y x 1,3 + y x 8,7 = 1,5. Tìm y
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\(8,3+1,7+1,5+8,5+1,3+8,7\)
\(=\left(8,3+1,7\right)+\left(1,5+8,5\right)+\left(1,3+8,7\right)\)
\(=10+10+10\)
\(=10\times3\)
\(=30\)
_HT_
8,3 + 1,7 + 1.5 + 8,5 + 1,3 + 8,7
= (8,3 + 1,7) + (1,5 + 8,5) + (8,7 + 1,3)
= 10 + 10 + 10
= 30
a) \(\left(x-1,3\right)^2=9\Leftrightarrow\left[{}\begin{matrix}x-1,3=3\\x-1,3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4,3\\x=-1,7\end{matrix}\right.\)
b) 24-x = 32
⇔ 24-x = 25
⇔ 4-x=5
⇔ x=-1
c) (x+1,5)2+(y-2,5)10=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\y-2,5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1,5\\y=2,5\end{matrix}\right.\)
\(a,\left(x-1,3\right)^2=9\\ \Leftrightarrow\left(x-1,3+9\right)\left(x-1,3-9\right)=0\\ \Leftrightarrow\left(x-7,7\right)\left(x-10,3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7,7=\dfrac{77}{10}\\x=10,3=\dfrac{103}{10}\end{matrix}\right.\)
\(b,2^{4-x}=32=2^5\\ \Leftrightarrow4-x=5\\ \Leftrightarrow x=-1\)
\(c,\left(x+1,5\right)^2+\left(y-2,5\right)^{10}=0\\ \Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\y-2,5=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-1,5=-\dfrac{3}{2}\\y=2,5=\dfrac{5}{2}\end{matrix}\right.\)
f(-8) = (-8)2 = 64
f(-1,3) = (-1,3)2 = 1,69
f(-0,75) = (-0,75)2 = 0,5625
f(1,5) = (1,5)2 = 2,25.
a) 8,7 - y = 5,3 + 2
8,7 - y = 7,3
y = 8,7 - 7,3
y = 1,4
b) y x 5,3 = 9,01 x 4
y x 5,3 = 36,04
y = 36,04 : 5,3
y = 6,8
k mk nha
\(7,75-\left(0,5\times y\div5-6,2\right)=5\)
\(0,5\times y\div5-6,2=7,75-5=2,75\)
\(0,5\div5\times y-6,2=2,75\)
\(0,1\times y=2,75+6,2=8,95\)
\(\dfrac{1}{10}y=8,95\)
\(y=8,95\times10=89,5\)
\(y\div6\times7,2+1,3\times y+y\div2+15=19,95\)
\(1,2\times y+1,3\times y+0,5y=19,95-15=4,95\)
\(y\left(1,2+1,3+0,5\right)=4,95\)
\(2y=4,95\)
\(y=4,95\div2=2,475\)
y x 1,3 + y x 8,7 = 1,5
y x ( 1,3 + 8,7 ) = 1,5
y x 10 = 1,5
y = 1,5 : 10
y = 0,15
y x (1,3+8,7)=1,5
y x 10 = 1,5
y = 1,5 : 10
y = 0,15