ĐKXĐ : x khác -3;-4;-5;-6

pt <=> 1/(x+3).(x+4) + 1/(x+4).(x+5) + 1/(x+5).(x+6) = 1/18

<=> 1/x+3 - 1/x+4 + 1/x+4 - 1/x+5 + 1/x+5 - 1/x+6 = 1/18

<=> 1/x+3 - 1/x+6 = 1/18

<=> x+6-x-3/(x+3).(x+6) = 1/18

<=> 3/x^2+9x+18 = 1/18

<=> x^2+9x+18 = 3 : 1/18 = 48

<=> x^2+9x+18-48 = 0

<=> x^2+9x-30 = 0

<=>(x+9/2)^2 = 201/4

<=> x = \(\frac{+-\sqrt{201}-9}{2}\)(tm)

k mk nha

                   \(\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}=\frac{1}{18}\)

\(\Leftrightarrow\)\(\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{18}\)

\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{18}\)

\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+6}=\frac{1}{18}\)

\(\Leftrightarrow\)\(\frac{3}{\left(x+3\right)\left(x+6\right)}=\frac{1}{18}\)

\(\Leftrightarrow\)(x + 3)(x + 6) = 54 = 6.9 = (-6).(-9)

Đến đây giải tiếp nha

pt <=> 1/(x+2).(x+3) + 1/(x+3).(x+4) + 1/(x+4).(x+5) + 1/(x+5).(x+6) = 1/8

<=> 1/x+2 - 1/x+3 + 1/x+3 - 1/x+4 + 1/x+4 - 1/x+5 + 1/x+5 - 1/x+6 = 1/8

<=> 1/x+2 - 1/x+6 = 1/8

<=> (x+6-x-2)/(x+2).(x+6) = 1/8

<=> 4/(x+2).(x+6) = 1/8

<=>(x+2).(x+6) = 4 : 1/8 = 32

<=>x^2 + 8x + 12 = 32

<=> x^2+8x+12-32=0

<=>x^2+8x-20=0

<=>(x-2).(x+10)=0

<=> x-2 =0 hoặc x+10 = 0

<=> x=2 hoặc x=-10

giang sinh an lanh $%###Xuyen gam cu chuoi###%$

hk biết

phân tích mẫu thành nhân tử r` tách ra rút gọn như kiểu bài tính của lớp 5 ấy

bài tương tự : Câu hỏi của Lê Phương Oanh - Toán lớp 8 | Học trực tuyến (https://h-o-c-24.vn/hoi-dap/question/179719.html)

A=1/(x-2)(x-3) + 1/(x-3)(x-4) + 1/(x-4)(x-5) + 1/(x-5)(x-6)=1/8 (ĐKXĐ: x#2,x#3,x#4,x#5,x#6)

A= 1/x-2 -1/x-3 + 1/x-3 -1/x-4 .....-1/x-6=1/8

=>1/x-2 -1/x-6=1/8

=>8(x-6)-8(x-2)=(x-2)(x-6)

=> 8x-48-8x+16=x^2-8x+12

=> x^2-8x-20=0

=> (x-10)(x+2)=0 => x=10,x=-2 thuộc ĐKXĐ

Có cần thế ko ạ ??? Shinichi

Điều kiện xác định \(\hept{\begin{cases}x\ne2\\x\ne\\x\ne4\end{cases}3}\)

                              \(\hept{\begin{cases}x\ne5\\x\ne6\end{cases}}\)

Ta có : \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)

\(x^2-7x+12=\left(x-3\right)\left(x-4\right)\)

\(x^2-9x+20=\left(x-4\right)\left(x-5\right)\)

\(x^2-11+30=\left(x-5\right)\left(x-6\right)\)

Phương trình đã tương đương với 

\(\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-6\right)}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}+\frac{1}{x-5}-\frac{1}{x-4}+\frac{1}{x-6}-\frac{1}{x-5}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x-6}-\frac{1}{x-2}=\frac{1}{8}\Leftrightarrow\frac{4}{\left(x-6\right)\left(x-2\right)}=\frac{1}{8}\)

\(\Leftrightarrow x^2-8x-20=0\Leftrightarrow\left(x-10\right)\left(x+2\right)=0\)

\(x-10=0\Leftrightarrow x=10\)

hoặc 

\(x+2=0\Leftrightarrow x=-2\)

\(\Leftrightarrow\orbr{\begin{cases}x=10\\x=-2\end{cases}}\)thỏa mãn điều kiện phương trình 

Phương trình có nghiệm \(x=10;x=-2\)

\(\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}=\frac{1}{8}\) (ĐKXĐ: x \(\ne\) -2; x \(\ne\) -3; x \(\ne\) -4; x \(\ne\) -5; x \(\ne\) -6)

\(\Leftrightarrow\) \(\frac{1}{x^2+2x+3x+6}+\frac{1}{x^2+3x+4x+12}+\frac{1}{x^2+4x+5x+20}+\frac{1}{x^2+5x+6x+30}=\frac{1}{8}\)

\(\Leftrightarrow\) \(\frac{1}{x\left(x+2\right)+3\left(x+2\right)}+\frac{1}{x\left(x+3\right)+4\left(x+3\right)}+\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}=\frac{1}{8}\)

\(\Leftrightarrow\) \(\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{8}\)

\(\Leftrightarrow\) \(\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{8}\)

\(\Leftrightarrow\) \(\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{8}\)

\(\Leftrightarrow\) \(\frac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\frac{1}{8}\)

\(\Leftrightarrow\) \(\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{1}{8}\)

\(\Leftrightarrow\) \(\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{4}{32}\)

\(\Rightarrow\) (x + 2)(x + 6) = 32

\(\Leftrightarrow\) (x + 2)(x + 6) - 32 = 0

\(\Leftrightarrow\) x² + 6x + 2x + 12 - 32 = 0

\(\Leftrightarrow\) x² + 8x - 20 = 0

\(\Leftrightarrow\) x² + 8x + 16 - 36 = 0

\(\Leftrightarrow\) (x + 4)² - 36 = 0

\(\Leftrightarrow\) (x + 4 - 6)(x + 4 + 6) = 0

\(\Leftrightarrow\) (x - 2)(x + 10) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TMĐK\right)\\x=-10\left(TMĐK\right)\end{matrix}\right.\)

Vậy S = {2; -10}

Chúc bn học tốt!!

thank you very very..... much!

OMG!

Đặt 

\(A=\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}+\frac{1}{x^2-11x+30}\) 

( ĐKXĐ : \(x\ne2,x\ne3,x\ne4,x\ne5,x\ne6\) )

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

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

\(=\frac{1}{x-2}-\frac{1}{x-6}\)

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

Để : \(A\ge0\Leftrightarrow\frac{-4}{\left(x-2\right)\left(x-6\right)}\ge0\)

\(\Leftrightarrow\left(x-2\right)\left(x-6\right)\le0\)

TH1 : \(\hept{\begin{cases}x-2\le0\\x-6\ge0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\le2\\x\ge6\end{cases}}\) ( vô lý )

TH2 : \(\hept{\begin{cases}x-2\ge0\\x-6\le0\end{cases}\Leftrightarrow2\le x\le6}\)kết hợp với ĐKXĐ

\(\Rightarrow2< x< 6\)

Vậy : \(2< x< 6\) thỏa mãn bất phương trình.

ai giúp mình câu (a) với ạ

ĐKXĐ: \(x\ne\pm\frac{3}{2}\)

\(\frac{1}{\left(2x-3\right)^2}+\frac{3}{\left(2x-3\right)\left(2x+3\right)}-\frac{4}{\left(2x+3\right)^2}=0\)

\(\Leftrightarrow\frac{1}{\left(2x-3\right)^2}-\frac{1}{\left(2x-3\right)\left(2x+3\right)}+\frac{4}{\left(2x-3\right)\left(2x+3\right)}-\frac{4}{\left(2x-3\right)^2}=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=2x-3\left(vn\right)\\2x+3=4\left(2x-3\right)\Rightarrow x=\frac{5}{2}\end{matrix}\right.\)