Giải dùm mình nha mn
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Bài 17:
a: Xét ΔABC có \(BC^2=AB^2+AC^2\)
nên ΔABC vuông tại A
d: Xét tứ giác AEDF có
\(\widehat{AED}=\widehat{AFD}=\widehat{FAE}=90^0\)
nên AEDF là hình chữ nhật
mà AD là tia phân giác của \(\widehat{FAE}\)
nên AEDF là hình vuông
PTHH: \(Zn+S\underrightarrow{t^o}ZnS\)
Từ đề bài, ta có: \(\left\{{}\begin{matrix}n_S=n_{ZnS}=0,15\left(mol\right)\\n_{Zn\left(dư\right)}=0,05\left(mol\right)\end{matrix}\right.\)
c) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
\(ZnS+2HCl\rightarrow ZnCl_2+H_2S\uparrow\)
Theo PTHH: \(\Sigma n_{HCl}=2n_{ZnS}+2n_{Zn\left(dư\right)}=0,4\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{0,4\cdot36,5}{20\%}=73\left(g\right)\)
Ta có : (x - 1)5 - 1 = 31
=> (x - 1)5 = 31 + 1
=> (x - 1)5 = 32
=> (x - 1)5 = 25
=> x - 1 = 2
=> x = 3
a: Xét ΔADC và ΔEDB có
DA=DE
\(\widehat{ADC}=\widehat{EDB}\)
DC=DB
Do đó: ΔADC=ΔEDB
Giải
Ta gọi T = (1^2+2^2+...+2005^2)-(1.3+2.4+3.5+...+2004.2006)
Đặt A = 1^2+2^2+3^2+...+2005^2
=> A = 1.1 + 2.2 +3.3 +...+ 2005.2005
=> A = 1.(2-1) + 2.(3-1) + 3.(4-1) +...+ 2005.(2006-1)
==> A = 1.2-1.1 + 2.3-1.2 + 3.4-1.3+...+2005.2006-1.2005
=> A = (1.2+2.3+3.4+...+2005.2006)-(1+2+3+...+2005)
Xét 1.2 +2.3+3.4+...+2005.2006
= 1/3.(1.2.3+2.3.3+...+2005.2006.3)
=1/3.[1.2.(3-0)+2.3.(4-1)+...+2005.2006.(2007-2004)]
=1/3.(1.2.3+2.3.4-1.2.3+...+2005.2006.2007-2004.2005.2006)
= 1/3 . 2005.2006.2007
= 2005.2006.2007/3 = 2690738070
Vậy A= 2690738070 - (1+3+5+...+2005)
=> A= 2690738070- [(2005-1):2+1].(2005+1)/2
=> A = 2690738070 - 1006009
=> A = 2689732061
Đắt B = 1.3+2.4+3.5+4.6+...+2003.2005 +2004.2006
=> B= (1.3+3.5+...+2003.2005)+(2.4+4.6+...+2004.2006)
=> 6B = (1.3.6+3.5.6+...+2003.2005.6)+(2.4.6+4.6.6+...+2004.2006.6)
=> 6B = [1.3.(5+1)+3.5.(7-1)+...+2003.2005.(2007-2001)] + [2.4.(6-0)+4.6.(8-2)+...+2004.2006.(2008-2002)]
=> 6B = (1.3.5+1.3.1+3.5.7-1.3.5+...+2003.2005.2007-2001.2003.2005)+(2.4.6+4.6.8-2.4.6+...+2004.2006.2008-2002.2004.2006)
=> 6B = 1.3.1+2003.2005.2007 + 2004.2006.2008
=> 6B = 16132350300
=> B = 16132350300/6 = 2688725050
Vì T = A - B = 2689732061-2688725050
=> T = 1007011
Ta có : (x - 1)5 - 1 = 36
=> (x - 1)5 = 37
=> (x - 1) + 5 ko thỏa mãn
mình sửa lại đề nha: (x-1)5 - 1 = 36
Mình xin lỗi mn nhìu lắm
a, \(4x^3-5x^2y-xy=x\left(4x^2-5xy-y\right)\)
b, \(x^3y^4\left(x^2-y^3\right)-x^3y^3\left(x^4-y^4\right)\)
\(=x^3y^3\left[y\left(x^2-y^3\right)-x^4+y^4\right]\)
\(=x^3y^3\left(x^2y-x^4\right)=x^5y^3\left(y-x^2\right)\)
N.
2. Mexico City is smaller than Tokyo, but New York is the smallest.
3. New York is wetter than Mexico City, but Tokyo is the wettest.
4. New York is drier than Tokyo, but Mexico City is the driest.
5. Tokyo is cheaper than New York, but Mexico City is the cheapest.
6. Tokyo is more expensive than Mexico City, but New York is the most expensive.
7. Mexico City is hotter than New York, but Tokyo is the hottest.
8. Mexico City is colder than Tokyo, but New York is the coldest.
9. New York is cleaner than Mexico City, but Tokyo is the cleanest.
10. New York is more polluted than Tokyo, but Mexico City is the most polluted.
O.
2. London is as warm as Edinburg.
3. London is not as cloudy as Edinburg.
4. Edinburg isn't as bright as London.
5. London isn't as wet as Edinburg.
Em cám ơn nhiều ạ