1. xác định a,b,c,d a) \(\dfrac{10x-4}{x^3-4x}\) = \(\dfrac{a}{x} \dfrac{b}{x-2} \dfrac{c}{x 2}\) b) \(\dfrac{1}{x^3-1}\) = \(\dfrac{a}{x-1} \dfrac{bx c}{x^2 x 1}\) c) \(\dfrac{x^3 2x}{x^4-1}\) = \...

\(B=\dfrac{4\left(x+3\right)^2}{\left(3x+5\right)^2-4x^2}-\dfrac{\left(x^2-25\right)}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{\left(4x+15\right)^2-x^2}\)

\(=\dfrac{4\left(x+3\right)^2}{\left(3x+5-2x\right)\left(3x+5+2x\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{\left(4x+15-x\right)\left(4x+15+x\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{5\left(x-5\right)\left(x+1\right)}-\dfrac{3\left(x+3\right)\left(x+1\right)}{15\left(x+5\right)\left(x+3\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{x+5}{5\left(x+1\right)}-\dfrac{x+1}{5\left(x+5\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+5\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+1\right)^2}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{4\left(x^2+6x+9\right)-\left(x^2+10x+25\right)-\left(x^2+2x+1\right)}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{4x^2+24x+36-x^2-10x-25-x^2-2x-1}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{2x^2+12x+10}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{2\left(x^2+6x+5\right)}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{2\left(x^2+5x+x+5\right)}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{2\left(x+5\right)\left(x+1\right)}{5\left(x+5\right)\left(x+1\right)}=\dfrac{2}{5}\)

Đúng(0)

LT

lê thị hương giang

10 tháng 12 2017

đc bn , nhg mà đề bài câu a b2 sao tự nhiên lại có " n "

bn xem lại đề đi

Đúng(0)

NH

NGo HOANG 2

11 tháng 3 2023

a.\(\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\) b.\(\dfrac{7}{x+2}=\dfrac{3}{x-5}\) c.\(\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)d.\(\dfrac{12x+1}{11x-4}+\dfrac{10x-4}{9}=\dfrac{20x+17}{18}\) e.\(\dfrac{11}{x}=\dfrac{9}{x+1}+\dfrac{2}{x-4}\) ...

Đọc tiếp

#Toán lớp 8

2

NH

Ngô Hải Nam

11 tháng 3 2023

bạn tách một câu vài câu hỏi chứ đừng gộp như thế này ko ai trả lời đâu

Đúng(0)

NL

Nguyễn Lê Phước Thịnh

11 tháng 3 2023

a: =>\(4x-5=2x-2+x=3x-2\)

=>x=3

b: \(\Leftrightarrow7x-35=3x+6\)

=>4x=41

=>x=41/4

c: =>(2x+5)(x+5)-2x^2=0

=>2x^2+10x+5x+25-2x^2=0

=>15x=-25

=>x=-5/3

e: \(\Leftrightarrow\dfrac{11}{x}=\dfrac{9x-36+2x+2}{\left(x+1\right)\left(x-4\right)}\)

=>11(x^2-3x-4)=x(11x-34)

=>11x^2-33x-44=11x^2-34x

=>x=44

Đúng(0)

LT

Lê Thu Hiền

21 tháng 2 2021

tìm điều kiện xác định của các phương trình sau\(a,3x^2-2x=0\) \(b,\dfrac{1}{x-1}=3\)\(c,\dfrac{2}{x-1}=\dfrac{x}{2x-4}\) \(d,\dfrac{2x}{x^2-9}=\dfrac{1}{x+3}\)\(e,2x=\dfrac{1}{x^2-2x+1}\) \(f,\dfrac{1}{x-2}=\dfrac{2x}{x^2-5x+6}\) giúp mik với , mik cần...

Đọc tiếp

#Toán lớp 8

2

NL

Nguyễn Lê Phước Thịnh

21 tháng 2 2021

a)\(x\in R\)

b)\(x\ne1\)

c) \(x\notin\left\{1;2\right\}\)

d) \(x\notin\left\{3;-3\right\}\)

e) \(x\ne1\)

f) \(x\notin\left\{2;3\right\}\)

Đúng(0)

LT

Lê Thu Hiền

21 tháng 2 2021

bạn trình bày rõ ràng hơn được hông??

Đúng(0)

ND

nguyễn đăng

19 tháng 12 2021

a,\(\dfrac{x+1}{x-3}+\dfrac{-2x^2+2x}{x^2-9}+\dfrac{x-1}{x+3}\)

b,\(\dfrac{1-2x}{6x^3y}+\dfrac{3+2y}{6x^3y}+\dfrac{2x-4}{6x^3y}\)

c,\(\dfrac{5}{2x^2y}+\dfrac{3}{5xy^2}+\dfrac{x}{3y^3}\)

d,\(\dfrac{5}{4\left(x+2\right)}+\dfrac{8-x}{4x^2+8x}\)

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

#Toán lớp 8

1

NH

Nguyễn Hoàng Minh

19 tháng 12 2021

\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)

Đúng(2)

T

ThanhNghiem

22 tháng 9 2023

Thực hiện phép tính: (câu nào khó quá bỏ qua)a) \(\dfrac{x^2+2}{x^3+1}\)-\(\dfrac{1}{x+1}\)b)\(\dfrac{x}{x^2-2x}\)-\(\dfrac{x^2+4x}{x^3-4x}\)-\(\dfrac{2}{x^2+2x}\)c)\(\dfrac{1}{2-2x}\)-\(\dfrac{3}{2+2x}\)+\(\dfrac{2x}{x^2-1}\)d) \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}\)+\(\dfrac{1}{\left(b-c\right)\left(c-a\right)}\)+\(\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)À mà nay sinh nhật tui...

Đọc tiếp

#Toán lớp 8

1

NL

Nguyễn Lê Phước Thịnh

10 tháng 11 2023

a:

ĐKXĐ: x<>-1

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

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

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

b: \(\dfrac{x}{x^2-2x}-\dfrac{x^2+4x}{x^3-4x}-\dfrac{2}{x^2+2x}\)

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

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

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

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

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

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

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

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

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

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

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

d:

\(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)

\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)

Đúng(0)

S

Sarah

25 tháng 2 2021

Tìm điều kiện của x để phân thức sau xác định;

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

b)\(\dfrac{x^3+2x}{4x^2-25}\)

c)\(\dfrac{2x^2+2x}{8x^3+27}\)

d)\(\dfrac{2x+1}{\left(2x+2\right)\left(4y^2-9\right)}\)

#Toán lớp 8

2

Y

Yeutoanhoc

25 tháng 2 2021

`a,ĐKXĐ:x-4 ne 0,2x+2 ne 0`

`<=>x ne 4,x me -1`

`b,ĐKXĐ:4x^2-25 ne 0`

`<=>(2x-5)(2x+5) ne 0`

`<=>x ne +-5/2`

`c,ĐKXĐ:8x^3+27 ne 0`

`<=>8x^3 ne -27`

`<=>2x ne -3`

`<=>x ne -3/2`

`d,2x+2 ne 0,4y^2-9 ne 0`

`<=>2x ne -2,(2y-3)(2y+3) ne 0`

`<=>x ne -1,y ne +-3/2`

Đúng(2)

NL

Nguyễn Lê Phước Thịnh

25 tháng 2 2021

b) ĐKXĐ: \(x\notin\left\{\dfrac{5}{2};-\dfrac{5}{2}\right\}\)

c) ĐKXĐ: \(x\ne-\dfrac{3}{2}\)

d) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\notin\left\{\dfrac{3}{2};-\dfrac{3}{2}\right\}\end{matrix}\right.\)

Đúng(0)

BT

Bach Tang Oni

2 tháng 12 2021

a) \(\dfrac{4x}{x^2+2x}\)+\(\dfrac{8}{x^2+2x}\) b) \(\dfrac{2x-3x}{x-2}\)-\(\dfrac{2x-4}{x-2}\) ...

Đọc tiếp

#Toán lớp 8

1

I

ILoveMath

2 tháng 12 2021

\(a,=\dfrac{4x+8}{x^2+2x}=\dfrac{4\left(x+2\right)}{x\left(x+2\right)}=\dfrac{4}{x}\\ b,=\dfrac{\left(2x-3\right)-\left(2x-4\right)}{x-2}=\dfrac{2x-3-2x+4}{x-2}=\dfrac{1}{x-2}\\ c,=\dfrac{2x-1-3x-2}{x+3}=\dfrac{-x-3}{x+3}=\dfrac{-\left(x+3\right)}{x+3}=-1\\ d,=\dfrac{11x-18+x}{2x-3}=\dfrac{12x-18}{2x-3}=\dfrac{6\left(2x-3\right)}{2x-3}=6\)

\(e,=\dfrac{3x-6-9x+3}{2x+1}=\dfrac{-6x-3}{2x+1}=\dfrac{-3\left(2x+1\right)}{2x+1}=-3\)

Đúng(3)

CG

co gai buong binh

19 tháng 11 2018

a) \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\) b) \(\dfrac{a+2b}{3a-b}+\dfrac{2a-5b}{b-3a}\) c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\) d) \(\dfrac{4x}{x^2-4}+\dfrac{x}{x+2}+\dfrac{2}{x-2}\) e) \(\dfrac{3x^2-x+3}{x^3-1}+\dfrac{1-x}{x^2+x+1}+\dfrac{2}{1-x}\) f) \(\dfrac{1}{x^2+3x+2}+\dfrac{1-x}{x^2+x+1}+\dfrac{2}{1-x}\) g) \(\dfrac{a^3}{\left(a-b\right)\left(a-c\right)}+\dfrac{b^3}{\left(b-a\right)\left(b-c\right)}+\dfrac{c^3}{\left(c-a\right)\left(c-b\right)}\) h)...

Đọc tiếp