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1, xy-2x+3y=9
<=> xy-2x+3y-9=0
<=> x(y-2) + 3(y-2)=0
<=>(y-2)(x+3)=0
<=>+) y-2=0 <=> y=2
+)x+3=0<=>x=-3
bài 2 :
Gọi UCLN ( n+3; 2n+5) là d
\(\Rightarrow n+3⋮d;2n+5⋮d\)
\(\Rightarrow2n+6⋮d;2n+5⋮d\)
\(\Rightarrow\left(2n+6\right)-\left(2n+5\right)⋮d\)
\(\Rightarrow2n+6-2n-5⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d\inƯ\left(1\right)=\left\{\pm1\right\}\)
mà 1 là UCLN(n+3;2n+5)
\(\Rightarrow d=1\)
Bài 1 tự làm!
Bài 2:
a, \(\left(3x-4\right)\left(x-1\right)^3=0\Rightarrow\left[{}\begin{matrix}3x-4=0\\\left(x-1\right)^3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=1\end{matrix}\right.\)
b, \(2^{2x-1}:4=8^3\Rightarrow2^{2x-1}:2^2=2^9\)
\(\Rightarrow2x-1-2=9\Rightarrow2x-3=9\Rightarrow2x-12\Rightarrow x=6\)
c, Đề chưa rõ
d, \(\left(x+2\right)^5=2^{10}\Rightarrow\left(x+2\right)^5=4^5\Rightarrow x+2=4\Rightarrow x=2\)
e, \(\left(3x-2^4\right).7^3=2.7^4\Rightarrow3x-2^4=2.7^4:7^3\Rightarrow3x-16=2.7=14\)
\(\Rightarrow3x=14+16=30\Rightarrow x=\dfrac{30}{3}=10\)
f, \(\left(x+1\right)^2=\left(x+1\right)^0\Rightarrow\left(x+1\right)^2=1\) (vì x0 = 1)
\(\Rightarrow x+1=1\Rightarrow x=0\)
a) \(\left(x^2-4\right)\left(x^2-9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x^2-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x^2=2^2\\x^2=3^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm2\\x=\pm3\end{matrix}\right.\)
b) \(\left(x^2-4\right)\left(x^2-9\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4\ge0;x^2-9\le0\\x^2-4\le0;x^2-9\ge0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2\ge4;x^2\le9\\x^2\le4;x^2\ge9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4\le x^2\le9\left(tm\right)\\9\le x^2\le4\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow x^2\in\left\{4;5;...;9\right\}\)
\(\Rightarrow x\in\left\{\pm2;\pm\sqrt{5};...;\pm3\right\}\).
a) ( x2 - 4 ) . ( x2 - 9 ) = 0
=> \(\left[{}\begin{matrix}x^2-4=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x^2=4\\x^2=9\end{matrix}\right.\)
= > \(\left[{}\begin{matrix}x=-2\\x=2\\x=3\\x=-3\end{matrix}\right.\)
Ta có: 3+2^x-0^1=2^x+3-0=2^x+3=11
=> 2^x=11-3=8=2^3
=> x=3