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X x 4,5 + 1,5 x X + X x 6
=X x(4,5+1,5+6)
=X x12
Còn lại mik chịu

Bai 1:
x = ko co so nao ; y = ko co so nao phu hop
Bai 2:
S= 1,5..
Chac the ko bit nua

-12.(x-5)+7.(3-x)=5
-12x-60+21-7x=5
(-12x-7x)-(60+21)=5
-19x-81=5
-19x=5-81
-19x=-76
x=-76:(-19)
x=4
x=
x=44:(-19)
\(-12\left(x-5\right)+7\left(3-x\right)=5\)
\(60-12x+21-7x=5\)
\(81-19x=5\)
\(19x=81-5\)
\(19x=76\)
\(x=4\)

ta có: L = \(\frac{7}{3}+\frac{11}{3^2}+\frac{15}{3^3}+...+\frac{403}{3^{100}}\)
<=> \(3L=7+\frac{11}{3}+\frac{15}{3^2} +..+\frac{403}{3^{99}}\)
=> \(3L-L=\left(7+\frac{11}{3}+\frac{15}{3^2}+...+\frac{403}{3^{99}}\right)-\left(\frac{7}{3}+\frac{11}{3^2}+...+\frac{403}{3^{100}}\right)\)
<=> \(2L=7+\left(\frac{11}{3}-\frac{7}{3}\right)+\left(\frac{15}{3^2}-\frac{11}{3^2}\right)+...+\left(\frac{403}{3 ^{99}}-\frac{399}{3^{99}}\right)-\frac{403}{3^{100}}\)
<=> \(2L=7+4\cdot\frac{1}{3}+4\cdot\frac{1}{3^2}+..+4\cdot\frac{1}{3^{99}}-\frac{403}{3^{100}}\)
<=> \(2L=7+4\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\frac{403}{3^{100}}\)
<=>\(2L=7+4\left[\frac{1}{2}\cdot\left(1-\frac{1}{3^{99}}\right)\right]-\frac{403}{3^{100}}\)
<=> \(2L=7+2-\frac{2}{3^{99}}-\frac{403}{3^{100}}\)
<=> \(L=3,5+1-\frac{1}{3^{99}}-\frac{403}{2\cdot3^{100}}\)
<=> \(L=4,5-\frac{1}{3^{99}}-\frac{403}{2\cdot3^{100}}<4,5\)
1 ĐÚNG NHÉ

0,48x700-2,4x45x2
=0,48x(700-5x45x2)
=0,48x(700-450)
=0,48x250
=120
4,5+3,5+2,5+1,5+0,5-4-3-2-1
=(4,5-4)+(3,5-3)+(2,5-2)+(1,5-1)+0,5
=0,5+0,5+0,5+0,5+0,5
=0,5x5
=2,5


\(2.x-15=11\)
\(2x=11+15\)
\(2x=26\)
\(x=26:2\)
\(x=13\)
\(b,4.\left(2-x\right)-5\left(3-x\right)=-12\)
\(8-4x-15+5x=-12\)
\(x-7=-12\)
\(x=-12+7\)
\(x=-5\)
\(c,\left|x-7\right|+13=25\)
\(\left|x-7\right|=25-13\)
\(\left|x-7\right|=12\)
\(\Rightarrow\orbr{\begin{cases}x-7=12\\x-7=-12\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=19\\x=-5\end{cases}}\)