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bài 3 : \(\left\{{}\begin{matrix}ab=2\\bc=3\\ca=54\end{matrix}\right.\)
hiển nhiên a;b;c =0 không phải nghiệm
\(\Leftrightarrow\left(abc\right)^2=2.3.54=18^2\)
\(\Leftrightarrow\left[{}\begin{matrix}abc=-18\\abc=18\end{matrix}\right.\)
abc=-18 => c=-9; a=-6; b=-1/3
abc=18 => c=9; a=6; b=1/3
Đặt:\(7a=3b=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{k}{7}\\b=\dfrac{k}{3}\end{matrix}\right.\)
\(\Rightarrow\dfrac{k}{7}.\dfrac{k}{3}=20\Rightarrow\dfrac{k^2}{21}=20\Rightarrow k^2=420\Rightarrow k=\pm\sqrt{420}\)
Xét: \(k=\sqrt{420}\)
\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{\sqrt{420}}{7}\\b=\dfrac{\sqrt{420}}{3}\end{matrix}\right.\)
Xét: \(k=-\sqrt{420}\)
\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{-\sqrt{420}}{7}\\b=\dfrac{-\sqrt{420}}{3}\end{matrix}\right.\)
b) Dựa vào tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)
\(=\dfrac{a+b-c}{2+3-4}=\dfrac{100}{1}=100\)
\(\Rightarrow\left\{{}\begin{matrix}a=100.2=200\\b=100.3=300\\c=100.4=400\end{matrix}\right.\)
c) Đặt: \(\dfrac{a}{4}=\dfrac{b}{7}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=4k\\b=7k\end{matrix}\right.\)
\(\Rightarrow4k.7k=112\)
\(\Rightarrow28k^2=112\)
\(k^2=4\Rightarrow k=\pm2\)
Xét: \(k=2\)
\(\Rightarrow\left\{{}\begin{matrix}a=2.4=8\\b=2.7=14\end{matrix}\right.\)
Xét:\(k=-2\)
\(\Rightarrow\left\{{}\begin{matrix}a=-2.4=-8\\c=-2.7=-14\end{matrix}\right.\)
\(\text{a) }7a=3b\text{ và }ab=20\\ \text{Đặt }7a=3b=k\Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{7}k\\b=\dfrac{1}{3}k\end{matrix}\right.\left(1\right)\\ \text{Từ }\left(1\right)\text{ suy ra : }\\ ab=20\\ \Leftrightarrow\left(\dfrac{1}{7}k\right)\left(\dfrac{1}{3}k\right)=20\\ \Leftrightarrow\left(\dfrac{1}{7}\cdot\dfrac{1}{3}\right)\left(k\cdot k\right)=20\\ \Leftrightarrow\dfrac{1}{21}k^2=20\\ \Leftrightarrow k^2=420\\ \Leftrightarrow k=\sqrt{420}\\ \text{Từ }k=\sqrt{420}\text{ suy ra : }\left\{{}\begin{matrix}a=\dfrac{1}{7}\cdot\sqrt{420}\\b=\dfrac{1}{3}\cdot\sqrt{420}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\dfrac{\sqrt{420}}{7}\\b=\dfrac{\sqrt{420}}{3}\end{matrix}\right.\\ \text{Vậy }a=\dfrac{\sqrt{420}}{7};b=\dfrac{\sqrt{420}}{3}\)
\(\text{b) }\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\text{ và }a+b-c=100\\ \text{ Theo bài ra ta có : }\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\\ a+b-c=100\\ \text{Áp dụng tính chất dãy tỉ số bằng nhau ta được : }\\ \dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{a+b-c}{2+3-4}=\dfrac{100}{1}=100\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=100\\\dfrac{b}{3}=100\\\dfrac{c}{4}=100\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=200\\b=300\\c=400\end{matrix}\right.\\ \text{Vậy }a=200;b=300;c=400\)
\(\text{c) }\dfrac{a}{4}=\dfrac{b}{7}\text{ và }ab=112\\ \text{Đặt }\dfrac{a}{4}=\dfrac{b}{7}=k\Rightarrow\left\{{}\begin{matrix}a=4k\\b=7k\end{matrix}\right.\left(1\right)\\ \text{Từ }\left(1\right)\text{ suy ra : }\\ ab=112\\ \Leftrightarrow4k\cdot7k=112\\ \Leftrightarrow28k^2=112\\ \Leftrightarrow k^2=4\\ \Leftrightarrow k=2\\ \text{Từ }k=2\Rightarrow\left\{{}\begin{matrix}a=4\cdot2\\b=7\cdot2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=8\\b=14\end{matrix}\right.\\ \text{Vậy }a=8;b=14\)
Ta có \(\frac{a}{b}=\frac{c}{d}=>\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=>\frac{a}{a-b}=\frac{c}{c-d} \)
\(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow2B=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{97}+\left(\dfrac{1}{2}\right)^{98}\)
\(\Rightarrow2B-B=1-\left(\dfrac{1}{2}\right)^{99}\)
\(B=1-\left(\dfrac{1}{2}\right)^{99}\)
\(2,\)
\(a,\dfrac{45^{10}.2^{10}}{75^{15}}\)
\(=\dfrac{5^{10}.9^{10}.2^{10}}{25^{15}.3^{15}}\)
\(=\dfrac{5^{10}.3^{20}.2^{10}}{5^{30}.3^{15}}\)
\(=\dfrac{5^{10}.3^{15}.\left(3^5.2^{10}\right)}{5^{10}.3^{15}.\left(5^{20}\right)}\)
\(=\dfrac{3^5.2^{10}}{5^{20}}\)
\(b,\dfrac{2^{15}.9^4}{6^3.8^3}\)
\(=\dfrac{2^{15}.3^8}{2^3.3^3.2^9}=\dfrac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5\)
\(c,\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{4^{10}.2^{10}+4^{10}}{4^4.2^4+4^4.4^7}=\dfrac{4^4.\left(4^6.2^{10}+4^6\right)}{4^4.\left(2^4+4^7\right)}\)
\(=\dfrac{4^{11}+4^6}{4^8.4^7}=\dfrac{4^6.\left(4^5+1\right)}{4^6.\left(4^2-4\right)}=\dfrac{1024+1}{16-4}=\dfrac{1025}{12}\)
\(d,\dfrac{81^{11}.3^{17}}{27^{10}.9^{15}}=\dfrac{3^{44}.3^{17}}{3^{30}.3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)
\(3,\)
\(a,\left(2x+4\right)^2=\dfrac{1}{4}\)
\(\left(2x+4\right)^2=\left(\dfrac{1}{2}\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x+4=\dfrac{1}{2}\\2x+4=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{2}-4=\dfrac{-7}{2}\\2x=\dfrac{-1}{2}-4=\dfrac{-9}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-7}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\)
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, \(5^{x+2}=625\)
\(5^{x+2}=5^4\)
\(x+2=4\)
\(x=2\)
2, a-b=ab => a=ab+b => a=b(a+1)
thay a=b(a+1) vào a:b ta có: => b:b(a+1)=a+1
Theo bài ra ta có: a:b=a-b
=> a+1=a-b
=>-b=1
=> b=-1
Thay b=-1 vào a-b=ab ta có : a-(-1)=-a
=> a +1=-a
=>a=-1/2
Vậy a=-1/2. b=-1
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)
\(\Leftrightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{c+a}{ca}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}\\\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{b}=\dfrac{1}{a}\end{matrix}\right.\)
\(\Leftrightarrow a=b=c\)
\(\Rightarrow P=1\)
ta có \(\left\{{}\begin{matrix}\dfrac{ab}{a+b}=\dfrac{ac}{a+c}\\\dfrac{ab}{a+b}=\dfrac{bc}{b+c}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a.\dfrac{b}{a+b}=a.\dfrac{c}{c+a}\\b.\dfrac{a}{a+b}=b.\dfrac{c}{b+c}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{a+b}=\dfrac{c}{c+a}\\\dfrac{a}{a+b}=\dfrac{c}{b+c}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}1+\dfrac{b}{a}=1+\dfrac{c}{a}\\1+\dfrac{a}{b}=1+\dfrac{c}{b}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{a}=\dfrac{c}{a}\\\dfrac{a}{b}=\dfrac{c}{b}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}b=c\\a=c\end{matrix}\right.\Rightarrow a=b=c\)
\(\Rightarrow P=\dfrac{ab^2+bc^2+ca^2}{a^3+b^3+c^3}=\dfrac{a^3+a^3+a^3}{a^3+a^3+a^3}=1\)