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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
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
a: \(=\left(-1\right)^{10}+\left(-1\right)^9+\left(-1\right)^8+...+\left(-1\right)^2+\left(-1\right)\)
\(=\left(1-1\right)+\left(1-1\right)+...+\left(1-1\right)\)
=0
b: \(=\left(-1\right)^{100}+\left(-1\right)^{99}+...+\left(-1\right)^2+\left(-1\right)\)
\(=\left(1-1\right)+...+\left(1-1\right)\)
=0
c: \(=1^{100}-1^{99}+1^{98}-1^{97}+...+1^2-1\)
=0
f: \(=3\cdot\sqrt{9-5}+7=3\cdot2+7=13\)
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
Giải:
Có: \(2^x=8^{y+1}\) và \(9^y=3^{x-9}\)
\(\Leftrightarrow\left\{{}\begin{matrix}2^x=2^{3y+3}\\3^{2y}=3^{x-9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3y+3\\2y=x-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3y+3\\y=\dfrac{x-9}{2}\end{matrix}\right.\)
\(\Leftrightarrow x+y=3y+3+\dfrac{x-9}{2}\)
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