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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.
\(\frac{1}{3}x^2y^5(-\frac{3}{5}x^3y)+x^5y^6=(\frac{1}{3}.\frac{-3}{5})(x^2.x^3)(y^5.y)+x^5y^6\)
\(=\frac{-1}{5}x^5y^6+x^5y^6=\frac{4}{5}x^5y^6\)
a) \(\left(1-\frac{2}{5}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{2}{9}\right)...\left(1-\frac{2}{99}\right)\)
\(=\frac{3}{5}.\frac{5}{7}.\frac{7}{9}...\frac{97}{99}\)
\(=\frac{3}{99}=\frac{1}{33}\)
b) Ta có: 2x = 8y+1 = (23)y+1 = 23y+3
=> x = 3y + 3 (1)
9y = 3x-9
=> (32)y = 3x-9
=> 32y = 3x-9
=> 2y = x - 9 (2)
Từ (1) và (2) => x + 2y = 3y + 3 + x - 9
=> x + y = 2y + x - 6
A=\(\left(-\frac{5}{4}\times\frac{2}{5}\right)\times\left(x^3x^2x^3\right)\times\left(yy^4\right)\)
A=\(-\frac{1}{2}x^8y^5\)
B=\(\left(-\frac{3}{4}\times-\frac{8}{9}\right)\times\left(x^5xx^2\right)\times\left(y^4y^2y^5\right)\)
B=\(\frac{2}{3}x^8y^{11}\)
2x = 8y+1 <=> 2x = ( 23 )y+1 = 23y+3
=> x = 3y + 3 (1)
9y = 3x-9 <=> 32.y = 3x-9
=> 2y = x - 9 => x = 2y + 9 (2)
Từ (1); (2) => 3y + 3 = 2y + 9
<=> 3y - 2y = 9 - 3=> y = 6
=> 2.6 = x - 9 <=> 12 = x - 9 => x = 21
=> x + y = 21 + 6 = 27
2) Theo đề được: \(\frac{3x}{15}=\frac{4y}{28}=\frac{2z}{18}=\frac{5x}{25}=\frac{3y}{21}\)
Áp dụng t/c dãy tỉ số = nhau được:
\(\frac{3x}{15}=\frac{4y}{28}=\frac{2z}{18}=\frac{3y}{21}=\frac{5x}{25}=\frac{3x-4y}{15-28}=\frac{3x-4y}{-13}\)
và \(\frac{3x}{15}=\frac{4y}{28}=\frac{2z}{18}=\frac{3y}{21}=\frac{5x}{25}=\frac{2z+3y-5x}{18+21-25}=\frac{2z+3y-5x}{14}\)
Vì \(\frac{3x-4y}{-13}=\frac{2z+3y-5x}{14}\) nên \(\frac{3x-4y}{2z+3y-5x}=\frac{-13}{14}\)
1) Ta có: \(\frac{x^3}{2^3}=\frac{y^3}{4^3}=\frac{z^3}{6^3}\) hay\(\left(\frac{x}{2}\right)^3=\left(\frac{y}{4}\right)^3=\left(\frac{z}{6}\right)^3\)
Do đó: \(\frac{x}{2}=\frac{y}{4}=\frac{z}{6}\)
=> \(\left(\frac{x}{2}\right)^2=\left(\frac{y}{4}\right)^2=\left(\frac{z}{6}\right)^2\) hay \(\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}\)
Áp dụng tính chất dãy tỉ số bằng nhau được:
\(\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}=\frac{x^2+y^2+z^2}{4+16+36}=\frac{14}{56}=\frac{1}{4}\)
=> \(\frac{x}{2}=\frac{y}{4}=\frac{z}{6}=\sqrt{\frac{1}{4}}=\frac{1}{2}\)
=> x=1 ; y=2 ; z=3
1, \(\left(xy\right)^2-\frac{1}{2}x^2y^2+3xy^2.\left(-\frac{1}{3}x\right)\)
\(=x^2y^2-\frac{1}{2}x^2y^2-x^2y^2\)
\(=-\frac{1}{2}x^2y^2\)
2, \(4.\left(-\frac{1}{2}x\right)^2-\frac{3}{2}x.\left(-x\right)+\frac{1}{3}x^2\)
\(=x^2+\frac{3}{2}x^2+\frac{1}{3}x^2\)
\(=\frac{17}{6}x^2\)
3, \(-4.\left(2x\right)^2y^3+\frac{1}{2}xy.\left(-2xy^2\right)+\frac{1}{4}x^2y^3\)
\(=-16x^2y^3-x^2y^3+\frac{1}{4}x^2y^3\)
\(=-\frac{67}{4}x^2y^3\)
4, \(\frac{1}{3}x^4y-\frac{5}{3}x^3.\left(\frac{5}{2}xy\right)+\frac{3}{4}x^4y\)
\(=\frac{1}{3}x^4y-\frac{25}{6}x^4y+\frac{3}{5}x^4y\)
\(=-\frac{97}{30}x^4y\)
5, \(\left(-2x^3y^4\right)^2-5x^2y.\left(\frac{3}{4}x^4y^7\right)-\frac{2}{3}x^6y^8\)
\(=4x^6y^8-\frac{15}{4}x^6y^8-\frac{2}{3}x^6y^8\)
\(=-\frac{5}{12}x^6y^8\)
Bài 1:
Bài 2:
\(\frac{4^x}{2^{x+y}}=8\Leftrightarrow4^x=8.2^{x+y}\Leftrightarrow\left(2^2\right)^x=2^3.2^{x+y}\Leftrightarrow2^{2x}=2^{x+y+3}\)<=>2x=x+y+3<=>x=y+3
\(\frac{9^{x+y}}{3^{5y}}=243\Leftrightarrow9^{x+y}=243.3^{5y}\Leftrightarrow\left(3^2\right)^{x+y}=3^5.3^{5y}\Leftrightarrow3^{2x+2y}=3^{5y+5}\)<=>2x+2y=5y+5
<=>2x=3y+5 mà x=y+3 => 2(y+3)=3y+5 <=> 2y+6=3y+5 <=> 6-5=3y-2y <=> y=1 <=> x=1+3=4
Vậy xy=4.1=4