1) Thuc hien phep tinh cong 2 phan thuc \(\dfrac{2x}{x^2-2x 1} \dfrac{x 1}{x-1}\) duoc ket qua la:A. \(\dfrac{x^2 2x 1}{\left(x-1\right)^2}\) B. \(\dfrac{x^2 2x-1}{\left(x-1\right)^2}\) ...

NT

Nguyễn Thái Sơn

14 tháng 12 2021

1) Thuc hien phep tinh cong 2 phan thuc \(\dfrac{2x}{x^2-2x+1}+\dfrac{x+1}{x-1}\) duoc ket qua la:A. \(\dfrac{x^2+2x+1}{\left(x-1\right)^2}\) B. \(\dfrac{x^2+2x-1}{\left(x-1\right)^2}\) C. \(\dfrac{x^2-x-1}{\left(x-1\right)^2}\) ...

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#Toán lớp 11

1

CD

Cù Đức Anh

14 tháng 12 2021

undefined

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NP

Nga Phạm

22 tháng 12 2017

thuc hien phep tinh a.\(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\) b.\(\left(\dfrac{1}{x^2+1}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+1-2\right)\) c.\(\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}.\left(\dfrac{1}{x^2-2x+1}+\dfrac{1}{1-x^2}\right)\) d.\(\left(\dfrac{x^2+xy}{x^3+x^2y+xy^2+y^3}+\dfrac{y}{x^2+y^2}\right):\left(\dfrac{1}{x-y}-\dfrac{2xy}{x^3-x^2y+xy^2-y^3}\right)\) ...

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#Toán lớp 8

1

NL

Nguyễn Lê Phước Thịnh

13 tháng 6 2022

a: \(=\dfrac{4x^2+4x+1-\left(4x^2-4x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)

\(=\dfrac{8x}{2x+1}\cdot\dfrac{5}{4x}=\dfrac{10}{2x+1}\)

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

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

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TH

Tran Ha

14 tháng 12 2021

bai 1 thuc hien phep tinh a)\(\dfrac{\left(\dfrac{1}{x^2+4x+4}-\dfrac{1}{x^2-4x+4}\right)}{\left(\dfrac{1}{x+2}+\dfrac{1}{x-2}\right)}\)

b)\(\left(\dfrac{5x+y}{x^2-5xy}+\dfrac{5x-y}{x^2+5xy}\right)\cdot\dfrac{x^2-25y^2}{x^2+y^2}\)

#Toán lớp 8

1

NH

Nguyễn Hoàng Minh

14 tháng 12 2021

\(a,=\dfrac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}:\dfrac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{-8x}{\left(x-2\right)^2\left(x+2\right)^2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{2x}=\dfrac{-4}{\left(x-2\right)\left(x+2\right)}\)

\(b,=\dfrac{5x^2+26xy+5y^2+5x^2-26xy+5y^2}{x\left(x-5y\right)\left(x+5y\right)}\cdot\dfrac{\left(x-5y\right)\left(x+5y\right)}{x^2+y^2}\\ =\dfrac{10\left(x^2+y^2\right)}{x\left(x^2+y^2\right)}=\dfrac{10}{x}\)

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NP

Nga Phạm

4 tháng 12 2017

thuc hien phep tinh

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

#Toán lớp 8

2

PT

Phương Trâm

4 tháng 12 2017

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

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

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

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

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

\(=\dfrac{3x^2+4x+1-\left(x^2-2x+1\right)-\left(x^2+2x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\)

\(=\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\)

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

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

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

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

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LT

lê thị hương giang

4 tháng 12 2017

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

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

\(=\dfrac{3x^2+4x+1-x^2+2x-1-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\)

\(=\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\)

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

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

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NT

Nguyễn Thế Phúc Anh

25 tháng 2 2018

Cho bieu thuc: \(Q=\left(\dfrac{x^2-2x}{2x^2+8}+\dfrac{2x^2}{x^2.\left(x-2\right)}\right).\left(\dfrac{x^2-x-2}{x^2}\right)\)

a, Rut gon bieu thuc Q

b, Tim gia tri ca x de Q co gia tri bang \(\dfrac{1}{4}\)

#Toán lớp 8

0

NT

nguyễn thái hồng duyên

29 tháng 6 2018

chứng minh rằng : a)...

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#Toán lớp 8

1

NL

Nguyễn Lê Phước Thịnh

13 tháng 7 2022

b: \(=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right]\cdot\dfrac{x}{x+1}\)

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

\(=\dfrac{2x+2+6x^2+4x-2}{3x\left(x+1\right)}\cdot\dfrac{x}{x+1}\)

\(=\dfrac{6x^2+6x}{3\left(x+1\right)}\cdot\dfrac{1}{x+1}\)

\(=\dfrac{6x\left(x+1\right)}{3\left(x+1\right)^2}=\dfrac{2x}{x+1}\)

c: \(VT=\left[\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}+\dfrac{1}{\left(x+1\right)^2}\cdot\dfrac{1+x^2}{x^2}\right]\cdot\dfrac{x^3}{x-1}\)

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

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

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

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SG

Sách Giáo Khoa

28 tháng 4 2017

Chứng minh đẳng thức...

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#Toán lớp 8

1

NT

Nguyen Thuy Hoa

24 tháng 6 2017

Phân thức đại số

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MD

Mai Diễm My

1 tháng 6 2018

rảnh vãi

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NL

Ngân Lê Bảo

20 tháng 1 2021

giải phương...

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#Toán lớp 8

2

ND

Nguyễn Duy Khang

20 tháng 1 2021

undefined

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ND

Nguyễn Duy Khang

20 tháng 1 2021

\(a,\dfrac{3}{2x-1}+1=\dfrac{2x-1}{2x+1};ĐKXĐ:x\ne\pm\dfrac{1}{2}\\ \Leftrightarrow\dfrac{3}{2x-1}-\dfrac{2x-1}{2x+1}+1=0\\ \Leftrightarrow\dfrac{3\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\dfrac{\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}=0\\ \Rightarrow3\left(2x+1\right)-\left(2x-1\right)^2+\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow6x+3-\left(4x^2-4x+1\right)+\left(4x^2-1\right)=0\\ \Leftrightarrow6x+3-4x^2+4x-1+4x^2-1=0\\ \Leftrightarrow10x+1=0\\ \Leftrightarrow10x=-1\\ \Leftrightarrow x=-\dfrac{1}{10}\)

Vậy \(x\in\left\{-\dfrac{1}{10}\right\}\)

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DV

Dung Vu

18 tháng 11 2021

Rút gọn các biểu thức sau : A = \(2x^2\left(-3x^3+2x^2+x-1\right)+2x\left(x^2-3x+1\right)\)B = \(2x:\dfrac{1}{2}x+x^2\)C = \(\left[1:\left(1+x\right)+2x:\left(1-x^2\right)\right]:\left(\dfrac{1}{x}-1\right)\)D = \(\dfrac{x^2-y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}+\dfrac{y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}\)E = \(\dfrac{\left|x-3\right|}{x^2-9}.\left(x^2+6x+9\right)\)F...

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#Toán lớp 8

0