(2016.x+3.y+1).(2016^x+2016.x+2016.x+y)=225

=2016+3+1.(x+y).........

...................

tk đi chỉ tiếp

X=0

Y=1

lời giải chi tiết vs ạ

`1/9(x-3)^2-1/25(x+5)^2=0`

`<=>(1/3x-1)^2-(1/5x+1)^2=0`

`<=>(1/3x-1-1/5x-1)(1/3x-1+1/5x+1)=0`

`<=>(2/15x-2). 8/15x=0`

`<=>2/15x-2=0` hoặc `8/15x=0`

`<=>x=15`         hoặc `x=0`

Vậy `S=`{`15;0`}

6/20:3=6/20x1/3=6/60=1/10

11h-8h30p=2h30p=2.5h

10 7/10-4 3/10

=107/10-43/10

=64/10=32/5

chúc bn học tốt!

cảm ơn cậu ạ

a) ta có x-2/5=3/8 => x=3/8+2/5 => x=31/40

b) ta có x^2/6=24/25 => x^2=144/25 => x= +_ 12/5

c) x-1/x-5 =6/7 với x # 5) => 7(x-1)=6(x-5) => 7x-7=6x-30 => 7x-6x=6-30 => x=-24 \

nghĩ thế !

\(\Leftrightarrow2\sqrt{x-4}=5\left(x\ge4\right)\\ \Leftrightarrow\sqrt{x-4}=\dfrac{5}{2}\\ \Leftrightarrow x-4=\dfrac{25}{4}\\ \Leftrightarrow x=\dfrac{41}{4}\left(tm\right)\)

cảm ơn nha

\(x^4+\sqrt{x^2+2016}=2016\)

\(\Leftrightarrow x^4+x^2+\frac{1}{4}=x^2+2016-\sqrt{x^2+2016}+\frac{1}{4}\)

\(\Leftrightarrow\left(x^2+\frac{1}{2}\right)^2=\left(\sqrt{x^2+2016}-\frac{1}{2}\right)^2\)

\(\Leftrightarrow x^2+\frac{1}{2}=\sqrt{x^2+2016}-\frac{1}{2}\text{ }\left(do\text{ }\sqrt{x^2+2016}-\frac{1}{2}>0\right)\)

\(\Leftrightarrow x^2+1=\sqrt{x^2+2016}\)

\(t=x^2\ge0\)

\(\rightarrow t+1=\sqrt{t+2016}\Leftrightarrow t^2+2t+1=t+2016\)

\(\Leftrightarrow t^2+t-2015=0\Leftrightarrow t=\frac{-1+\sqrt{8061}}{2}\text{ }\left(do\text{ }t\ge0\right)\)

\(\Leftrightarrow x=\pm\sqrt{\frac{-1+\sqrt{8061}}{2}}\)