a) \(P=\dfrac{3m+\sqrt{9m}-3}{m+\sqrt{m}-2}-\dfrac{\sqrt{m}-2}{\sqrt{m}-1}+\dfrac{1}{\sqrt{m}+2}-1\)

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

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

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

b) Đk: \(\left\{{}\begin{matrix}m\ge0\\m\ne1\end{matrix}\right.\)

\(\left|P\right|=2\Leftrightarrow\left|\dfrac{\sqrt{m}+1}{\sqrt{m}-1}\right|=2\Leftrightarrow\left[{}\begin{matrix}\dfrac{\sqrt{m}+1}{\sqrt{m}-1}=-2\\\dfrac{\sqrt{m}+1}{\sqrt{m}-1}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{m}+1=-2\sqrt{m}+2\\\sqrt{m}+1=2\sqrt{m}-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3\sqrt{m}=1\\\sqrt{m}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{1}{9}\left(N\right)\\m=9\left(N\right)\end{matrix}\right.\)

c) \(P=\dfrac{\sqrt{m}+1}{\sqrt{m}-1}=\dfrac{\sqrt{m}-1+2}{\sqrt{m}-1}=1+\dfrac{2}{\sqrt{m}-1}\)

\(P\in N\Rightarrow\dfrac{2}{\sqrt{m}-1}\in Z\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{m}-1=-2\\\sqrt{m}-1=-1\\\sqrt{m}-1=1\\\sqrt{m}-1=2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{m}=-1\left(VN\right)\\\sqrt{m}=0\left(1\right)\\\sqrt{m}=2\left(VN,m\ne N\right)\\\sqrt{m}=3\left(2\right)\end{matrix}\right.\)

(1) \(\Leftrightarrow m=0\left(loại,P\notin N\right)\)

(2) \(\Leftrightarrow m=9\left(N\right)\)

Kl: a) \(P=\dfrac{\sqrt{m}+1}{\sqrt{m}-1}\)

b) m=1/9 , m = 9

c) m =9

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

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

\(=1+\dfrac{1}{x}-\dfrac{1}{x+1}\)

\(\Rightarrow f\left(1\right).f\left(2\right)...f\left(2020\right)=5^{1+1-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+...+1+\dfrac{1}{2020}-\dfrac{1}{2021}}\)

\(=5^{2021-\dfrac{1}{2021}}\)

\(\Rightarrow\dfrac{m}{n}=2021-\dfrac{1}{2021}=\dfrac{2021^2-1}{2021}\)

\(\Rightarrow m-n^2=2021^2-1-2021^2=-1\)

x tiến đến đâu bạn, điều kiện của m và n nữa, mình nghĩ m,n>=2 mới hợp lý

gúp mình với nha

Câu hỏi của Linh Suzu - Toán lớp 7 | Học trực tuyến, nhớ tìm trước khi hỏi, lần sau t ko tìm đâu

\(\lim\limits_{x\rightarrow0}f\left(x\right)=\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{x+1}-x-2}{x^2}=\lim\limits_{x\rightarrow0}\dfrac{\left(2\sqrt{x+1}\right)^2-\left(x+2\right)^2}{x^2\left(2\sqrt{x+1}+x+2\right)}=\lim\limits_{x\rightarrow0}\dfrac{4x+4-x^2-4x-4}{x^2\left(2\sqrt{x+1}+x+2\right)}=\lim\limits_{x\rightarrow0}\dfrac{-1}{2\sqrt{x+1}+x+2}=-\dfrac{1}{4}\)

\(f\left(0\right)=2-9m\)

De ham so lien tuc tai x=0

\(\Rightarrow f\left(0\right)=\lim\limits_{x\rightarrow0}f\left(x\right)\Leftrightarrow2-9m=-\dfrac{1}{4}\Rightarrow m=\dfrac{1}{4}\)

Giả sử trong 4 số a;b;c;d không tồn tại 2 số bằng nhau

Không mất tính tổng quát ta giả sử a < b < c < d

=> a² < b² < c² < d² (do a;b;c;d nguyên dương)

=> \(\frac{1}{a^2}>\frac{1}{b^2}>\frac{1}{c^2}>\frac{1}{d^2}\)

\(\Rightarrow\frac{4}{a^2}>\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}=1\)

=> a² < 4

=> a < 2 (1)

Lại có: \(\frac{1}{a^2}\)< 1 (theo đê bai)

=> a² > 1

=> a > 1 (do a nguyên dương) (2)

Từ (1) và (2) => 1 < a < 2, mâu thuẫn với đề là a nguyên dương

Như vậy trong 4 số đã cho luôn tồn tại ít nhất 2 số bằng nhau (đpcm)