac{-\left(x+y\right)}{x-2}\)

\(\dfrac{-y}{y-4}=\dfrac{--y}{4-y}=\dfrac{y}{4-y}\)

9 tháng 11 2021

mik cam on bn

23 tháng 7 2023

\(a,=\dfrac{4y.5x^3}{3x^2.2y^3}=\dfrac{20x^3y}{6x^2y^3}=\dfrac{10x}{3y^2}\\ b,=\dfrac{\left(x-1\right)^2.x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2.x.\left(x+1\right)}{\left(x-1\right)^2.\left(x+1\right)}=x\)

\(c,=\dfrac{x\left(2+x\right).3\left(x^3+1\right)}{\left(x^2-x+1\right).3.\left(x+2\right)}=\dfrac{3x.\left(x+2\right).\left(x+1\right)\left(x^2-x+1\right)}{\left(x^2-x+1\right).3\left(x+2\right)}=x\left(x+1\right)\)

3 tháng 12 2021

\(1,\\ a,=xy^2-\dfrac{3}{2}y^3+\dfrac{5}{4}x^2\\ b,=\left(x-7\right)\left(x+7\right):\left(x-7\right)=x+7\\ 2,\dfrac{1}{a^2}-ab=\dfrac{1-a^3b}{a^2};\dfrac{1}{a^2}\text{ giữ nguyên}\\ 3,=\dfrac{-7}{t}\\ 4,=\dfrac{1-x+1-y}{x-y}=\dfrac{2-x-y}{x-y}\)

3 tháng 12 2021

Bài 1:

\(a,\left(16x^3y^2-24x^2y^3+20x^4\right):16x^2=16x^2\left(xy^2-\dfrac{3}{2}y^3+\dfrac{5}{4}x^2\right):16x^2=xy^2-\dfrac{3}{2}y^3+\dfrac{5}{4}x^2\)

\(b,\left(x^2-49\right):\left(x-7\right)=\left[\left(x-7\right)\left(x+7\right)\right]:\left(x-7\right)=x+7\)

Bài 2:

\(\dfrac{1}{a^2}-ab=\dfrac{1-a^2b}{a^2}\)

\(\dfrac{1}{a^2}\)

Bài 3:

\(\dfrac{7\left(t-z\right)}{t\left(z-t\right)}=\dfrac{-7\left(z-t\right)}{t\left(z-t\right)}=\dfrac{-7}{t}\)

Bài 4:

\(\dfrac{x-1}{y-x}+\dfrac{1-y}{x-y}=\dfrac{x-1}{y-x}-\dfrac{1-y}{y-x}=\dfrac{x-1-1+y}{y-x}=\dfrac{x+y-2}{y-x}\)

22 tháng 7 2023

\(a,\dfrac{x+2}{x-1}-\dfrac{x-3}{x-1}-\dfrac{x-4}{1-x}\\ =\dfrac{x+2}{x-1}-\dfrac{x-3}{x-1}+\dfrac{x-4}{x-1}\\ =\dfrac{x+2-x+3+x-4}{x-1}\\ =\dfrac{x+1}{x-1}\)

\(b,\dfrac{1}{x+5}-\dfrac{1}{x-5}+\dfrac{2x}{x^2-25}\\ =\dfrac{1}{x+5}-\dfrac{1}{x-5}+\dfrac{2x}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{x-5-x-5+2x}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{2x-10}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{2\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{2}{x+5}\)

\(c,x+\dfrac{2y^2}{x+y}-y\\ =\dfrac{x\left(x+y\right)+2y^2-y\left(x+y\right)}{x+y}\\ =\dfrac{x^2+xy+2y^2-xy-y^2}{x+y}\\ =\dfrac{x^2+y^2}{x+y}\)

`a, a/(a-3) - 3/(a+3) = (a(a+3) - 3(a-3))/(a^2-9)`

`= (a^2+9)/(a^2-9)`

`b, 1/(2x) + 2/x^2 = x/(2x^2) + 4/(2x^2) = (x+4)/(2x^2)`

`c, 4/(x^2-1) - 2/(x^2+x) = (4x)/(x(x-1)(x+1)) - (2(x-1))/(x(x+1)(x-1))`

`= (2x+2)/(x(x-1)(x+1)`

`= 2/(x(x-1))`

18 tháng 2 2022

\(\left(\dfrac{x}{x+1}+\dfrac{x-1}{x}\right):\left(\dfrac{x}{x+1}-\dfrac{x-1}{x}\right)\) \(\left(đk:x\ne0;-1\right)\)

\(=\dfrac{x^2+\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}:\left(\dfrac{x^2-\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}\right)\)

\(=\dfrac{x^2+x^2-1}{x\left(x+1\right)}.\dfrac{x\left(x+1\right)}{x^2-x^2+1}\)

\(=\dfrac{\left(2x^2-1\right)x\left(x+1\right)}{x\left(x+1\right)}=2x^2-1\)