3x+2y=6
2x-2y=14
giải pt bằng pt cộng đại số ạ
giúp mik với có vài đoạn mik chưa hiểu lắm ý
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\(\left\{{}\begin{matrix}0,5x-1,5y=1\\-x+3y=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-3y=2\\-x+3y=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3y+2\\-\left(3y+2\right)+3y=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3y+2\\-3y-2+3y=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3y+2\\-2=2\end{matrix}\right.\)
=> Hpt vô nghiệm
0,5x - 1,5y = 1 (1)
-x + 3y = 2 (2)
Từ (2) ta có:
x = 3y - 2 (3)
Thế (3) vào (1), ta có:
0,5(3y - 2) - 1,5y = 1
1,5y - 1 - 1,5y = 1
0y = 1 + 1
0y = 2 (vô lý)
Vậy
\(\left(\dfrac{-4}{9}\right)^2=\dfrac{16}{81}\Rightarrow x=2\)
\(\left(-\dfrac{1}{3}\right)^3=\dfrac{-1}{27}\Rightarrow x=1\)
\(\left(-\dfrac{1}{3}\right)^4=\dfrac{1}{81}\Rightarrow x=1\)
\(\left(-\dfrac{4}{9}\right)^x=\dfrac{16}{81}\\ \left(-\dfrac{4}{9}\right)^x=\left(\dfrac{4}{9}\right)^2\\ \left(-\dfrac{4}{9}\right)^x=\left(-\dfrac{4}{9}\right)^2\\ x=2\\ -----------\\ \left(-\dfrac{1}{3}\right)^{2x+1}=-\dfrac{1}{27}\\ \left(-\dfrac{1}{3}\right)^{2x+1}=\left(-\dfrac{1}{3}\right)^3\\ 2x+1=3\\ 2x=3-1=2\\ x=\dfrac{2}{2}=1\\ -----------\\ \left(-\dfrac{1}{3}\right)^{3x+1}=\dfrac{1}{81}\\\left(-\dfrac{1}{3}\right)^{3x+1}=\left(\dfrac{1}{3}\right)^4\\ \left(-\dfrac{1}{3}\right)^{3x+1}=\left(-\dfrac{1}{3}\right)^4\\ 3x+1=4\\ 3x=4-1=3\\ x=\dfrac{3}{3}=1\)
\(\left(\dfrac{7}{5}\right)^x=\dfrac{49}{25}\Leftrightarrow\left(\dfrac{7}{5}\right)^x=\left(\dfrac{7}{5}\right)^2\Leftrightarrow x=2\)
\(7A=\dfrac{7^{2022}+14}{7^{2022}+2}=1+\dfrac{12}{7^{2022}+2}\)
\(7B=\dfrac{7^{2024}+14}{7^{2024}+2}=1+\dfrac{12}{7^{2024}+2}\)
\(7^{2022}+2< 7^{2024}+2\)
=>\(\dfrac{12}{7^{2022}+2}>\dfrac{12}{7^{2024}+2}\)
=>\(\dfrac{12}{7^{2022}+2}+1>\dfrac{12}{7^{2024}+2}+1\)
=>7A>7B
=>A>B
\(a,32< 2^n< 128\)
\(=>2^5< 2^n< 2^7\)
\(=>n=6\)
Vậy...
\(b,2.16\ge2^n>4\)
\(=>2^5\ge2^n>2^2\)
\(=>n\in\left\{3;4;5\right\}\)
Vậy...
\(c,3^2.3^n=3^5\)
\(3^n=3^5:3^2\)
\(3^n=3^3\)
\(=>n=3\)
Vậy...
\(d,\left(2^2:4\right).2^n=4\)
\(\left(2^2:2^2\right).2^n=4\)
\(1.2^n=4\)
\(2^n=4:1\)
\(2^n=4\)
\(=>2^n=2^2\)
\(=>n=2\)
Vậy ...
\(e,\dfrac{1}{9}.3^4.3^n=3^7\)
\(\dfrac{1}{9}.81.3^n=3^7\)
\(3^2.3^n=3^7\)
\(3^n=3^7:3^2\)
\(3^n=3^5\)
\(=>n=5\)
Vậy...
\(g,\dfrac{1}{2}.2^n+4.2^n=9.2^5\)
\(\left(\dfrac{1}{2}+4\right).2^n=9.2^5\)
\(\dfrac{9}{2}.2^n=9.32\)
\(\dfrac{9}{2}.2^n=288\)
\(2^n=288:\dfrac{9}{2}\)
\(2^n=2^6\)
\(=>n=6\)
Vậy...
a) \(32< 2^n< 128\\ \Rightarrow2^5< 2^n< 2^7\\ \Rightarrow5< n< 7\)
Mà: \(n\inℕ^∗\)
\(\Rightarrow n=6\)
b) \(2.16\ge2^n>4\\ \Rightarrow2^1.2^4\ge2^n>2^2\\ \Rightarrow2^5\ge2^n>2^2\\ \Rightarrow5\ge n>2\)
Mà: \(n\inℕ^∗\)
\(\Rightarrow n\in\left\{5;4;3\right\}\)
c) \(3^2.3^n=3^5\\ \Rightarrow3^{n+2}=3^5\\ \Rightarrow n+2=5\\ \Rightarrow n=3\left(nhận\right)\)
\(\left\{{}\begin{matrix}3x+2y=6\\2x-2y=14\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x+2y=6\\3x+2x=14+6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x+2y=6\\5x=20\end{matrix}\right. \\ \Leftrightarrow\left\{{}\begin{matrix}3\cdot4+2y=6\\x=\dfrac{20}{5}=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2y=6-12=-6\\x=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{6}{2}=-3\\x=4\end{matrix}\right.\)
Vậy: ...