\(\Rightarrow\left(x-1\right)^2-\left(2x-3\right)^2=0\\ \Rightarrow\left(x-1-2x+3\right)\left(x-1+2x-3\right)=0\\ \Rightarrow\left(2-x\right)\left(3x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{4}{3}\end{matrix}\right.\)

Nguyễn Hoàng Minh  thank you

đề?

c1 x²-3x-1

c2 3x²-5x-2

1:Ta có: \(x^2-3x-1\)

\(=x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{13}{4}\)

\(=\left(x-\dfrac{3}{2}\right)^2-\dfrac{13}{4}\ge-\dfrac{13}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)

cần gấp xin cảm ơn ạ

\(\left\{{}\begin{matrix}4x^2-12x+9\\x^2+10x+25\end{matrix}\right.\)

=2x² - 2.2x.3 + 3²

=4x²-12x+ 9

b) =x²+ 2.x.5+5²

=x²+ 10x+25

=(1-2x)(1+2x)

\(1-4x^2=1^2-\left(2x\right)^2=\left(1-2x\right)\left(1+2x\right)\)

\(1-4x^2\\ =1^2-\left(2x\right)^2\\ =\left(1-2x\right)\left(1+2x\right)\)

\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-....+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-6}=\dfrac{1}{10}\Leftrightarrow\dfrac{x-6-x+1}{\left(x-1\right)\left(x-6\right)}=\dfrac{1}{10}\)

\(\Leftrightarrow x^2-7x+56=0\Leftrightarrow x^2-2.\dfrac{7}{2}x+\dfrac{49}{4}+\dfrac{175}{4}=\left(x-\dfrac{7}{2}\right)^2+\dfrac{175}{4}>0\)

Vậy phương trình vô nghiệm 

ĐKXĐ: \(x\notin\left\{1;2;3;4;5;6\right\}\)

Ta có: \(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x-4}+\dfrac{1}{x-3}+\dfrac{1}{x-5}-\dfrac{1}{x-4}+\dfrac{1}{x-6}-\dfrac{1}{x-5}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{1}{x-6}-\dfrac{1}{x-1}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{10\left(x-1\right)}{10\left(x-6\right)\left(x-1\right)}-\dfrac{10\left(x-6\right)}{10\left(x-1\right)\left(x-6\right)}=\dfrac{\left(x-1\right)\left(x-6\right)}{10\left(x-1\right)\left(x-6\right)}\)

Suy ra: \(x^2-7x+6=10x-10-10x+60\)

\(\Leftrightarrow x^2-7x+6=50\)

\(\Leftrightarrow x^2-7x-44=0\)

\(\Leftrightarrow x^2-11x+4x-44=0\)

\(\Leftrightarrow x\left(x-11\right)+4\left(x-11\right)=0\)

\(\Leftrightarrow\left(x-11\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-11=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

Vậy: S={11;-4}

ĐKXĐ : \(x\notin\left\{1;2;...;6\right\}\)

\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+...+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{\left(x-1\right)-\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}+\dfrac{\left(x-2\right)-\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+...+\dfrac{\left(x-5\right)-\left(x-6\right)}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+...+\dfrac{1}{x-6}-\dfrac{1}{x-5}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{1}{x-6}-\dfrac{1}{x-1}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{5}{\left(x-1\right)\left(x-6\right)}=\dfrac{5}{50}\\ \Rightarrow\left(x-1\right)\left(x-6\right)=50\\ \Leftrightarrow x^2-7x-44=0\\ \Leftrightarrow\left(x-11\right)\left(x+4\right)=0\\ \Leftrightarrow\begin{matrix}x=-4\\x=11\end{matrix}\left(t.m\right)\)

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

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

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

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

\(\Leftrightarrow\frac{3}{x^2+11x+28}=\frac{1}{18}\) 

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

\(\Leftrightarrow x^2+11x=26\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-13\end{cases}}\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{x+2}+\frac{1}{x+5}+\frac{1}{x+7}-\frac{1}{x+1}-\frac{1}{x+3}-\frac{1}{x+4}-\frac{1}{x+6}=0\)

\(\Leftrightarrow\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{\left(x+4\right)\left(x+5\right)}-\frac{1}{\left(x+6\right)\left(x+7\right)}=0\)

\(\Leftrightarrow\frac{8x+20}{x\left(x+1\right)\left(x+4\right)\left(x+5\right)}+\frac{8x+36}{\left(x+2\right)\left(x+3\right)\left(x+6\right)\left(x+7\right)}=0\).Đến đây mk chịu