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Đặt z = x3y5
=> z + 3z + 5z + ... + (2k - 1).z = 3249 . z
=> z . (1 + 3 + 5 + ... + 2k - 1) = z . 3249
=> 1 + 3 + 5 + ... + 2k - 1 = 3249
Số số hạng (vế trái) là:
(2k - 1 - 1) : 2 + 1 = (2k - 2) : 2 + 1 = 2.(k - 1) : 2 + 1 = k - 1 + 1 = k
=> (2k - 1 + 1) . k : 2 = 3249
=> 2k2 = 3249 . 2
=> k2 = 3249
=> k2 = 572 = (-57)2
Mà k thuộc N => k = 57.
x3y5+3x3y5+5x3y5+...+(2k-1)x3y5 =3249x3y5
x3y5.[1+3+5+...+(2k-1)]=3249x3y5
=>1+3+5+...+(2k-1)=3249
\(\frac{\left(2k-1+1\right).\left[\left(2k-1-1\right):2\right]}{2}=3249\)
\(\frac{2k.\left[\left(2k-2\right):2+1\right]}{2}=3249\)
\(\frac{2k.\left(k-1+1\right)}{2}=3249\)
\(k^2=3249\)
\(k=57\)
c: \(=\dfrac{7}{23}\cdot\left(\dfrac{-4}{3}-\dfrac{5}{2}\right)=\dfrac{7}{23}\cdot\dfrac{-8-15}{6}\)
\(=\dfrac{7}{23}\cdot\dfrac{-23}{6}=-\dfrac{7}{6}\)
d: \(=\dfrac{5}{7}\left(23+\dfrac{1}{4}-13-\dfrac{1}{4}\right)=\dfrac{5}{7}\cdot10=\dfrac{50}{7}\)
e: \(=\dfrac{2^5\cdot3^3\cdot5^3}{2^3\cdot3^3\cdot2^2\cdot5^2}=5\)
i: \(=\dfrac{1}{3^{10}}\cdot3^{50}-\dfrac{2^{10}}{3^{10}}:\dfrac{4^5}{3^{10}}\)
\(=3^{40}-1\)
c: \(=\dfrac{7}{23}\cdot\left(\dfrac{-4}{3}-\dfrac{5}{2}\right)=\dfrac{7}{23}\cdot\dfrac{-8-15}{6}\)
\(=\dfrac{7}{23}\cdot\dfrac{-23}{6}=-\dfrac{7}{6}\)
d: \(=\dfrac{5}{7}\left(23+\dfrac{1}{4}-13-\dfrac{1}{4}\right)=\dfrac{5}{7}\cdot10=\dfrac{50}{7}\)
e: \(=\dfrac{2^5\cdot3^3\cdot5^3}{2^3\cdot3^3\cdot2^2\cdot5^2}=5\)
i: \(=\dfrac{1}{3^{10}}\cdot3^{50}-\dfrac{2^{10}}{3^{10}}:\dfrac{4^5}{3^{10}}\)
\(=3^{40}-1\)
1.
\(-3x^5y^4+3x^2y^3-7x^2y^3+5x^5y^4\)
\(=(-3x^5y^4+5x^5y^4)+(3x^2y^3-7x^2y^3)\)
\(=2x^5y^4-4x^2y^3\)
2.
\(\frac{1}{2}x^4y-\frac{3}{2}x^3y^4+\frac{5}{3}x^4y-x^3y^4\)
\(=(\frac{1}{2}x^4y+\frac{5}{3}x^4y)-(\frac{3}{2}x^3y^4+x^3y^4)\)
\(=\frac{13}{6}x^4y-\frac{5}{2}x^3y^4\)
3.
\(5x-7xy^2+3x-\frac{1}{2}xy^2\)
\(=(5x+3x)-(7xy^2+\frac{1}{2}xy^2)\)
\(=8x-\frac{15}{2}xy^2\)
4.
\(\frac{-1}{5}x^4y^3+\frac{3}{4}x^2y-\frac{1}{2}x^2y+x^4y^3\)
\(=(\frac{-1}{5}x^4y^3+x^4y^3)+(\frac{3}{4}x^2y-\frac{1}{2}x^2y)\)
\(=\frac{4}{5}x^4y^3+\frac{1}{4}x^2y\)
5.
\(\frac{7}{4}x^5y^7-\frac{3}{2}x^2y^6+\frac{1}{5}x^5y^7+\frac{2}{3}x^2y^6\)
\(=(\frac{7}{4}x^5y^7+\frac{1}{5}x^5y^7)+(-\frac{3}{2}x^2y^6+\frac{2}{3}x^2y^6)\)
\(=\frac{39}{20}x^5y^7-\frac{5}{6}x^2y^6\)
6.
\(\frac{1}{3}x^2y^5(-\frac{3}{5}x^3y)+x^5y^6=(\frac{1}{3}.\frac{-3}{5})(x^2.x^3)(y^5.y)+x^5y^6\)
\(=\frac{-1}{5}x^5y^6+x^5y^6=\frac{4}{5}x^5y^6\)
Đặt z = x3y5
=> z + 3z + 5z + ... + (2k - 1).z = 3249 . z
=> z . (1 + 3 + 5 + ... + 2k - 1) = z . 3249
=> 1 + 3 + 5 + ... + 2k - 1 = 3249
Số số hạng (vế trái) là:
(2k - 1 - 1) : 2 + 1 = (2k - 2) : 2 + 1 = 2.(k - 1) : 2 + 1 = k - 1 + 1 = k
=> (2k - 1 + 1) . k : 2 = 3249
=> 2k2 = 3249 . 2
=> k2 = 3249
=> k2 = 572 = (-57)2
Mà k thuộc N => k = 57.
\(x^3y^5+3x^3y^5+5x^3y^5+...+\left(2k-1\right)x^3y^5=3249x^3y^5\)
\(x^3y^5\left(1+3+5+..+2k-1\right)=3249x^3y^5\)
\(1+3+5+...+\left(2k-1\right)=3249\)
Số số hạng:
(2k - 1 - 1) : 2 + 1 = (2k - 2) : 2 + 1 = 2(k - 1) : 2 + 1 = k - 1 + 1 = k (số hạng)
Tổng trên là:
\(\frac{k\left(2k-1+1\right)}{2}=\frac{2k^2}{2}=k^2\)
=> k2 = 3249
=> k = 57