Hỏi bài Online

a) \(n_{H_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

\(n_{Cl}=n_{HCl}=2n_{H_2}=2.0,06=0,12\left(mol\right)\)

m=m_{kim loại}+m_Cl=1,965+0,12.35,5=6,225(gam)

b) \(H_2+CuO\overset{t^0}{\rightarrow}Cu+H_2O\left(1\right)\)

\(yH_2+Fe_xO_y\overset{t^0}{\rightarrow}xFe+yH_2O\left(2\right)\)

Fe+H₂SO₄\(\rightarrow FeSO_4+H_2\left(3\right)\)

\(m_{Cu}=1,28\left(g\right)\rightarrow n_{Cu}=\dfrac{1,28}{64}=0,02\left(mol\right)\)

-Theo(1): \(n_{H_2\left(1\right)}=n_{CuO}=n_{Cu}=0,02\left(mol\right)\)

\(\rightarrow m_{CuO}=0,02.80=1,6\left(g\right)\rightarrow m_{Fe_xO_y}=3,92-1,6=2,32\left(g\right)\)

\(\rightarrow n_{H_2\left(2\right)}=n_{H_2}-n_{H_2\left(1\right)}=0,06-0,02=0,04\left(mol\right)\)

-Theo(2): \(n_{Fe_xO_y}=\dfrac{1}{y}n_{H_2}=\dfrac{0,04}{y}mol\rightarrow M_{Fe_xO_y}=\dfrac{2,32}{\dfrac{0,04}{y}}=58y\)

\(\rightarrow56x+16y=58y\rightarrow56x=42y\rightarrow\dfrac{x}{y}=\dfrac{42}{56}=\dfrac{3}{4}\)

\(\rightarrow Fe_3O_4\)

%CuO=\(\dfrac{1,6}{3,92}.100\approx40,8\%\)

%Fe₃O₄=100%-40,8%=59,2%

Đề sai sửa lại là:

\(x=\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\dfrac{125}{27}}}\)

\(\Leftrightarrow x=\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}+\sqrt[3]{3-\sqrt{9+\dfrac{125}{27}}}\)

\(\Leftrightarrow x^3=3+\sqrt{9+\dfrac{125}{27}}+3-\sqrt{9+\dfrac{125}{27}}+3.\left(\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}+\sqrt[3]{3-\sqrt{9+\dfrac{125}{27}}}\right)\left(\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}.\sqrt[3]{3-\sqrt{9+\dfrac{125}{27}}}\right)\)

\(\Leftrightarrow x^3=6+3x.\left(\dfrac{-5}{3}\right)\)

\(\Leftrightarrow x^3+5x-6=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0\)

\(\Leftrightarrow x=1\)

Vậy x là số nguyên

Lời giải:

Xét modulo $3$ cho $n$ thôi . Ở đây mình xét cụ thể TH $n=3k$. TH \(n=3k+1,3k+2\) ta hoàn toàn làm tương tự

TH1: \(n=3k\)

Ta có :

\(11^3\equiv 1\pmod 7\Rightarrow 11^n=11^{3k}\equiv 1\pmod 7\Rightarrow 11^{n+2}\equiv 11^2\equiv 2\pmod 7\)

\(12^6\equiv 1\pmod 7\Rightarrow 12^{2n}=12^{6k}\equiv 1\pmod 7\Rightarrow 12^{2n+1}\equiv 12\pmod 7\)

\(\Rightarrow 11^{n+2}+12^{2n+1}\equiv 14\equiv 0\pmod 7\) $(1)$

Lại có:

\(11^3\equiv 1\pmod {19}\Rightarrow 11^n=11^{3k}\equiv 1\pmod {19}\Rightarrow 11^{n+2}\equiv 7\pmod {19}\)

\(12^6\equiv 1\pmod {19}\Rightarrow 12^{2n}=12^{6k}\equiv 1\pmod {19}\Rightarrow 12^{2n+1}\equiv 12\pmod {19}\)

\(\Rightarrow 11^{n+2}+12^{2n+1}\equiv 19\equiv 0\pmod {19}\) $(2)$

Từ \((1),(2)\) kết hợp với \((7,19)=1\) suy ra \(11^{n+2}+12^{2n+1}\vdots (7.19=133)\) (đpcm)

Xét modulo $3$ cho $n$ thôi . Ở đây mình xét cụ thể TH $n=3k$. TH $n=3k+1,3k+2$ ta hoàn toàn làm tương tự

TH1: $n=3k$

Ta có :

${11}^3\equiv 1\left(\mathrm{mod}7\right)\Rightarrow {11}^n={11}^{3k}\equiv 1\left(\mathrm{mod}7\right)\Rightarrow {11}^{n+2}\equiv {11}^2\equiv 2\left(\mathrm{mod}7\right)$

${12}^6\equiv 1\left(\mathrm{mod}7\right)\Rightarrow {12}^{2n}={12}^{6k}\equiv 1\left(\mathrm{mod}7\right)\Rightarrow {12}^{2n+1}\equiv 12\left(\mathrm{mod}7\right)$

$\Rightarrow {11}^{n+2}+{12}^{2n+1}\equiv 14\equiv 0\left(\mathrm{mod}7\right)$ $\left(1\right)$

Lại có:

${11}^3\equiv 1\left(\mathrm{mod}19\right)\Rightarrow {11}^n={11}^{3k}\equiv 1\left(\mathrm{mod}19\right)\Rightarrow {11}^{n+2}\equiv 7\left(\mathrm{mod}19\right)$

${12}^6\equiv 1\left(\mathrm{mod}19\right)\Rightarrow {12}^{2n}={12}^{6k}\equiv 1\left(\mathrm{mod}19\right)\Rightarrow {12}^{2n+1}\equiv 12\left(\mathrm{mod}19\right)$

$\Rightarrow {11}^{n+2}+{12}^{2n+1}\equiv 19\equiv 0\left(\mathrm{mod}19\right)$ $\left(2\right)$

Từ $\left(1\right),\left(2\right)$ kết hợp với $(7,19)=1$ suy ra ${11}^{n+2}+{12}^{2n+1}⋮(7.19=133)$ (đpcm)

\(A=5^{n+2}+26.5^n+8^{2n+1}\)

\(A=5^n\left(5^2+26\right)+\left(8^2\right)^n.8\)

\(A=5^n.51+64^n.8\)

\(A=5^n.59-5^n.8+64^n.8\)

\(A=5^n.59+8.\left(-5^n+64^n\right)\)

Ta có: \(\left(5^n.59\right)⋮59\left(1\right)\)

mà \(\left(-5^n+64^n\right)\) luôn chia hết cho \(\left(-5+64\right)=59\Leftrightarrow8.\left(-5^n+64^n\right)⋮59\left(2\right)\)

Từ (1)(2)⇒ A\(⋮\)59

a) \(\left(x^2+x\right)^2+4x^2+4x-12\)

\(=x^4+2x^3+x^2+4x^2+4x-12\)

\(=x^4+2x^3+5x^2+4x-12\)

\(=\left(x^4+2x^3\right)+\left(5x^2+10x\right)-\left(6x+12\right)\)

\(=x^3\left(x+2\right)+5x\left(x+2\right)-6\left(x+2\right)\)

\(=\left(x+2\right)\left(x^3+5x-6\right)\)

\(=\left(x+2\right)\left(x^3-x+6x-6\right)\)

\(=\left(x+2\right)\left[x^2\left(x-1\right)+6\left(x-1\right)\right]\)

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

b) \(x^3+3x^2-4\)

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

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

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

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

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

\(x^3+3x^2-4=x^3-1+3x^2-3\)

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

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

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

a) Ta có:

\(2x^3-5x^2+8x-3\)

\(=2x^3-x^2-4x^2+2x+6x-3\)

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

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

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

b) Ta có:

\(\left(x^2+x\right)^2+4x^2+4x-12\)

\(=x^4+2x^3+x^2+4x^2+4x-12\)

\(=x^4+2x^3+5x^2+4x-12\)

\(=x^4+2x^3+5x^2+10x-6x-12\)

\(=x^3\left(x+2\right)+5x\left(x+2\right)-6\left(x+2\right)\)

\(=\left(x+2\right)\left(x^3+5x-6\right)\)

\(=\left(x+2\right)\left(x^3-x^2+x^2-x+6x-6\right)\)

\(=\left(x+2\right)\left[x^2\left(x-1\right)+x\left(x-1\right)+6\left(x-1\right)\right]\)

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

3,5Ω

Bài 2:

Gọi S1 ,S2 lần lượt là tiết diện của dây ni kê lin và dây constantan .

Theo bài ra ta có R =l1.p1/S1 =l2.p2/S2

<=> \(\dfrac{10.0,4.10^{-6}}{S1}\)=\(\dfrac{32.0,5.10^{-6}}{S2}\)

<=> 1/S1=4/S2 => S2=4S1

Mà S =r^2. pi

=>r1^2/r2^2=S1/S2=1/4

=>r1/r2=\(\sqrt{\dfrac{1}{4}}\)=\(\dfrac{1}{2}\)

Vậy .....

Chúc bạn học tốt !

- Hình a : tiếng nói

- Hình b : chữ viết

- Hình c : cử chỉ

- Hình d : biển cấm

\(B=\left(x^2+x\right)^2+4x^2+4x-12\)

\(=\left(x^2+x\right)^2+4\left(x^2+x\right)-12\)

Đặt \(x^2+x=t\), ta có:

\(B=t^2+4t-12\)

\(=t^2+6t-2t-12\)

\(=t\left(t+6\right)-2\left(t+6\right)\)

\(=\left(t+6\right)\left(t-2\right)\)

\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)

Chúc bạn học tốt!

sorry thiếu dầu =

A=\(\left[\dfrac{\dfrac{42}{31}.\dfrac{31}{7}-\left(15-\dfrac{2}{3}\right)}{\dfrac{29}{6}+\dfrac{1}{6}.\dfrac{20}{3}}\right].\dfrac{31}{50}\)

= \(\left(\dfrac{6-\dfrac{43}{3}}{\dfrac{29}{6}+\dfrac{10}{9}}\right).\dfrac{31}{50}\)=\(\left(\dfrac{\dfrac{-25}{3}}{\dfrac{107}{18}}\right).\dfrac{31}{50}\)=\(\dfrac{-150}{107}.\dfrac{31}{50}\)=\(\dfrac{-93}{107}\)

mình vẽ hơi xấu xin bạn bỏ qua

mình vẽ hơi xấu bạn thông cảm nha