Tim a,b biet
a/27=-5/9=-45/b
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\(\frac{a}{27}=\frac{-5}{9}\)
=>ax9=-5x27=>a=\(\frac{-5.27}{9}=-15\)
\(\frac{-5}{9}=\frac{-45}{b}\)
b x -5=-45x9=\(\frac{-45.9}{-5}=81\)
a: \(\left(x+5\right)^2>=0\forall x\)
\(\left(2y-8\right)^2>=0\forall y\)
Do đó: \(\left(x+5\right)^2+\left(2y-8\right)^2>=0\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+5=0\\2y-8=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-5\\y=4\end{matrix}\right.\)
b: \(\left(x+3\right)\left(2y-1\right)=5\)
=>\(\left(x+3\right)\left(2y-1\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)
=>\(\left(x+3;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(-2;3\right);\left(2;1\right);\left(-4;-2\right);\left(-8;0\right)\right\}\)
Giải:
\(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)
\(\Rightarrow\dfrac{a}{27}=\dfrac{-5}{9}\)
\(\Rightarrow a=\dfrac{-5.27}{9}=-15\)
\(\Rightarrow\dfrac{-45}{b}=\dfrac{-5}{9}\)
\(\Rightarrow b=\dfrac{-45.9}{-5}=81\)
\(-\frac{15}{27}=-\frac{5}{9}=-\frac{45}{81}\)
=> a= (-15)
b=81
\(\dfrac{-5}{9}\)=\(\dfrac{-45}{b}\)
⇒ b= [9. (-45)] : -5
⇒ b= -405 : -5
⇒ b= 81
⇒ \(\dfrac{-45}{81}\)
\(\dfrac{a}{27}\)= \(\dfrac{-5}{9}\)
⇒ a= [ 27 .(-5) ] : 9
⇒ a= -135 : 9
⇒ a= -15
⇒ \(\dfrac{-15}{27}\)
⇒ \(\dfrac{-15}{27}\)=\(\dfrac{-5}{9}\)=\(\dfrac{-45}{81}\)
\(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)
⇒\(a=\dfrac{-5.27}{9}=-15\)
⇒\(b=\dfrac{-45.9}{5}=-81\)
Ta có:
a/27=-5/9=-45/b
(*1)-5/9=a/27
=>-15/27=a/27
=>a=-15
(*2)-5/9=-45/b
=>-45/81=-45/b
=>b=81