tìm x,biết:a)\(\frac{2}{\left(x 2\right)\left(x 4\right)} \frac{4}{\left(x 4\right)\left(x 8\right)} \frac{6}{\left(x 8\right)\left(x 14\right)}=\frac{x}{\left(x 2\right)\left(x 14\right)}\)b)\(\frac{...

HK

Huyền Kelly

1 tháng 8 2016 - olm

tìm x,biết:a)\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)b)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)c)\(\left(x+2\right)^2=\frac{38}{25}+\frac{9}{10}-\frac{11}{15}+\frac{13}{21}-\frac{15}{28}+\frac{17}{36}-...+\frac{197}{4851}-\frac{199}{4950}\)giúp tớ...

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#Toán lớp 7

0

LT

Lê Thùy Nhi

14 tháng 9 2016

tìm x: \(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

#Toán lớp 7

1

IM

Isolde Moria

14 tháng 9 2016

\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}-\frac{1}{x+8}-\frac{1}{x+16}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+16}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{\left(x+16\right)-\left(x+2\right)}{\left(x+2\right)\left(x+16\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow x+16-x-2=x\)

\(\Rightarrow x=14\)

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LT

lê THỊ LINH CHI

19 tháng 7 2016 - olm

\(\frac{2}{\left(x+2\right).\left(x+4\right)}+\frac{4}{\left(x+4\right).\left(x+8\right)}+\frac{6}{\left(x+8\right).\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)

#Toán lớp 7

0

NV

Nguyễn Vi Thảo Nhi

6 tháng 7 2016 - olm

\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

#Toán lớp 7

1

LT

Le Thi Khanh Huyen

6 tháng 7 2016

\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(=\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(=\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(=\frac{x+16}{\left(x+2\right)\left(x+14\right)}-\frac{x+2}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(=\frac{8}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow x=8\)

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SK

Shinichi Kudo

18 tháng 6 2017 - olm

Tìm x,

\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

#Toán lớp 6

2

N

nghia

18 tháng 6 2017

\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Leftrightarrow\frac{x}{\left(x+2\right)\left(x+14\right)}=\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}-\frac{1}{x+8}-\frac{1}{x+14}\)

\(\Leftrightarrow\frac{x}{\left(x+2\right)\left(x+14\right)}=\frac{1}{x+2}-\frac{1}{x+14}\)

\(\Leftrightarrow\frac{x}{\left(x+2\right)\left(x+14\right)}=\frac{\left(x+14\right)-\left(x+2\right)}{\left(x+2\right)\left(x+14\right)}\)

\(\Leftrightarrow x=\left(x+14\right)-\left(x+2\right)\)

\(\Leftrightarrow x=x+14-x-2\)

\(\Leftrightarrow x=\left(x-x\right)+\left(14-2\right)\)

\(\Leftrightarrow x=0+12\)

\(\Leftrightarrow x=12\)

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HH

Hattori Hejji

18 tháng 6 2017

x=12

Áp dụng tổng đặc biệt đấy

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G6

GT 6916

18 tháng 8 2018 - olm

Tìm x

\(\frac{2}{\left(x+2\right).\left(x+4\right)}+\frac{4}{\left(x+4\right).\left(x+8\right)}+\frac{6}{\left(x+8\right).\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)

#Toán lớp 7

1

TT

Trần Thanh Phương

18 tháng 8 2018

\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\frac{12}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow x=12\)

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G6

GT 6916

29 tháng 7 2018 - olm

Tìm x

\(\frac{2}{\left(x+2\right).\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)

#Toán lớp 7

2

S

ST

29 tháng 7 2018

=>\(\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

=>\(\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

=>\(\frac{x+14-x-2}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

=>\(\frac{12}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

=>x=12

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D

Doraemon

31 tháng 8 2018

Ta có: \(\frac{2}{\left(x+2\right).\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{x+14-x-2}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{12}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow x=12\)

Vậy \(x=12\)

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DT

Đỗ Thị Ngọc Khánh

14 tháng 6 2015 - olm

Tim x : \(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

#Toán lớp 7

1

NT

nguyen truong giang

14 tháng 6 2015

đây là toán lớp 6 sao ?

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PT

Pham Thi Hoa

7 tháng 6 2015 - olm

Tìm x biết \(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+4\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

#Toán lớp 8

1

LZ

login zeus

31 tháng 8 2017

mình cũng ko biết giải

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NT

Nguyễn Thị Thanh Trúc

7 tháng 6 2015 - olm

Tìm x thuộc Q biết: \(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)

#Toán lớp 7

3

AM

Ác Mộng

7 tháng 6 2015

ĐKXĐ:\(x\ne\left\{-2;-4;-8;-14\right\}\)

\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)

\(\Leftrightarrow2\left(x+8\right)\left(x+14\right)+4\left(x+2\right)\left(x+14\right)+6\left(x+2\right)\left(x+4\right)=x\left(x+8\right)\left(x+14\right)\)

\(\Leftrightarrow2x^2+44x+224+4x^2+64x+112+6x^2+36x+48=x^3+22x^2+112x\)

\(\Leftrightarrow12x^2+144x+384=x^3+22x^2+112x\)

\(\Leftrightarrow x^3+22x^2-12x^2+112x-144x-384=0\)

\(\Leftrightarrow x^3+10x^2-32x-384=0\)

\(\Leftrightarrow\left(x-6\right)\left(x^2+16x+64\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+8\right)^2=0\)

\(\Leftrightarrow x=6\)(x=-8 loại vì x=-8 thì PT không xác định)

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S

Sagittarus

7 tháng 6 2015

x=6

cần lời giải ko

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HP

Hân Phan

10 tháng 9 2017 - olm

\(\frac{2}{\left(x+2\right)\left(x+4\right)}\)+\(\frac{4}{\left(x+4\right)\left(x+8\right)}\)+\(\frac{6}{\left(x+8\right)\left(x+14\right)}\)=\(\frac{x}{\left(x+2\right)\left(x+14\right)}\)

với x ko thuộc {-2;-4;-6;-8;-14}

#Toán lớp 7

1

S

ST

10 tháng 9 2017

\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{x+14}{\left(x+2\right)\left(x+14\right)}-\frac{x+2}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{x+14-x+2}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{12}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

=> x = 12

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