dài vãi

\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{13.15}\right).x=\dfrac{-26}{45}\\ \Leftrightarrow\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{13.15}\right).x=\dfrac{-52}{45}\\ \Leftrightarrow\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{13}-\dfrac{1}{15}\right).x=\dfrac{-52}{45}\\ \Leftrightarrow\left(1-\dfrac{1}{15}\right).x=\dfrac{-52}{45}\\ \Leftrightarrow\dfrac{14}{15}.x=\dfrac{-52}{45}\\ \Leftrightarrow x=-\dfrac{26}{21}\)

(11.3+13.5+...+113.15).x=−2645⇔(21.3+23.5+...+213.15).x=−5245⇔(1−13+13−15+...+113−115).x=−5245⇔(1−115).x=−5245⇔1415.x=−5245⇔x=−2621

3. Tóm tắt:

\(m_1=1,5kg\)

\(t_1=100^oC\)

\(m_2=200g=0,2kg\)

\(m_3=2,5kg\)

\(t_2=20^oC\)

\(c_1=380J/kg.K\)

\(c_2=880J/kg.K\)

\(c_3=4200J/kg.K\)

===========

a) \(t=?^oC\)

b) \(t'=70^oC\)

\(Q'=?J\)

Giải:

a) Nhiệt độ cuối cùng của nước:

\(Q_1=Q_2+Q_3\)

\(\Leftrightarrow m_1.c_1.\left(t_1-t\right)=\left(t-t_2\right)\left(m_2.c_2+m_3.c_3\right)\)

\(\Leftrightarrow1,5.380.\left(100-t\right)=\left(t-20\right)\left(0,2.880+2,5.4200\right)\)

\(\Leftrightarrow57000-570t=10676t-213520\)

\(\Leftrightarrow57000+213520=10676t+570t\)

\(\Leftrightarrow270520=11246t\)

\(\Leftrightarrow t=\dfrac{270520}{11246}=24,05^oC\)

b) Nhiệt lượng cần cung cấp để đun nóng nước:

\(Q'=\left(m_1.c_1+m_2.c_2+m_3.c_3\right).\left(t'-t\right)\)

\(\Leftrightarrow Q'=\left(1,5.880+0,2.880+2,5.4200\right)\left(70-24,05\right)\)

\(\Leftrightarrow Q'=11246.45,95\)

\(\Leftrightarrow Q'=516753,7J\)

1. Tóm tắt:

\(m=300kg\)

\(\Rightarrow P=10m=3000N\)

\(h=1,5m\)

\(s=6m\)

\(F=1250N\)

==========

a) \(A_i=?J\)

b) \(A_{tp}=?J\)

c) \(H=?\%\)

\(F_{ms}=?N\)

Giải:

a) Công có ích thực hiện được:

\(A_i=P.h=3000.1,5=4500J\)

b) Công toàn phần thực hiện được:

\(A_{tp}=F.s=1250.6=7500J\)

c) Hiệu suất mặt phẳng nghiêng:

\(H=\dfrac{A_i}{A_{tp}}.100\%=\dfrac{4500}{7500}.100\%=60\%\)

Công của lực ma sát:

\(A_{ms}=A_{tp}-A_i=7500-4500=3000J\)

Độ lớn của lực ma sát:

\(A_{ms}=F_{ms}.s\Rightarrow F_{ms}=\dfrac{A_{ms}}{s}=\dfrac{3000}{6}=500N\)

2. Tóm tắt:

\(\text{℘}=1500W\)

\(m=120kg\)

\(\Rightarrow P=10m=1200N\)

\(h=16m\)

\(t=20s\)

===========

a) \(A_{tp}=?J\)

b) \(H=?J\)

Giải:

a) Công thực hiện được trong thời gian nâng vật:

\(\text{℘}=\dfrac{A_{tp}}{t}\Rightarrow A_{tp}=\text{℘}.t=1500.20=30000J\)

b) Công có ích thực hiện được:

\(A_t=P.h=1200.16=19200J\)

Hiệu suất của máy:

\(H=\dfrac{A_i}{A_{tp}}.100\%=\dfrac{19200}{30000}.100\%=64\%\)

\(2xR+yO_2\underrightarrow{^{^{t^0}}}2R_xO_y\)

\(2KMnO_4+16HCl_{\left(đ\right)}\underrightarrow{^{^{t^0}}}2KCl+2MnCl_2+5Cl_2+8H_2O\)

\(C_nH_{2n+2}+\dfrac{3n+1}{2}O_2\underrightarrow{^{^{t^0}}}nCO_2+\left(n+1\right)H_2O\)

\(8Al+30HNO_3\rightarrow8Al\left(NO_3\right)_3+3N_2O+15H_2O\)

Bài nào ạ. Ảnh bị lỗi.

\(a,A=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ A=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\\ b,A=\dfrac{2\left(\sqrt{x}+1\right)-3}{\sqrt{x}+1}=2-\dfrac{3}{\sqrt{x}+1}\in Z\\ \Leftrightarrow\sqrt{x}+1\inƯ\left(3\right)=\left\{1;3\right\}\left(\sqrt{x}+1\ge1\right)\\ \Leftrightarrow\sqrt{x}\in\left\{0;2\right\}\\ \Leftrightarrow x\in\left\{0;4\right\}\left(tm\right)\)

a) \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)

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

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

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

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

\(\Rightarrow A=\dfrac{\left(2x-2\sqrt{x}\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Rightarrow A=\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Rightarrow A=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Rightarrow A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)