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\(\left(x^2+11x+12\right)\left(x^2+9x+20\right)\left(x^2+13x+42\right)=36\left(x^2+11x+30\right)\left(x^2+11x+31\right)\)
\(\Leftrightarrow\left(x^2+11x+12\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)=36\left(x^2+11x+30\right)\left(x^2+11x+31\right)\)
\(\Leftrightarrow\left(x^2+11x+12\right)\left(x^2+11x+28\right)\left(x^2+11x+30\right)=36\left(x^2+11x+30\right)\left(x^2+11x+31\right)\)
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 \(a=56\Leftrightarrow x^2+11x+30=56\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=2\end{matrix}\right.\)
đề sai rồi bạn , phải là ( x2+11x + 12)(x2+9x+20 ) = 36(x2+11x+30)(x2+11x+31)
\(\left(3x-2\right)\left(4x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\4x+5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{5}{4}\end{cases}}\)
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...
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: \(x\ne-4;-5;-6;-7\)
\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow x^2+11x-26=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)
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.
\(\Rightarrow\left[{}\begin{matrix}x_1=\dfrac{-11-\sqrt{105}}{2}\\x_2=\dfrac{-11+\sqrt{105}}{2}\end{matrix}\right.\)
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. \(\left(3x-2\right)\left(4x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-\frac{5}{4}\end{matrix}\right.\)
À tớ thiếu ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-4\\x\ne-5\\x\ne-6\\x\ne-7\end{matrix}\right.\)
Ta có: \(\left|x+\frac{1}{2}\right|\ge0\left|x+\frac{1}{6}\right|\ge0;...;\left|x+\frac{1}{110}\ge0\right|\)
\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{100}\right|\ge0\)
\(\Rightarrow11x\ge0\Rightarrow x\ge0\)
\(\Rightarrow x+\frac{1}{2}>0;x+\frac{1}{6}>0;...;x+\frac{1}{100}>0\)
\(\Rightarrow\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{6}\right|=x+\frac{1}{6};...;\left|x+\frac{1}{100}\right|=x+\frac{1}{110}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{110}\right)=11x\)
\(\Rightarrow10x+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)=11x\)
\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=11x\)
\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=11x\)
\(\Rightarrow10x+\frac{10}{11}=11x\)
\(\Rightarrow x=\frac{10}{11}\)
vì |x+1/2| ; |x+1/6| ; ............ ; |x+110| lớn hơn hoặc bằng 0=> 11x lớn hớn hoặc bằng 0=> x lớn hớn hoặc bằng 0
=>x+1/2 ; x+1/6 ; ............ ; x+110 lớn hơn hoặc bằng 0
ta có: x+1/2+x+1/6+x+1/12+...+x+1/110=11x
(x+x+...+x)+(1/1.2+1/2.3+1/3.4+...+1/10.11)=11x
10x+(1-1/10)=11x
x= 1/9
à mình bỏ dấu" | " vì khi mà lớn hơn hoặc bằng 1 rồi thfi bỏ ra nó vẫn có giá trị bằng giá trị trị lúc ban đầu
Vì \(\left|x+\frac{1}{2}\right|\ge0;\left|x+\frac{1}{6}\right|\ge0;\left|x+\frac{1}{12}\right|\ge0;...;\left|x+\frac{1}{110}\right|\ge0\)
\(\Rightarrow11x\ge0\)
\(\Rightarrow x\ge0\)
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 \(x=\frac{10}{11}\)
a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)
\(=8x^5+2x^4-6x^3-14x^2\)
b: \(=2x^3-3x^2-5x+6x^2-9x-15\)
\(=2x^3+3x^2-14x-15\)
c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)
d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)
e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)
=2x^2-5x+1
