2^4.x-3^2.x=145-255:51
`2^4 x-3^2 x=145-255:51`
`16x-9x=140`
`7x=140`
`x=20`
\(2^4.x-3^2.x=145-255:51\)
\(16 . x - 9. x = 145 - 5\)
\(( 16 - 9 ). x = 140\)
\(7 . x = 140\)
\(x = 140 : 7 = 20\)
16 . x - 9. x = 145 - 5
( 16 - 9 ). x = 140
7. x = 140
x = 140 : 7
x = 20
a)3a2- 6ab + 3b2 - 12c2
=(\(\sqrt{3}\)a - \(\sqrt{3}\)b)2 - (\(\sqrt{12}\)c)2
=[(\(\sqrt{3}\)a - \(\sqrt{3}\)b) - \(\sqrt{12}\)c ) x (\(\sqrt{3}\)a - \(\sqrt{3}\)b) + \(\sqrt{12}\)c]
b) x2 - 25 + y2 + 2xy
=(x + y)2 - 52
=[(x + y) - 5] x [(x + y) + 5]
c) x2y - x3 - 9y +9x
= y(x2-9) -x(x2-9)
= (x2-9) x (y - x)
\(219-7(x+1)=100\)
\(7(x+1)=219-100\)
\(7(x+1)=119\)
\(x+1=119:7\)
\(x+1=17\)
\(x=17-1=16\)
7(x+1)=219-100
7(x+1)=119
x+1=119:7
x+1=17
x=17-1
x=16
\(5(x-3)-x^2+3x=0\)
\(<=>(x-3)(5-x)=0\)
\(< =>\left[{}\begin{matrix}\text{x − 3 = 0}\\\text{5 − x = 0 }\end{matrix}\right.< =>\left[{}\begin{matrix}\text{x = 3}\\\text{x = 5 }\end{matrix}\right.\)
`=> 5x - 15 - x^2 + 3x = 0`
`<=> -(x^2 - 8x + 15) = 0`
`<=> -(x-5)(x-3) = 0`
`<=>` \(\left[{}\begin{matrix}x-5=0\\x-3=0\end{matrix}\right.\)
`<=>` \(\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)
x là [Cu2+]
với cả \(\left[Cu\left(NH_3\right)_4^{2+}\right]=0,001-x\) nhé :)