Kq đúng rồi nhưng cách trình bày sao cho hợp lý ấy
\(\Leftrightarrow\)6354=a65+6300-63a
\(\Leftrightarrow\)2a=54
\(\Leftrightarrow\)a=27
vậy a=27
Cho đa thức \(P\left(x\right)=x^4+ãx^3+bx^2+cx+d\)có P(1)=7, P(2)=28,P(3)=63 TÍnh\(A=\frac{P\left(100\right)+P\left(-96\right)}{8}\)
Hóa :v
\(mHCl_{\left(dd\right)}=\frac{a\cdot C}{100}\Leftrightarrow nHCl_{\left(dd\right)}=\frac{a\cdot C}{100\cdot36,5}=\frac{a\cdot C}{3650}\)
\(\Leftrightarrow mH2O_{\left(dd\right)}=mdd-mHCl_{\left(dd\right)}=a-\frac{a\cdot C}{100}=\frac{\left(100-C\right)a}{100}\)
\(nH2=\frac{0,05a}{2}=0,025a\)
PT: 2Na + 2HCl ---> 2NaCl +H2 (1)
PT: Fe +2HCl ---->FeCl2+H2 (2)
PT: Na dư +H2O ----> NaPH +2 (3)
Theo (2) (3) => nH2=nHCl/2\(=\frac{a\cdot C}{3650\cdot2}=\frac{a\cdot C}{7300}\) (*)
Theo (3)(*) => \(\frac{\left(100-C\right)a}{1800}+\frac{aC}{7300}=nH2=0,025a\)
\(\Leftrightarrow\frac{\left(100-C\right)}{1800}+\frac{C}{7300}=0,025\)
=> C=73 (chẵn số)
=> C%= 73%
\(A=1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\)
\(\frac{A}{2}=\frac{1}{2}+\frac{3}{2^4}+\frac{4}{2^5}+....+\frac{100}{2^{101}}\)\(A-\frac{A}{2}=\left(1+\frac{3}{2^3}+....+\frac{100}{2^{100}}\right)-\left(\frac{1}{2}+\frac{3}{2^4}+.....+\frac{100}{2^{101}}\right)\)
\(\frac{A}{2}=\frac{1}{2}+\frac{3}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+....+\frac{1}{2^{100}}-\frac{100}{2^{101}}\)
\(\frac{A}{2}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+....+\frac{1}{2^{100}}-\frac{1}{2^{101}}\)
\(\frac{A}{2}=\left(1-\left(\frac{1}{2}\right)^{101}\right).2-\frac{100}{2^{101}}\)
\(\frac{A}{2}=\frac{2^{101}-1}{2^{100}}-\frac{100}{2^{101}}\)
\(A=\frac{2^{101}-1}{2^{99}}-\frac{100}{2^{100}}\)
tính
\(\left(\frac{1}{100}+\frac{99}{2}+\frac{98}{3}+...+100\right)\div\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-2\)
\(\left(100+\frac{99}{2}+\frac{98}{3}+...+\frac{1}{100}\right):\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-2\)
câu 1: giải hệ phương trình
\(\left(x+y\right)^2+\left(y+z\right)^4+....+\left(x+z\right)^{100}=-\left(y+z+x\right)\)
\(\left(xy\right)^2+2\left(yz\right)^4+....+100\left(zx\right)^{100}=-[\left(x+y+z\right)+2\left(yz+zx+xy\right)+......+99\left(x+y+z\right)]\)\(\left(\frac{1}{x}+\frac{1}{y}\right)^2+\left(\frac{1}{y^2}+\frac{1}{z^2}\right)^2+...+\left(\frac{1}{x^{99}}+\frac{1}{z^{99}}\right)^2=-\frac{1}{\left(xy\right)^2+2\left(yz\right)^2+.....+99\left(zx\right)^2}\)
tìm x,y,z
Tính \(\left(100+\frac{99}{2}+\frac{98}{3}+...+\frac{1}{100}\right)\div\left(\frac{1}{2}+\frac{1}{3}+..+\frac{1}{101}\right)-2\)
Giai hpt \(\hept{\begin{cases}\frac{101}{x}+\frac{100}{y}=A-100\\x+\frac{7200y}{101}=A\end{cases}}\left(A\inℕ^∗\right)\)
Tính \(\left(100+\frac{99}{2}+\frac{98}{3}+...+\frac{1}{100}\right):\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-2\)
Lời giải rõ ràng nha
