\(\left(x+1\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=2\)

\(\Leftrightarrow x^2+4x+3-x^2-3x+10=2\)

\(\Leftrightarrow x=-11\)

7/2 nhé bạn

    \(\frac{29}{12}:\frac{1}{2}-\frac{5}{12}:\frac{1}{2}-\frac{1}{2}\)

=  \(\left(\frac{29}{12}-\frac{5}{12}\right):\frac{1}{2}-\frac{1}{2}\)

=   \(2:\frac{1}{2}-\frac{1}{2}\)

=  \(4-\frac{1}{2}\)

=\(\frac{8}{2}-\frac{1}{2}\)

=\(\frac{7}{2}\)

Cảm ơn các bạn nhiều nhé

\(D=\dfrac{1}{3-\sqrt{7}}-\dfrac{1}{3+\sqrt{7}}\)

\(=\dfrac{3+\sqrt{7}-3+\sqrt{7}}{2}\)

\(=\sqrt{7}\)

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

Điều kiện: x≠ \(1\)

Ta có:

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

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

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

\(=3\left[\left(\dfrac{1}{x-1}\right)^2+2.\dfrac{1}{x-1}.\dfrac{1}{2}+\dfrac{1}{4}\right]+\dfrac{1}{4}\)

\(=3\left(\dfrac{1}{x-1}+\dfrac{1}{2}\right)^2+\dfrac{1}{4}\) ≥ \(\dfrac{1}{4}\) (Vì \(3\left(\dfrac{1}{x-1}+\dfrac{1}{2}\right)^2\text{≥}0\) )

Min P=\(\dfrac{1}{4}\) ⇔\(x=-1\)

1:

a: =>(|x|+4)(|x|-1)=0

=>|x|-1=0

=>x=1; x=-1

b: =>x^2-4>=0

=>x>=2 hoặc x<=-2

d: =>|2x+5|=2x-5

=>x>=5/2 và (2x+5-2x+5)(2x+5+2x-5)=0

=>x=0(loại)

\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(A=7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{69}-\frac{1}{70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(A=7\cdot\frac{3}{35}=\frac{21}{35}\)

\(A=\frac{7}{10\cdot11}+\frac{7}{11\cdot12}+\frac{7}{12\cdot13}+...+\frac{7}{69\cdot70}\)

\(A=7\left(\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+\frac{1}{12\cdot13}+...+\frac{1}{69\cdot70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)=7\cdot\frac{3}{35}=\frac{3}{5}\)

\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\)

\(B=\frac{1}{2}\left(\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+\frac{2}{29\cdot31}+...+\frac{2}{73\cdot75}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)

\(C=\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{2008\cdot2010}\)

\(C=\frac{4}{2}\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)

\(C=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(C=2\left(\frac{1}{2}-\frac{1}{2010}\right)=2\cdot\frac{502}{1005}=\frac{1004}{1005}\)

\(B=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

\(2B=\frac{2}{25.27}+\frac{2}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

\(2B=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)

\(2B=\frac{1}{25}-\frac{1}{75}=\frac{2}{75}\)

Ta có: \(A=\left(x-3\right)^2+\left(11-x\right)^2\)

\(=x^2-6x+9+x^2-22x+121\)

\(=2x^2-28x+130\)

\(=2\left(x^2-14x+49+16\right)\)

\(=2\left(x-7\right)^2+32\ge32\forall x\)

Dấu '=' xảy ra khi x=7