Thực hiện phép tinh:a)$\frac{4}{x 2}$ $\frac{2}{x-2}$ $\frac{5x-6}{4-x^2}$b)$\frac{1-3x}{2x}$ $\frac{3x-2}{2x-1}$ $\frac{3x-2}{2x-4x^2}$c)$\frac{1}{x^2 6x 9}$ $\frac{1}{6x-x^2-9}$ \(\...

1

DD

Đinh Đức Hùng

6 tháng 7 2017

a ) $\frac{4}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}=\frac{4\left(x-2\right)+2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{6-5x}{\left(x+2\right)\left(x-2\right)}=\frac{6x-4+6-5x}{\left(x+2\right)\left(x-2\right)}$

$=\frac{x+2}{\left(x+2\right)\left(x-2\right)}=\frac{1}{x+2}$

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

$=\frac{-6x^2+5x-1+6x^2-4x+2-3x}{2x\left(2x-1\right)}=\frac{-2x+1}{2x\left(2x-1\right)}=\frac{-1}{2x}$

c ) $\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}=\frac{1}{\left(x+3\right)^2}+\frac{1}{-\left(x-3\right)^2}+\frac{x}{\left(x+3\right)\left(x-3\right)}$

$=\frac{\left(x-3\right)^2-\left(x+3\right)^2+x\left(x+3\right)\left(x-3\right)}{\left(x+3\right)^2\left(x-3\right)^2}=\frac{-12x+x^3-9x}{\left(x+3\right)^2\left(x-3\right)^2}=\frac{x^3-21x}{x^4-18x^2+81}$

d ) $\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{1-x}=\frac{x^2+2+2\left(x-1\right)-\left(x^2+x+1\right)}{x^3-1}=\frac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}$

$=\frac{1}{x^2+x+1}$

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

$=\frac{2x}{x+2y}$

Bài 1: Thực hiện các phép tính...

HL

Hoàng Linh Nhi

11 tháng 8 2020

#Giáo dục công dân lớp 8

0

HL

Hoàng Linh Nhi

12 tháng 8 2020

Mình biết rồi, cảm ơn bạn

S

Sakura서 정

27 tháng 3 2020

thực hiện các phép tính sau
a) $\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}$
b) $\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}$

c) $\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}$

d) $\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}$

2

NL

Nguyễn Lê Phước Thịnh

27 tháng 3 2020

a) Ta có: $\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}$

$=\frac{x+1}{2\left(x+3\right)}+\frac{2x+3}{x\left(x+3\right)}$

$=\frac{x\left(x+1\right)}{2x\left(x+3\right)}+\frac{2\cdot\left(2x+3\right)}{2x\left(x+3\right)}$

$=\frac{x^2+x+4x+6}{2x\left(x+3\right)}$

$=\frac{x^2+5x+6}{2x\left(x+3\right)}$

$=\frac{x^2+2x+3x+6}{2x\left(x+3\right)}$

$=\frac{x\left(x+2\right)+3\left(x+2\right)}{2x\left(x+3\right)}$

$=\frac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}=\frac{x+2}{2x}$

b) Ta có: $\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}$

$=\frac{3}{2x+6}-\frac{x-6}{x\left(2x+6\right)}$

$=\frac{3x}{x\left(2x+6\right)}-\frac{x-6}{x\left(2x+6\right)}$

$=\frac{3x-x+6}{x\left(2x+6\right)}=\frac{2x+6}{x\left(2x+6\right)}=\frac{1}{x}$

c) Ta có: $\frac{5x+10}{4x-8}\cdot\frac{4-2x}{x+2}$

$=\frac{5\left(x+2\right)\cdot2\cdot\left(2-x\right)}{4\cdot\left(x-2\right)\cdot\left(x+2\right)}$

$=\frac{5\cdot2\cdot\left(2-x\right)}{-4\left(2-x\right)}=\frac{5\cdot2}{-4}=\frac{-5}{2}$

d) Ta có: $\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}$

$=\frac{\left(1-2x\right)\left(1+2x\right)\cdot3x}{x\left(x+4\right)\cdot2\left(2-x\right)}$

$=\frac{\left(1-2x\right)\left(1+2x\right)\cdot3}{2\left(x+4\right)\cdot\left(2-x\right)}=\frac{3\left(1-4x^2\right)}{2\left(-x^2-2x+8\right)}$

$=\frac{3-12x^2}{-2x^2-4x+16}$

JS

Jeong Soo In

27 tháng 3 2020

a) $\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}$

$=\frac{x+1}{2\left(x+3\right)}+\frac{2x+3}{x\left(x+3\right)}$ $\left(ĐKXĐ:x\ne-3;x\ne0\right)$

$=\frac{x^2+x}{2x\left(x+3\right)}+\frac{4x+6}{2x\left(x+3\right)}$

$=\frac{x^2+2x+3x+6}{2x\left(x+3\right)}=\frac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}=\frac{x+2}{2x}$

b) $\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}$ $\left(ĐKXĐ:x\ne0;x\ne-3\right)$

$=\frac{3x}{2x\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}=\frac{2\left(x+3\right)}{2x\left(x+3\right)}=\frac{1}{x}$

c) $\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\frac{2\left(2-x\right)}{x+2}$ $\left(ĐKXĐ:x\ne\pm2\right)$

$=\frac{-5\left(x-2\right)}{2\left(x-2\right)}=\frac{-5}{2}$

Bài 1: Thực hiện phép tính a. $\frac{11x+10}{3x-3}+\frac{15x+13}{4-4x}$ b. $\frac{5x+3}{x^2-3x}+\frac{9-x}{9-3x}$ c. $\frac{4xy-1}{5x^2y}-\frac{2xy-1}{5x^2y}$ d. $\frac{x+8}{x^2-16}-\frac{2}{x^2+4x}$ e. $\frac{x^2-49}{2x+1}.\frac{3}{7-x}$ f. $\frac{3x^2-2x}{x^2-1}.\frac{1-x^4}{\left(2-3x\right)^3}$ g. $\frac{5xy}{2x-3}:\frac{15xy^3}{12-8x}$ h....

PT

Phạm Tú Uyên

12 tháng 8 2020

3

AH

Akai Haruma

Giáo viên

12 tháng 8 2020

f)

$\frac{3x^2-2x}{x^2-1}.\frac{1-x^4}{(2-3x)^3}$

$=\frac{2x-3x^2}{x^2-1}.\frac{x^4-1}{(2-3x)^3}=\frac{x(2-3x)(x^2-1)(x^2+1)}{(x^2-1)(2-3x)^3}$

$=\frac{x(x^2+1)}{(2-3x)^2}$
g)

$\frac{5xy}{2x-3}:\frac{15xy^3}{12-8x}=\frac{5xy}{2x-3}.\frac{12-8x}{15xy^3}$

$=\frac{5xy}{2x-3}.\frac{-4(2x-3)}{15xy^3}=\frac{-4}{3y^2}$

h)

$\frac{x^2+2x}{3x^2-6x+3}:\frac{2x+4}{5x-5}=\frac{x(x+2)}{3(x-1)^2}:\frac{2(x+2)}{5(x-1)}$

$=\frac{x(x+2)}{3(x-1)^2}.\frac{5(x-1)}{2(x+2)}$

$=\frac{5x}{6(x-1)}$

AH

Akai Haruma

Giáo viên

12 tháng 8 2020

d)

$\frac{x+8}{x^2-16}-\frac{2}{x^2+4x}=\frac{x+8}{(x-4)(x+4)}-\frac{2}{x(x+4)}$

$=\frac{x(x+8)}{x(x-4)(x+4)}-\frac{2(x-4)}{x(x+4)(x-4)}$

$=\frac{x^2+8x-2(x-4)}{x(x+4)(x-4)}=\frac{x^2+6x+8}{x(x+4)(x-4)}$

$=\frac{(x+2)(x+4)}{x(x+4)(x-4)}=\frac{x+2}{x(x-4)}$
e)

$\frac{x^2-49}{2x+1}.\frac{3}{7-x}=\frac{(x-7)(x+7)}{2x+1}.\frac{-3}{x-7}$

$=\frac{-3(x+7)}{2x+1}$

Thực hiện phép tínha) $\frac{\text{x + 9}}{x^2 - 9}-\frac{\text{3}}{\text{x^2 + 3x}}$b) $\frac{\text{3x + 5 }}{\text{x^2 - 5x }}+\frac{\text{ 25 - x }}{\text{25 - 5x }}$c) $\frac{\text{3 }}{\text{2x }}+\frac{\text{3x - 3 }}{\text{2x - 1 }}+\frac{ 2x^2 + 1 }{\text{4x^2 - 2x }}$d) $\frac{\text{1}}{\text{3x - 2 }}-\frac{1}{\text{3x + 2 }}- \frac{\text{3x - 6}}{\text{4 - 9x^2}}$e) \(\frac{\text{18 }}{\text{(x - 3)(x^2 - 9) }}-\frac{\text{3 }}{\text{x^2 - 6x + 9...

NT

Nguyen Thuy Dung

4 tháng 12 2017

0

NT

Nguyễn Thị Anh Thư

18 tháng 11 2019

1. Thực hiện phép tính:

a) $\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4}$

b) $\frac{1}{x^2+6x+y}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}$

c) $\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{1-x}$

d) $\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}$

1

CE

Chitanda Eru (Khối kiến thức)

18 tháng 11 2019

a)có khả năng sai đề bài

b)Liệu có sai đề bài không

c)$=\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}$(phân số cuối có âm vì (1-x)=-(x-1)

$=\frac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}$(Hơi tắt)

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

d)$=\frac{x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{4xy}{\left(x-2y\right)\left(x+2y\right)}$

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

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

BT

bùi thị phương uyên

16 tháng 12 2019

1, Thực hiện tính cộng, trừ, nhân, chia các phân thức sau:

a,$\frac{2x-7}{10x-4}-\frac{3x+5}{4-10x}$

b,$\frac{2x+3}{4x^2y^2}:\frac{6x+9}{10x^2y}$

c,$\frac{x^2-y^2}{6x^2y^2}:\frac{x+y}{3xy}$

d,$\left(\frac{3x}{1-3x}+\frac{2x}{3x+1}\right):\frac{6x^2+10x}{1-6x+9x^2}$

1

𝑮𝒊𝒂 𝑯𝒖𝒚

17 tháng 12 2019

a) $\frac{2x-7}{10x-4}-\frac{3x+5}{4-10x}$

$=\frac{2x-7}{10x-4}-\frac{-\left(3x+5\right)}{-\left(4-10x\right)}$

$=\frac{2x-7}{10x-4}-\frac{5-3x}{10x-4}$

$=\frac{2x-7-\left(5-3x\right)}{10x-4}$

$=\frac{2x-7-5+3x}{10x-4}$

$=\frac{5x-12}{10x-4}$

K

Kaijo

15 tháng 3 2020

5,Thực hiện phép tính

1,$\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}$

2,$\frac{1}{1-x}+\frac{2x}{x^2-1}$

3,$\frac{1}{xy-x^2}-\frac{1}{y^2-xy}$

4,$\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}$

5,$\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}$

3

U

Uchiha

15 tháng 3 2020

1,$\frac{3}{2x+6}-\frac{x-6}{x\left(2x+6\right)}$

=$\frac{3x}{x\left(2x+6\right)}+\frac{x-6}{x\left(2x+6\right)}$

=$\frac{3x+x-6}{x\left(2x+6\right)}$=$\frac{4x-6}{x\left(2x+6\right)}=\frac{2\left(2x-3\right)}{x\left(2x+6\right)}$

U

Uchiha

15 tháng 3 2020

2, $\frac{1}{1-x}-\frac{2x}{1-x^2}$=$\frac{1+x}{\left(1-x\right)\left(1+x\right)}+\frac{2x}{\left(1-x\right)\left(1+x\right)}$=$\frac{1+x+2x}{\left(1-x\right)\left(1+x\right)}=\frac{3x+1}{\left(1-x\right)\left(1+x\right)}$

K

Kaijo

16 tháng 3 2020

7,Thực hiện phép...

3

AT

Ánh Tuyết

16 tháng 3 2020

a, $\frac{4x+1}{2}-\frac{3x+2}{3}=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{12x+3-6x-4}{6}=\frac{6x-1}{6}$

b, $\frac{x+3}{x^2-1}-\frac{1}{x^2+x}=\frac{x+3}{\left(x-1\right)\left(x+2\right)}-\frac{1}{x\left(x+1\right)}$

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

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

$=\frac{x+1}{x\left(x-1\right)}$

2

๖²⁴ʱTú❄⁀ᶦᵈᵒᶫ

16 tháng 3 2020

$\frac{4x+1}{2}-\frac{3x+2}{3}$

$=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{6x-1}{6}$

tương tự đến hết nha a hay cj gì đps !

K

Kaijo

16 tháng 3 2020

1,Thực hiện phép tính

a,$\frac{3}{2x^2+2x}+\frac{2x-1}{x^2-1}-\frac{2}{x}$

b,$\frac{3x}{5x+5y}-\frac{x}{10x-10y}$

c,$\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}$

d,$\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}$

3

2

๖²⁴ʱTú❄⁀ᶦᵈᵒᶫ

16 tháng 3 2020

$\frac{3x}{5x+5y}-\frac{x}{10x-10y}$

$=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x+y\right)}$

$=\frac{30x\left(x-y\right)-5x\left(x+y\right)}{5\left(x+y\right).10\left(x+y\right)}$

$=\frac{5x\left(5x-7y\right)}{50\left(x+y\right)\left(x-y\right)}$

$=\frac{x\left(5x-7y\right)}{\left(x+y\right)\left(x-y\right)}$

2

๖²⁴ʱTú❄⁀ᶦᵈᵒᶫ

16 tháng 3 2020

chỗ cuối tớ sai

$=\frac{x\left(5x-7y\right)}{10\left(x+y\right)\left(x-y\right)}$

đây nha , e xin lỗi