tk

She mentioned having seen her boyfriend at her cousin's wedding. 

She didn't mention doing homework yesterday

imbalance

Với \(n=0\) thì đpcm thành \(0⋮30\), luôn đúng.

Với \(n=1\) thì đpcm thành \(x^5-x⋮30\). Ta thấy:

\(VT=x^5-x=x\left(x^4-1\right)=x\left(x^2-1\right)\left(x^2+1\right)=x\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\)

Ta thấy \(x\left(x-1\right)\left(x+1\right)\) là tích của 3 số liên tiếp nên nó chia hết cho 6 \(\Rightarrow VT⋮6\)    (1)

 Nếu \(x⋮5\Rightarrow VT⋮5\)

 Nếu \(x\equiv\pm1\left[5\right]\) thì \(x-1\) hoặc \(x+1\) chia hết cho 5 \(\Rightarrow VT⋮5\)

 Nếu \(x\equiv\pm2\left[5\right]\) thì \(x^2+1⋮5\Rightarrow VT⋮5\)

 Vậy với mọi \(x\) thì \(VT⋮5\)    (2)

 Do \(ƯCLN\left(5,6\right)=1\) nên từ (1) và (2) \(\Rightarrow x^5-x⋮30\)

 Vậy với \(n=1\) thì khẳng định đúng.

 Giả sử khẳng định đúng đến \(n=k\ge0\). Ta cần chứng minh khẳng định đúng với \(n=k+1\)

 Với \(n=k+1\), ta có: 

 \(x^{4n+1}-x\) \(=x^{4\left(k+1\right)+1}-x\)

\(=x^{4k+5}-x\)

\(=x^4.x^{4k+1}-x^5+x^5-x\)

\(=x^4\left(x^{4k+1}-x\right)+\left(x^5-x\right)\)

Mà theo giả thiết quy nạp, \(x^{4k+1}-x⋮30\) và theo cmt thì \(x^5-x⋮30\)

\(\Rightarrow x^{4n+1}-x=x^4\left(x^{4k+1}-x\right)+\left(x^5-x\right)⋮30\). Như vậy, khẳng định đúng với \(n=k+1\).

 Theo nguyên lí quy nạp, ta có đpcm.

b) Để ý rằng phương trình của trục Ox là \(y=0\). Do đó pt hoành độ giao điểm của Ox và d là \(\left(m^2+1\right)x_A-2m=0\Leftrightarrow x_A=\dfrac{2m}{m^2+1}\)

 Mà \(OA=\left|x_A\right|=\left|\dfrac{2m}{m^2+1}\right|=\dfrac{2\left|m\right|}{m^2+1}\) , \(OA=\dfrac{4}{5}\)

\(\Rightarrow\dfrac{2\left|m\right|}{m^2+1}=\dfrac{4}{5}\) 

\(\Leftrightarrow2m^2-5\left|m\right|+2=0\)

Xét \(m\ge0\), khi đó \(2m^2-5m+2=0\Leftrightarrow\left[{}\begin{matrix}m=2\\m=\dfrac{1}{2}\end{matrix}\right.\) (nhận)

Xét \(m< 0\), khi đó \(2m^2+5m+2=0\Leftrightarrow\left[{}\begin{matrix}m=-\dfrac{1}{2}\\m=-2\end{matrix}\right.\) (nhận)

Vậy \(m\in\left\{\pm2;\pm\dfrac{1}{2}\right\}\) thỏa mãn ycbt.

c) Theo câu b), ta có \(OA=\dfrac{2\left|m\right|}{m^2+1}\). d cắt Oy tại \(B\left(0,-2m\right)\)

\(\Rightarrow OB=\left|-2m\right|=2\left|m\right|\)

Có \(OA=2OB\Leftrightarrow\dfrac{2\left|m\right|}{m^2+1}=4\left|m\right|\)

\(\Leftrightarrow\left|m\right|\left(2-\dfrac{1}{m^2+1}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=0\\2m^2+1=0\left(vôlý\right)\end{matrix}\right.\)

Vậy \(m=0\) thỏa mãn ycbt.

d) Gọi \(h\) là khoảng cách từ O đến d thì khi đó:

\(\dfrac{1}{h^2}=\dfrac{1}{OA^2}+\dfrac{1}{OB^2}\)

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

\(=\dfrac{m^4+2m^2+1}{4m^2}+\dfrac{1}{4m^2}\)

\(=\dfrac{m^4+2m^2+2}{4m^2}\)

\(\Rightarrow h^2=\dfrac{4m^2}{m^4+2m^2+2}\)

Đặt \(t=m^2\left(t>0\right)\) thì ta có \(h^2=\dfrac{4t}{t^2+2t+2}=P\)

\(\Leftrightarrow Pt^2+2\left(P-2\right)t+2P=0\)    (*)

Có \(\Delta'=\left(P-2\right)^2-2P^2=P^2-4P+4-2P^2=-P^2-4P+4\)

\(\Delta'\ge0\Leftrightarrow-2-2\sqrt{2}\le P\le-2+2\sqrt{2}\)

Ta thấy \(P=\dfrac{2P}{P}=2>0\) nên để pt đã cho có 1 nghiệm dương thì \(S>0\Leftrightarrow-2\left(P-2\right)>0\Leftrightarrow P< 2\) 

 Kết hợp 2 điều kiện, ta được \(-2-2\sqrt{2}\le P\le-2+2\sqrt{2}\)

 Vậy \(maxP=-2+2\sqrt{2}\). Dấu "=" xảy ra khi \(t=\dfrac{-2\left(-2+2\sqrt{2}-2\right)}{2\left(-2+2\sqrt{2}\right)}=\sqrt{2}\) 

\(\Leftrightarrow m^2=\sqrt{2}\Leftrightarrow m=\pm\sqrt[4]{2}\)

Vậy \(m=\pm\sqrt[4]{2}\) thỏa mãn ycbt.

Ta có: n_FeSO4 = n_CuSO4 = x (mol)

- Khi nhúng M vào FeSO₄:

\(2M+nFeSO_4\rightarrow M_2\left(SO_4\right)_n+nFe\)

2x/n_______x_________________x (mol)

⇒ m _tăng = m_Fe - m_M = 56x - 2x.M_M/n = 16 (1)

- Khi nhúng vào CuSO₄:

\(2M+nCuSO_4\rightarrow M_2\left(SO_4\right)_n+nCu\)

2x/n________x_________________x (mol)

⇒ m _tăng = m_Cu - m_M = 64x - 2x.M_M/n = 20 (2)

Từ (1) và (2) ⇒ x = 0,5 (mol)

⇒ M_M = 12n

Với n = 2 thì M_M = 24 (g/mol)

→ M là Mg.

ĐKXĐ: \(x\notin\left\{0;-20\right\}\)

\(\dfrac{100}{x}-\dfrac{100}{x+20}=\dfrac{5}{12}\)

=>\(\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{12}\)

=>\(\dfrac{20\left(x+20\right)-20x}{x\left(x+20\right)}=\dfrac{1}{12}\)

=>\(\dfrac{400}{x\left(x+20\right)}=\dfrac{1}{12}\)

=>\(x\left(x+20\right)=400\cdot12=4800\)

=>\(x^2+20x-4800=0\)

=>(x+80)(x-60)=0

=>\(\left[{}\begin{matrix}x+80=0\\x-60=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-80\left(nhận\right)\\x=60\left(nhận\right)\end{matrix}\right.\)

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne1\end{matrix}\right.\)

\(\dfrac{2}{\sqrt{x}-1}+\dfrac{2\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}+\dfrac{x-10\sqrt{x}+3}{\sqrt{x^3}-1}\)

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

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

\(=\dfrac{3x-8\sqrt{x}+5+2x-2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{5x-8\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)\left(5\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{5\sqrt{x}-3}{x+\sqrt{x}+1}\)

\(\dfrac{2}{\sqrt{x}-1}+\dfrac{2\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}+\dfrac{x-10\sqrt{x}+3}{\sqrt{x^3}-1}\left(x\ne1,x>=0\right)\\ =\dfrac{2}{\sqrt{x}-1}+\dfrac{2\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}+\dfrac{x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{2\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{2x+2\sqrt{x}+2+2\left(x-1\right)+x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{5x-8\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{5x-5\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(5\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{5\sqrt{x}-3}{x+\sqrt{x}+1}\)

Tùy nha bạn

Lên lớp 10 cái đó được tính vào phần toán đại đấy

a: \(\left\{{}\begin{matrix}x+3y=11\\3x-y=9-2y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+3y=11\\3x+y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3y=11\\9x+3y=27\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}9x+3y-x-3y=27-11\\x+3y=11\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}8x=16\\3y=11-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{11-x}{3}=\dfrac{11-2}{3}=\dfrac{9}{3}=3\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}5\left(x+2y\right)=3x-1\\2x+4=3\left(x-5y\right)-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+10y-3x=-1\\2x+4-3x+15y=-12\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+10y=-1\\-x+15y=-16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\-2x+30y=-32\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+10y-2x+30y=-1+\left(-32\right)\\x-15y=16\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}40y=-33\\x=15y+16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{33}{40}\\x=15\cdot\dfrac{-33}{40}+16=\dfrac{29}{8}\end{matrix}\right.\)

a) 

\(\left\{{}\begin{matrix}x+3y=11\\3x-y=9-2y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+3y=11\\3x+y=9\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x+9y=33\\3x+y=9\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}8y=24\\3x+y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\3x+3=9\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=3\\x=\dfrac{6}{3}=2\end{matrix}\right.\)

b) 

\(\left\{{}\begin{matrix}5\left(x+2y\right)=3x-1\\2x+4=3\left(x-5y\right)-12\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5x+10y=3x-1\\2x+4=3x-15y-12\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\x-15y=16\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\2x-30y=32\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}40y=-33\\x-15y=16\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{33}{40}\\x+\dfrac{99}{8}=16\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{33}{40}\\x=16-\dfrac{99}{8}=\dfrac{29}{8}\end{matrix}\right.\)

\(\dfrac{x+100}{4}+\dfrac{x+99}{5}=\dfrac{x+98}{6}+\dfrac{x+97}{7}\)

=>\(\left(\dfrac{x+100}{4}+1\right)+\left(\dfrac{x+99}{5}+1\right)=\left(\dfrac{x+98}{6}+1\right)+\left(\dfrac{x+97}{7}+1\right)\)

=>\(\dfrac{x+104}{4}+\dfrac{x+104}{5}=\dfrac{x+104}{6}+\dfrac{x+104}{7}\)

=>\(\left(x+104\right)\left(\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{7}\right)=0\)

=>x+104=0

=>x=-104

\(\dfrac{x+100}{4}+\dfrac{x+99}{5}=\dfrac{x+98}{6}+\dfrac{x+97}{7}\\ \dfrac{x+100}{4}+\dfrac{x+99}{5}-\dfrac{x+98}{6}-\dfrac{x+97}{7}=0\\ \left(\dfrac{x+100}{4}+1\right)+\left(\dfrac{x+99}{5}+1\right)-\left(\dfrac{x+98}{6}+1\right)-\left(\dfrac{x+97}{7}+1\right)=0\\ \dfrac{x+104}{4}+\dfrac{x+104}{5}-\dfrac{x+104}{6}-\dfrac{x+104}{7}=0\\ \left(x+104\right)\cdot\left(\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{7}\right)=0\)

Vì \(\left(\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{7}\right)\ne0\) nên:

\(x+104=0\\ x=-104\)

Vậy \(x=-104\)