\(\dfrac{1}{1-\dfrac{1}{1-2^{-1}}}+\dfrac{1}{1+\dfrac{1}{2^{-1}}}\)

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

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

Đề sai rồi nha em

Tìm 2 số tự nhiên chẵn nên tổng của 2 số này phải là 1 số chẵn, mà đề lại cho số lẻ.

Câu thứ 3, là b chứ k phải \(\sqrt{y}\) nha:)

\(\dfrac{5}{2\sqrt{5}}=\dfrac{5\sqrt{5}}{2.5}=\dfrac{\sqrt{5}}{2}\)

\(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{4+2\sqrt{2}}{5.2}=\dfrac{2+\sqrt{2}}{5}\)

\(\dfrac{y+b\sqrt{y}}{b\sqrt{y}}=\dfrac{y\sqrt{y}+by}{by}=\dfrac{\sqrt{y}+b}{b}\left(y>0;b\ne0\right)\)

Tỉ số chiều dài với chiều rộng là: \(1+\dfrac{4}{5}=\dfrac{9}{5}\)

Tổng số phần bằng nhau là:

\(9+5=14\)

Chiều dài là:

\(140:14.9=90\left(m\right)\)

Chiều rộng là:

\(140-90=50\left(m\right)\)

Diện tích là:

\(90.50=4500\left(m^2\right)\)

a) \(\sqrt{-x^2+x+4}=x-3\left(đk:x\ge3\right)\)

\(-x^2+x+4=x^2-6x+9\)

\(2x^2-7x-5=0\)

\(\Delta=49-4.2.\left(-5\right)=89\)

\(\left[{}\begin{matrix}x=\dfrac{7+\sqrt{89}}{4}\left(TM\right)\\x=\dfrac{7-\sqrt{89}}{4}\left(L\right)\end{matrix}\right.\)

b) \(\sqrt{-2x^2+6}=x-1\left(đk:x\ge1\right)\)

\(-2x^2+6=x^2-2x+1\)

\(3x^2-2x-5=0\)

\(\Delta=4+4.3.5=64\)

\(\left[{}\begin{matrix}x=\dfrac{2-8}{6}=-1\left(L\right)\\x=\dfrac{2+8}{6}=\dfrac{5}{3}\left(TM\right)\end{matrix}\right.\)

c) \(\sqrt{x+2}=1+\sqrt{x-3}\left(Đk:x\ge3\right)\)

\(x+2=1+x-3+2\sqrt{x-3}\)

\(\sqrt{x-3}=2\)

\(x-3=4\)

\(x=7\)

a) \(\left(3+\sqrt{2}\right)^2=9+6\sqrt{2}+2=11+6\sqrt{2}\)

b) \(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)

\(=\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}\)

\(=3+\sqrt{2}+3-\sqrt{2}=6\)

c) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)

\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

\(=\sqrt{7}-1-\sqrt{7}-1=-2\)

d) \(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)

\(=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)

\(=3\sqrt{5}-2-3\sqrt{5}-2=-4\)

\(B=\dfrac{y}{y-9}+\dfrac{1}{\sqrt{y}+3}+\dfrac{1}{\sqrt{y}-3}\left(Đk:y\ge0;y\ne9\right)\)

\(=\dfrac{y+\sqrt{y}-3+\sqrt{y}+3}{y-9}\)

\(=\dfrac{y+2\sqrt{y}}{y-9}\)

\(B>1\Rightarrow\dfrac{y+2\sqrt{y}}{y-9}>1\)

\(\Rightarrow\dfrac{y+2\sqrt{y}-y+9}{y-9}>0\)

\(\Leftrightarrow\dfrac{2\sqrt{y}+9}{y-9}>0\)

\(\Leftrightarrow y>9\left(2\sqrt{y}+9>0\right)\)

Gọi \(d=\left(2n+1;3n+1\right)\)

\(\Rightarrow\left\{{}\begin{matrix}2n+1⋮d\\3n+1⋮d\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}6n+3⋮d\\6n+2⋮d\end{matrix}\right.\)

\(\Rightarrow\left(6n+3\right)-\left(6n+2\right)⋮d\)

\(\Rightarrow1⋮d\Rightarrow d=1\)

\(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}+\dfrac{a-12}{a-4}\)(\(a>2\))

\(=\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)+a-12}{a-4}\)

\(=\dfrac{a+5\sqrt{a}+6-a+3\sqrt{a}-2+a-12}{a-4}\)

\(=\dfrac{a+8\sqrt{a}-8}{a-4}\)