Rút gọn các biểu thức:\(\frac{x 1}{x^2-2x}.\frac{4-x}{x^2 x}\)\(\frac{x^3}{x 2006}.\frac{2x 1975}{x 1} \frac{x^3}{x 2006}.\frac{31-x}{x 1}\)\(\frac{19x 8}{x-7}.\frac{5x-9}{x 2006}-\frac{19x 8}{x-7}.\f...

LT

Lê Thanh Ngọc

16 tháng 11 2016

Rút gọn các biểu thức:

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

\(\frac{x^3}{x+2006}.\frac{2x+1975}{x+1}+\frac{x^3}{x+2006}.\frac{31-x}{x+1}\)

\(\frac{19x+8}{x-7}.\frac{5x-9}{x+2006}-\frac{19x+8}{x-7}.\frac{4x-2}{x+2006}\)

#Toán lớp 8

0

ND

Nguyễn Đức Trường

15 tháng 5 2020

Giải phương trình sau :

a) \(\frac{x^2-2x+1}{x^2-2x+2}+\frac{x^2-2x+2}{x^2-2x+3}=\frac{7}{6}\)

b) \(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)

#Toán lớp 8

0

TM

Trần Minh Thắng

22 tháng 8 2020

...

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#Toán lớp 6

3

XO

Xyz OLM

22 tháng 8 2020

1) Ta có : \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)

=> x + 1 = 0

=> x = - 1

b) \(\frac{x+4}{2006}+\frac{x+3}{2007}=\frac{x+2}{2008}+\frac{x+1}{2009}\)

=> \(\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+1}{2009}+1\right)\)

=> \(\frac{x+2010}{2006}+\frac{x+2010}{2007}=\frac{x+2010}{2008}+\frac{x+2010}{2009}\)

=> \(\left(x+2010\right)\left(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)

Vì \(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\ne0\)

=> x + 2010 = 0

=> x = -2010

c) \(\frac{x+1945}{45}+\frac{x+1954}{54}=\frac{x+1975}{75}+\frac{x+1969}{69}\)

\(\Rightarrow\left(\frac{x+1945}{45}-1\right)+\left(\frac{x+1954}{54}-1\right)=\left(\frac{x+1975}{75}-1\right)+\left(\frac{x+1969}{69}-1\right)\)

=> \(\frac{x+1900}{45}+\frac{x+1900}{54}=\frac{x+1900}{75}+\frac{x+1900}{69}\)

=> \(\left(x+1900\right)\left(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\right)=0\)

=> \(x+1900=0\left(\text{Vì }\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\ne0\right)\)

=> x = -1900

d) \(\frac{x+2008}{10}+\frac{x+2010}{9}=\frac{x+2012}{8}+\frac{x+2014}{7}\)

=> \(\left(\frac{x+2008}{10}+2\right)+\left(\frac{x+2010}{9}+2\right)=\left(\frac{x+2012}{8}+2\right)+\left(\frac{x+2014}{7}+2\right)\)

=> \(\frac{x+2028}{10}+\frac{x+2028}{9}=\frac{x+2028}{8}+\frac{x+2028}{7}\)

=> \(\left(x+2028\right)\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\right)=0\)

=> x + 2028 = 0 \(\left(\text{Vì }\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\ne0\right)\)

=> x = -2028

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TD

Tạ Đức Hoàng Anh

22 tháng 8 2020

1) Ta có: \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

+ TH₁: \(x+1=0\)\(\Leftrightarrow\)\(x=-1\)

+ TH₂: \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}=0\)

Vì \(\hept{\begin{cases}\frac{1}{10}>\frac{1}{13}\\\frac{1}{11}>\frac{1}{14}\\\frac{1}{12}>0\end{cases}}\)\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}>\frac{1}{13}+\frac{1}{14}\)

\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0\)

mà \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}=0\)

\(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=-1\)

2) Ta có: \(\frac{x+4}{2006}+\frac{x+3}{2007}=\frac{x+2}{2008}+\frac{x+1}{2009}\)

\(\Leftrightarrow\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+3}{2007}+1\right)-\left(\frac{x+2}{2008}+1\right)-\left(\frac{x+1}{2009}+1\right)=0\)

\(\Leftrightarrow\frac{x+2010}{2006}+\frac{x+2010}{2007}-\frac{x+2010}{2008}-\frac{x+2010}{2009}=0\)

\(\Leftrightarrow\left(x+2010\right).\left(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)

+ TH₁: \(x+2010=0\)\(\Leftrightarrow\)\(x=-2010\)

+ TH₂: \(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}=0\)

Vì \(\hept{\begin{cases}\frac{1}{2006}>\frac{1}{2008}\\\frac{1}{2007}>\frac{1}{2009}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{2006}+\frac{1}{2007}>\frac{1}{2008}+\frac{1}{2009}\)

\(\Rightarrow\)\(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}>0\)

mà \(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}=0\)

\(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=-2010\)

3) Ta có: \(\frac{x+1945}{45}+\frac{x+1954}{54}=\frac{x+1975}{75}+\frac{x+1969}{69}\)

\(\Leftrightarrow\left(\frac{x+1945}{45}-1\right)+\left(\frac{x+1954}{54}-1\right)-\left(\frac{x+1975}{75}-1\right)-\left(\frac{x+1969}{69}-1\right)=0\)

\(\Leftrightarrow\frac{x+1900}{45}+\frac{x+1900}{54}-\frac{x+1900}{75}-\frac{x+1900}{69}=0\)

\(\Leftrightarrow\left(x+1900\right).\left(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\right)=0\)

+ TH₁: \(x+1900=0\)\(\Leftrightarrow\)\(x=-1900\)

+ TH₂: \(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}=0\)

Vì \(\hept{\begin{cases}\frac{1}{45}>\frac{1}{75}\\\frac{1}{54}>\frac{1}{69}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{45}+\frac{1}{54}>\frac{1}{75}+\frac{1}{69}\)

\(\Rightarrow\)\(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}>0\)

mà \(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}=0\)

\(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=-1900\)

4) Ta có: \(\frac{x-99}{5}+\frac{x-97}{7}=\frac{x-95}{9}+\frac{x-93}{11}\)

\(\Leftrightarrow\left(\frac{x-99}{5}-1\right)+\left(\frac{x-97}{7}-1\right)-\left(\frac{x-95}{9}-1\right)-\left(\frac{x-93}{11}-1\right)=0\)

\(\Leftrightarrow\frac{x-104}{5}+\frac{x-104}{7}-\frac{x-104}{9}-\frac{x-104}{11}=0\)

\(\Leftrightarrow\left(x-104\right).\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)=0\)

+ TH₁: \(x-104=0\)\(\Leftrightarrow\)\(x=104\)

+ TH₂: \(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}=0\)

Vì \(\hept{\begin{cases}\frac{1}{5}>\frac{1}{7}\\\frac{1}{9}>\frac{1}{11}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{5}+\frac{1}{7}>\frac{1}{9}+\frac{1}{11}\)

\(\Rightarrow\)\(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}>0\)

mà \(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}=0\)

\(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=104\)

5) Ta có: \(\frac{x+2008}{10}+\frac{x+2010}{9}=\frac{x+2012}{8}+\frac{x+2014}{7}\)

\(\Leftrightarrow\left(\frac{x+2008}{10}+2\right)+\left(\frac{x+2010}{9}+2\right)-\left(\frac{x+2012}{8}+2\right)-\left(\frac{x+2014}{7}+2\right)=0\)

\(\Leftrightarrow\frac{x+2028}{10}+\frac{x+2028}{9}-\frac{x+2028}{8}-\frac{x+2028}{7}=0\)

\(\Leftrightarrow\left(x+2028\right).\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\right)=0\)

+ TH₁: \(x+2028=0\)\(\Leftrightarrow\)\(x=-2028\)

+ TH₂: \(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}=0\)

Vì \(\hept{\begin{cases}\frac{1}{10}< \frac{1}{8}\\\frac{1}{9}< \frac{1}{7}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{9}< \frac{1}{8}+\frac{1}{7}\)

\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}< 0\)

mà \(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}=0\)

\(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=-2028\)

Chúc bn hok tốt nha

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NK

Nguyễn Khánh Linh

10 tháng 3 2020

Bài 3 : Xét dấu biểu thức sau : 1 , \(f\left(x\right)=\frac{x-7}{4x^2-19x+12}\) 2 , \(f\left(x\right)=\frac{11x+3}{-x^2+5x-7}\) 3 , \(f\left(x\right)=\frac{3x-2}{x^3-3x^2+2}\) 4 , \(f\left(x\right)=\frac{x^2+4x-12}{\sqrt{6}x^2+3x+\sqrt{2}}\) 5 , \(f\left(x\right)=\frac{x^2-3x-2}{-x^2+x-1}\) 6 , \(f\left(x\right)=\frac{x^3-5x+4}{x^4-4x^3+8x-5}\) 7 , \(f\left(x\right)=\frac{\left(x+3\right)\left(x-2\right)\left(-2x^2+x-1\right)}{\left(2x-5\right)\left(x^2+3x-10\right)}\) 8 ,...

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#Toán lớp 10

5

NV

Nguyễn Việt Lâm

Giáo viên

14 tháng 3 2020

1.

\(f\left(x\right)=\frac{x-7}{\left(x-4\right)\left(4x-3\right)}\)

Vậy:

\(f\left(x\right)\) ko xác định tại \(x=\left\{\frac{3}{4};4\right\}\)

\(f\left(x\right)=0\Rightarrow x=7\)

\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{3}{4}< x< 4\\x>7\end{matrix}\right.\)

\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3}{4}\\4< x< 7\end{matrix}\right.\)

2.

\(f\left(x\right)=\frac{11x+3}{-\left(x-\frac{5}{2}\right)^2-\frac{3}{4}}\)

Vậy:

\(f\left(x\right)=0\Rightarrow x=-\frac{3}{11}\)

\(f\left(x\right)>0\Rightarrow x< -\frac{3}{11}\)

\(f\left(x\right)< 0\Rightarrow x>-\frac{3}{11}\)

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NV

Nguyễn Việt Lâm

Giáo viên

14 tháng 3 2020

3.

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

Vậy:

\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{3}\right\}\)

\(f\left(x\right)=0\Rightarrow x=\frac{2}{3}\)

\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< 1-\sqrt{3}\\\frac{2}{3}< x< 1\\x>1+\sqrt{3}\end{matrix}\right.\)

\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}1-\sqrt{3}< x< \frac{2}{3}\\1< x< 1+\sqrt{3}\end{matrix}\right.\)

4.

\(f\left(x\right)=\frac{\left(x-2\right)\left(x+6\right)}{\sqrt{6}\left(x+\frac{\sqrt{6}}{4}\right)^2+\frac{8\sqrt{2}-3\sqrt{6}}{8}}\)

Vậy:

\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)

\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)

\(f\left(x\right)< 0\Rightarrow-6< x< 2\)

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LS

Lenkin san

21 tháng 7 2019

Giair pt:

c, x ( 3x-1) (3x+1) (3x+2) =8

d, (x+1) (2x+3) (2x+5) (x+3)=45

e,x⁴+ 3x³- 15x²- 19x + 3 = 0

f, \(\frac{1}{x^2+x}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}=\frac{1}{3}\)

h,\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

#Toán lớp 8

2

S

svtkvtm

21 tháng 7 2019

https://i.imgur.com/jTzVBzQ.jpg

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S

svtkvtm

21 tháng 7 2019

https://i.imgur.com/1Xvpjty.jpg

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H2

Hiếu 2k6

5 tháng 3 2020

Ai giúp vs !!!
\(a.\frac{3x-7}{5}=\frac{2x-1}{3}\\ b.\frac{4x-7}{12}-x=\frac{3x}{8}\\ c.\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\\ d.\frac{5x-8}{3}=\frac{1-3x}{2}\\ e.\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\\ f.\frac{x-1}{\frac{2}{5}}-3-\frac{3x-2}{\frac{5}{4}}-2=1\)

#Toán lớp 8

8

KN

Kiệt Nguyễn

5 tháng 3 2020

\(\frac{3x-7}{5}=\frac{2x-1}{3}\)

\(\Leftrightarrow9x-21=10x-5\)

\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)

\(\frac{4x-7}{12}-x=\frac{3x}{8}\)

\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow-56-64x=36x\)

\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)

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KN

Kiệt Nguyễn

5 tháng 3 2020

\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)

\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)

Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0

Vậy x = 2019

\(\frac{5x-8}{3}=\frac{1-3x}{2}\)

\(\Leftrightarrow10x-16=3-9x\)

\(\Leftrightarrow19x=19\Leftrightarrow x=1\)

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DN

Đặng Nguyễn Thảo Nguyên

22 tháng 12 2019

Rút gọn biểu thức:

\(\left(\frac{x-1}{x^3}-\frac{x-1}{x^3-x^2}+\frac{3}{x^3-2x^2+x}\right):\frac{16x^2+16x-8}{x^5-2x^4+x^3}\)

#Toán lớp 8

0

KN

Kẹo Ngọt Cây

15 tháng 4 2020

1)2x(25x-4)-(5x-2)(5x+1)=8 / 5)\(2\left(x-2\right)-3\left(3x-1\right)=\left(x-3\right)\) 2)x(4x-3)-(2x-2)(2x-1)=5 / 6)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\) 3)\(\frac{5}{2x+3}+\frac{3}{9-x^2}=\frac{8}{7\left(x=3\right)}\) / 7)\(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\) 4)\(\frac{2}{3\left(x-2\right)}+\frac{5}{12-3x^2}=\frac{3}{4\left(x+2\right)}\) /...

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#Toán lớp 10

1

KN

Kẹo Ngọt Cây

15 tháng 4 2020

Đây là lớp 8 nha các b giúp mk với

Do mk viết nhầm

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DQ

Dragneel Quốc Việt

31 tháng 1 2017

...

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#Toán lớp 8

6

CW

Cold Wind

31 tháng 1 2017

a) \(\frac{x+1}{4}-\frac{x+2}{5}+\frac{x+4}{7}-\frac{x+5}{8}+\frac{x+7}{10}-\frac{x+9}{12}=0\)

\(\Leftrightarrow\left(\frac{x+4-3}{4}+\frac{x+7-3}{7}+\frac{x+10-3}{10}\right)-\left(\frac{x+5-3}{5}+\frac{x+8-3}{8}+\frac{x+12-3}{12}\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{4}+\frac{1}{7}+\frac{1}{10}\right)-\left(x-3\right)\left(\frac{1}{3}+\frac{1}{8}+\frac{1}{12}\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}\right)=0\)

\(\Leftrightarrow x=3\)

Vậy x=3

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CW

Cold Wind

31 tháng 1 2017

d) \(\frac{x^2-15x+1}{x+17}=x-2\Leftrightarrow x^2-15x+1=x^2+15x-34\)

\(\Leftrightarrow30x=35\Leftrightarrow x=\frac{7}{6}\)

vậy x=7/6

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TU

Tsukino Usagi

30 tháng 11 2016

Rút gọn biểu thức:

a, \(\frac{x^4+15x+7}{2x^3+2}.\frac{x}{14x^2+1}.\frac{4x^3+4}{x^4+15x+7}\)

b, \(\frac{x^7+3x^2+2}{x^3-1}.\frac{3x}{x+1}.\frac{x^2+x+1}{x^7+3x^2+2}\)

#Toán lớp 8

1

TH

Trần Huỳnh Cẩm Hân

30 tháng 11 2016

a. 2x

b.\({3x}\over x^2-1\)

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