Bài 7: Tỉ lệ thức

\(\frac{a}{b}\) = \(\frac{c}{d}\) => a . d = b . c => a . b . d = b . c . d

hay => a = c

và từ a . d = b . c => a² . d² = b² . c²

ta có a = c => b = d

hay => a² . d² + 4b² = b² . c² + 4b²

ta có b = d => a² . d² + 4b² = b² . c² + 4d²

ta có b = d => b² = d²

=> a² . d² + 4b² . d² = b² . c² + 4d² . b²

= d² . ( a² + 4b² ) = b² . ( c² + 4d² )

=> \(\frac{a^2+4b^2}{b^2}\) = \(\frac{c^2+4d^2}{d^2}\)

thank bn ak

đặt \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=k => a=bk; b=dk

\(\dfrac{a}{3b}\)=\(\dfrac{bk}{3b}=\dfrac{k}{3}\)

\(\dfrac{c}{3d}=\)\(\dfrac{dk}{3d}=\)\(\dfrac{k}{3}\)

=>\(\dfrac{a}{3b}=\)\(\dfrac{c}{3d}\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=> \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

ta có : \(\dfrac{a}{3b}=\dfrac{bk}{3b}=\dfrac{k}{3}\left(1\right)\)

lại có \(\dfrac{c}{3d}=\dfrac{dk}{3d}=\dfrac{k}{3}\left(2\right)\)

từ (1) và (2) => \(\dfrac{a}{3b}=\dfrac{c}{3d}\)

\(\dfrac{x}{y}=\dfrac{3}{5}\)\(\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{2x}{10}=\dfrac{3y}{9}\)

Theo t/c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{2x}{10}=\dfrac{3y}{9}=\dfrac{2x-3y}{10-9}=\dfrac{54}{1}=54\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=54\\\dfrac{y}{3}=54\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=270\\y=162\end{matrix}\right.\)

\(\dfrac{x}{y}=\dfrac{3}{5}\&2x-3y=54\)

\(\dfrac{x}{y}=\dfrac{3}{5}\Leftrightarrow5x=3y\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{5}\)

\(\Leftrightarrow\dfrac{2x}{2.3}=\dfrac{3y}{3.5}\Leftrightarrow\dfrac{2x}{6}=\dfrac{3y}{15}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{2x}{6}=\dfrac{3y}{15}=\dfrac{2x-3y}{6-15}=\dfrac{54}{-9}=-6\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=-6\\\dfrac{y}{5}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-18\\y=-30\end{matrix}\right.\)

\(\dfrac{x}{y}=\dfrac{3}{5}\) và 2x-3y=54

Ta có: \(\dfrac{x}{y}=\dfrac{3}{5}\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{5}\)

+) \(\dfrac{x}{3}=\dfrac{2x}{6}\)(điều 1)

+) \(\dfrac{y}{5}=\dfrac{3y}{15}\)(điều 2 )

\(\Rightarrow\)\(\dfrac{2x}{6}=\dfrac{3y}{15}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{2x}{6}=\dfrac{3y}{15}=\dfrac{2x-3y}{6-15}=\dfrac{54}{-9}=-6\)

\(\Rightarrow x=-18;y=-30\)

Chúc bạn học tốt

\(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\)

\(\Rightarrow2B=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{97}+\left(\dfrac{1}{2}\right)^{98}\)

\(\Rightarrow2B-B=1-\left(\dfrac{1}{2}\right)^{99}\)

\(B=1-\left(\dfrac{1}{2}\right)^{99}\)

\(2,\)

\(a,\dfrac{45^{10}.2^{10}}{75^{15}}\)

\(=\dfrac{5^{10}.9^{10}.2^{10}}{25^{15}.3^{15}}\)

\(=\dfrac{5^{10}.3^{20}.2^{10}}{5^{30}.3^{15}}\)

\(=\dfrac{5^{10}.3^{15}.\left(3^5.2^{10}\right)}{5^{10}.3^{15}.\left(5^{20}\right)}\)

\(=\dfrac{3^5.2^{10}}{5^{20}}\)

\(b,\dfrac{2^{15}.9^4}{6^3.8^3}\)

\(=\dfrac{2^{15}.3^8}{2^3.3^3.2^9}=\dfrac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5\)

\(c,\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{4^{10}.2^{10}+4^{10}}{4^4.2^4+4^4.4^7}=\dfrac{4^4.\left(4^6.2^{10}+4^6\right)}{4^4.\left(2^4+4^7\right)}\)

\(=\dfrac{4^{11}+4^6}{4^8.4^7}=\dfrac{4^6.\left(4^5+1\right)}{4^6.\left(4^2-4\right)}=\dfrac{1024+1}{16-4}=\dfrac{1025}{12}\)

\(d,\dfrac{81^{11}.3^{17}}{27^{10}.9^{15}}=\dfrac{3^{44}.3^{17}}{3^{30}.3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)

\(3,\)

\(a,\left(2x+4\right)^2=\dfrac{1}{4}\)

\(\left(2x+4\right)^2=\left(\dfrac{1}{2}\right)^2=\left(\dfrac{-1}{2}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x+4=\dfrac{1}{2}\\2x+4=\dfrac{-1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{2}-4=\dfrac{-7}{2}\\2x=\dfrac{-1}{2}-4=\dfrac{-9}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-7}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{-7}{4};\dfrac{-9}{4}\right\}\)

\(b,\left(2x-3\right)^2=36\)

\(\left(2x-3\right)^2=6^2=\left(-6\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=6+3=9\\2x=-6+3=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{9}{2};\dfrac{-3}{2}\right\}\)

\(c,5^{x+2}=628\)

\(5^{x+2}=5^4\)

\(\Rightarrow x+2=4\)

\(\Rightarrow x=4-2=2\)

Vậy \(x=2\)

\(d,\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)

\(\Rightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)

\(\Rightarrow\left(x-1\right)^{x+2}.\left[\left(x-1\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)

Vậy \(x\in\left\{0;1;2\right\}\)

Bài 1:

B= \(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{99}\)

2B= \(2.[\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}]\)

2B= \(1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{98}\)

⇒2B-B= \(1-\left(\dfrac{1}{2}\right)^{99}\)

B= 1

Vậy B=1

Bài 2:

a, \(\dfrac{45^{10}.2^{10}}{75^{15}}\)= \(\dfrac{\left(3^2.5\right)^{10}.2^{10}}{\left(3.5^2\right)^{15}}=\dfrac{3^{20}.5^{10}.2^{10}}{3^{15}.5^{30}}=\dfrac{3^5.2^{10}}{5^{20}}\)

b, \(\dfrac{2^{15}.9^4}{6^3.8^3}=\dfrac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^3.\left(2^3\right)^3}=\dfrac{2^{15}.3^8}{2^3.3^3.2^9}=\dfrac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5\)

c,\(\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2.4\right)^{10}+4^{10}}{\left(2.4\right)^4+4^{11}}=\dfrac{2^{10}.4^{10}+4^{10}}{2^4.4^4+4^{11}}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6+4^6.4^5}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6.\left(4^5+1\right)}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6.\left(2^{10}+1\right)}=4^4=256\)

d, \(\dfrac{81^{11}.3^{17}}{27^{10}.9^{15}}=\dfrac{\left(3^4\right)^{11}.3^{17}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\dfrac{3^{44}.3^{17}}{3^{30}.3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)

Bài 3:

a, \(\left(2x+4\right)^2=\dfrac{1}{4}\)

\(\left(2x+4\right)^2=\left(\dfrac{1}{2}\right)^2\)

\(2x+4=\dfrac{1}{2}\)

\(2x=\dfrac{1}{2}-4\)

\(2x=-\dfrac{7}{2}\)

\(x=-\dfrac{7}{2}:2\)

\(x=-\dfrac{7}{2}.\dfrac{1}{2}\)

\(x=-\dfrac{7}{4}\)

b, \(\left(2x-3\right)^2=36\)

\(\left(2x-3\right)^2=6^2\)

\(2x-3=6\)

\(2x=9\)

\(x=\dfrac{9}{2}\)

c, \(5^{x+2}=625\)

\(5^{x+2}=5^4\)

\(x+2=4\)

\(x=2\)