K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

a)

ĐKXĐ: x khác -4;-5;-6;-7

\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+20}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{8}\\ \Rightarrow x^2+11x+28=24\\ \Leftrightarrow x^2+11x+4=0\)

ta có: \(\Delta=11^2-4.1.4=105>0\) nên phương trình có 2 nghiệm phân biệt.

\(\Rightarrow\left[{}\begin{matrix}x_1=\dfrac{-11-\sqrt{105}}{2}\\x_2=\dfrac{-11+\sqrt{105}}{2}\end{matrix}\right.\)

9 tháng 2 2018

Giải:

a) \(8\left(3x-2\right)-13x=5\left(12-3x\right)+7x\)

\(\Leftrightarrow24x-16-13x=60-15x+7x\)

\(\Leftrightarrow24x-13x+15x-7x=60+16\)

\(\Leftrightarrow19x=76\)

\(\Leftrightarrow x=\dfrac{76}{19}=4\)

Vậy ...

b) \(\dfrac{5x}{x+2}-\dfrac{3}{x-2}+\dfrac{3x^2+6}{\left(x-2\right)\left(x+2\right)}=0\) (1)

ĐKXĐ: \(x\ne\pm2\)

\(\left(1\right)\Leftrightarrow\dfrac{5x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{3x^2+6}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow5x\left(x-2\right)-3\left(x+2\right)+3x^2+6=0\)

\(\Leftrightarrow5x^2-10x-3x-6+3x^2+6=0\)

\(\Leftrightarrow8x^2-13x=0\)

\(\Leftrightarrow x\left(8x-13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\8x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=\dfrac{13}{8}\left(TM\right)\end{matrix}\right.\)

Vậy ...

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

ĐKXĐ: \(x\ne-1;x\ne3\)

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

\(\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\)

\(\Leftrightarrow x\left(x+1+x-3\right)=4x\)

\(\Leftrightarrow x\left(2x-2\right)=4x\)

\(\Leftrightarrow2x-2=4\)

\(\Leftrightarrow x=3\)

Vậy ...

10 tháng 12 2023

Sửa đề:

 \(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)

ĐKXĐ: \(x\notin\left\{1;3;8;20\right\}\)

PT=>\(-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-8}-\dfrac{1}{x-8}+\dfrac{1}{x-20}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)

=>\(-\dfrac{1}{x-4}=-\dfrac{3}{4}\)

=>\(x-1=\dfrac{4}{3}\)

=>\(x=\dfrac{4}{3}+1=\dfrac{7}{3}\)(nhận)

a: \(=\dfrac{4xy+x^2-2xy+y^2}{2\left(x+y\right)\left(x-y\right)}\cdot\dfrac{2x}{x+y}-\dfrac{y}{x-y}\)

\(=\dfrac{x}{x-y}-\dfrac{y}{x-y}=1\)

b: \(=\dfrac{x^2+x-2-2x^2-2x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{3\left(x+1\right)}{x}+\dfrac{4x^2+x+7}{x\left(x-1\right)}\)

\(=\dfrac{-x^2-x-2}{\left(x-1\right)}\cdot\dfrac{3}{x}+\dfrac{4x^2+x+7}{x\left(x-1\right)}\)

\(=\dfrac{4x^2+x+7-3x^2-3x-6}{x\left(x-1\right)}=\dfrac{x^2-2x+1}{x\left(x-1\right)}=\dfrac{x-1}{x}\)

c: \(=\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}\)

\(=\dfrac{x+7-x-4}{\left(x+7\right)\left(x+4\right)}=\dfrac{3}{\left(x+4\right)\left(x+7\right)}\)

 

12 tháng 2 2018

giải hết đống này chắc @@ quá,để tối đi,giờ t đi làm mấy bài ngắn ngắn

12 tháng 2 2018

tuỳ bạn, qua Tết mik đăng lại

19 tháng 12 2023

  loading...  

loading...  loading...  

15 tháng 2 2018

a) điều kiện xác định : \(x\ne0\)

ta có : \(A=\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\)

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

\(\Leftrightarrow A=\dfrac{x^3-x^2+x+x^2-x+1-\left(x^3+x^2+x-x^2-x-1\right)}{x^4-x^3+x^2+x^3-x^2+x+x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\) \(\Leftrightarrow A=\dfrac{x^3-x^2+x+x^2-x+1-x^3-x^2-x+x^2+x+1}{x^4+x^2+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\) \(\Leftrightarrow A=\dfrac{2}{x^4+x^2+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\Leftrightarrow\left(x^4+x^2+1\right)A=2=\dfrac{3}{x}\) \(\Leftrightarrow2x=3\Leftrightarrow x=\dfrac{3}{2}\left(tmđk\right)\) vậy \(x=\dfrac{3}{2}\)
15 tháng 2 2018

b) điều kiện : \(x\notin\left\{-4;-5;-6;-7\right\}\)

\(B=\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)

\(\Leftrightarrow B=\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow B=\dfrac{\left(x+6\right)\left(x+7\right)+\left(x+4\right)\left(x+7\right)+\left(x+4\right)\left(x+5\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow B=\dfrac{x^2+13x+42+x^2+11x+28+x^2+9x+20}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow B=\dfrac{3x^2+33x+90}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\) \(\Leftrightarrow B=\dfrac{3\left(x+5\right)\left(x+6\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\) \(\Leftrightarrow B=\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\) \(\Leftrightarrow54=\left(x+4\right)\left(x+7\right)\)

\(\Leftrightarrow54=x^2+11x+28\Leftrightarrow x^2+11x+28-54=0\)

\(\Leftrightarrow x^2+11x-26=0\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+13=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\left(tmđk\right)\)

vậy \(x=2;x=-13\)

4 tháng 12 2018

a) 1/x(x + 1) + 1/(x + 1)(x + 2) + 1/(x + 2)(x + 3) + 1/(x + 3)(x + 4)

( 1/x - 1/x+1) + (1/x+1 - 1/x+2) + (1/x+2 - 1/ x+3) + 1/(x+3 - 1/x+4)

(1/x +1/x+4) - ( 1/x+2 - 1/x+2) - ( 1/x+3 - 1/x+3)

1/x +1/x+4

2x+4/x(x+4)

4 tháng 12 2018

Câu b bạn tách các mẫu thành nhân tử rồi làm như câu a nhé

b: \(=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right]\cdot\dfrac{x}{x+1}\)

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

\(=\dfrac{2x+2+6x^2+4x-2}{3x\left(x+1\right)}\cdot\dfrac{x}{x+1}\)

\(=\dfrac{6x^2+6x}{3\left(x+1\right)}\cdot\dfrac{1}{x+1}\)

\(=\dfrac{6x\left(x+1\right)}{3\left(x+1\right)^2}=\dfrac{2x}{x+1}\)

c: \(VT=\left[\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}+\dfrac{1}{\left(x+1\right)^2}\cdot\dfrac{1+x^2}{x^2}\right]\cdot\dfrac{x^3}{x-1}\)

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

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

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