K
Khách

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

30 tháng 3 2017

\(\left(x^2+11x+12\right)\left(x^2+9x+20\right)\left(x^2+13x+42\right)=36\left(x^2+11x+30\right)\left(x^2+11x+31\right)\)

\(\Leftrightarrow\left(x^2+11x+12\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)=36\left(x^2+11x+30\right)\left(x^2+11x+31\right)\)

\(\Leftrightarrow\left(x^2+11x+12\right)\left(x^2+11x+28\right)\left(x^2+11x+30\right)=36\left(x^2+11x+30\right)\left(x^2+11x+31\right)\)

Đặt \(x^2+11x+30=a\)

\(\Leftrightarrow\left(a-18\right)\left(a-2\right)a=36a\left(a+1\right)\)

\(\Leftrightarrow a^3-56a^2=0\)

\(\Leftrightarrow a^2\left(a-56\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=0\\a=56\end{matrix}\right.\)

Với \(a=0\Leftrightarrow x^2+11x+30=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-6\end{matrix}\right.\)

Với \(a=56\Leftrightarrow x^2+11x+30=56\)

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

29 tháng 3 2017

đề sai rồi bạn , phải là ( x2+11x + 12)(x2+9x+20 ) = 36(x2+11x+30)(x2+11x+31)

27 tháng 6 2016

oho

12 tháng 7 2023

Mày nhìn cái chóa j

24 tháng 4 2019

\(\left(3x-2\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\4x+5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{5}{4}\end{cases}}\)

24 tháng 4 2019

ĐKXĐ: x khác -4;-5;-6;-7

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

\(\Rightarrow\frac{1}{\left(x+4\right).\left(x+5\right)}+\frac{1}{\left(x+5\right).\left(x+6\right)}+\frac{1}{\left(x+6\right).\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{x+7-x-4}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}\Rightarrow3.18=x^2+11x+28\)

\(\Rightarrow x^2+11x-26=0\)

\(\Rightarrow\left(x-2\right).\left(x+13\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-13\end{cases}\left(tm\right)}\)

Vậy...

NV
10 tháng 3 2019

a/ Đặt \(6x+7=a\Rightarrow\left\{{}\begin{matrix}6x+8=a+1\\6x+6=a-1\end{matrix}\right.\)

\(\Rightarrow\left(a-1\right)\left(a+1\right)a^2-72=0\)

\(\Leftrightarrow\left(a^2-1\right)a^2-72=0\)

\(\Leftrightarrow a^4-a^2-72=0\)

\(\Leftrightarrow\left(a^2-9\right)\left(a^2+8\right)=0\)

\(\Leftrightarrow a^2=9\) (do \(a^2+8>0\))

\(\Rightarrow\left[{}\begin{matrix}a=3\\a=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}6x+7=3\\6x+7=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=\frac{-5}{3}\end{matrix}\right.\)

b/ ĐKXĐ: \(x\ne-4;-5;-6;-7\)

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

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow x^2+11x-26=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)

24 tháng 4 2019

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

\(=x\left(x+4\right)+5\left(x+4\right)=\left(x+4\right)\left(x+5\right)\)

Tương tự: \(x^2+11x+30=\left(x+5\right)\left(x+6\right)\)

\(x^2+13x+42=\left(x+6\right)\left(x+7\right)\)

\(\Rightarrow PT=\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(=\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(=\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(=18\left(x+7\right)-18\left(x+4\right)=\left(x+7\right)\left(x+4\right)\)

\(=x^2+11x+28=54\)

\(=x^2+11x-26=0\)

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

\(=x\left(x-2\right)+13\left(x-2\right)=0\)

\(=\left(x+13\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-13\\x=2\end{matrix}\right.\)

b. \(\left(3x-2\right)\left(4x+5\right)=0\)

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

24 tháng 4 2019

À tớ thiếu ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-4\\x\ne-5\\x\ne-6\\x\ne-7\end{matrix}\right.\)

26 tháng 2 2022

hic, mk chx học

2 tháng 3 2019

a)\(\left(x^2+1\right)\left(x^2-4x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x^2-4x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=-1\left(vn\right)\\\left(x-2\right)^2=0\end{cases}\Rightarrow}x=2}\)

b)\(\left(3x-2\right)\left(\frac{2x+6}{7}-\frac{4x-3}{5}\right)=0\\ \Rightarrow\left(3x-2\right)\left(\frac{10x+30-28x+21}{35}\right)=0\)

\(\Rightarrow\left(3x-2\right)\left(\frac{-18x+51}{35}\right)=0\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{17}{6}\end{cases}}\)

c)\(\left(3,3-11x\right)\left(\frac{21x+6+10-30x}{15}\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{10}\\x=\frac{16}{9}\end{cases}}\)

21 tháng 10 2016

\(\left|x+\frac{1}{2}\right|\ge0;\left|x+\frac{1}{6}\right|\ge0;\left|x+\frac{1}{12}\right|\ge0;...;\left|x+\frac{1}{110}\right|\ge0\)

\(\Rightarrow11x\ge0\)

\(\Rightarrow x\ge0\)

Với \(x\ge0\) ta có:

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+\left(x+\frac{1}{12}\right)+...+\left(x+\frac{1}{110}\right)=11x\)

\(\Rightarrow\left(x+\frac{1}{1.2}\right)+\left(x+\frac{1}{2.3}\right)+\left(x+\frac{1}{3.4}\right)+...+\left(x+\frac{1}{10.11}\right)=11x\)

\(\Rightarrow\left(x+x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=11x\)

10 số x

\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=11x\)

\(\Rightarrow1-\frac{1}{11}=11x-10x\)

\(\Rightarrow x=\frac{10}{11}\)

Vậy \(x=\frac{10}{11}\)

24 tháng 3 2020

Phép nhân và phép chia các đa thứcPhép nhân và phép chia các đa thức