\(12\times4+12\times2+12+12\)

\(=12\times4+12\times2+12\times1+12\times1\)

\(=12\times\left(4+2+1+1\right)\)

\(=12\times8\)

\(=96\)

=(5/12+7/12)*2/3+1/3

=1*2/3+1/3

=2/3+1/3

=1

5/12 x 2/3 + 2/3 x 7/12 + 1/3

= 2/3 x ( 5/12 + 7/12 ) + 1/3

= 2/3 x 1 + 1/3

= 2/3 + 1/3

= 1

5/12x2/3+2/3x7/12+1/3

=2/3x(5/12+7/12)+1/3

=2/3x1+1/3

=2/3+1/3

=1

k tui nha @@@

\(\frac{5}{12}\)x \(\frac{2}{3}\)+ \(\frac{2}{3}\)x \(\frac{7}{12}\)+ \(\frac{1}{3}\)

= \(\frac{2}{3}\)x ( \(\frac{5}{12}\)+ \(\frac{7}{12}\) ) + \(\frac{1}{3}\)

= \(\frac{2}{3}\)x 1 + \(\frac{1}{3}\)

= \(\frac{2}{3}\)+ \(\frac{1}{3}\)

= 1

\(a,=1,6^2+2\cdot1,6\cdot3,4+3,4^2=\left(1,6+3,4\right)^2=5^2=25\\ b,Sửa:x^4-12x^3+12x^2-12x+11\\ =x^4-11x^3-x^3+11x^2+x^2-11x-x+11=x^3\left(x-11\right)-x^2\left(x-11\right)+x\left(x-11\right)-\left(x-11\right)\\ =\left(x-11\right)\left(x^3-x^2+x-1\right)=\left(x-11\right)\left(x-1\right)\left(x^2+1\right)\\ c,=\left(x^2+3\right)^2-\left(x^2-4\right)\left(x^2+12\right)\\ =x^4+6x^2+9-x^4-8x^2+48=-2x^2+57\)

2/3x(5/12+7/12)+1/3

2/3x       1        +1/3

2/3+1/3

=1              k mik nhaaaaa

5/12×2/3+2/3×7/12+1/3

=2/3×(5/12+7/12)+1/3

=2/3×          1          +1/3

=              2/3           +1/3

=                       1

\(\frac{5}{12}.\frac{2}{3}+\frac{2}{3}.\frac{7}{12}+\frac{1}{3}\)

\(=\frac{2}{3}.\left(\frac{5}{12}+\frac{7}{12}\right)+\frac{1}{3}\)

\(=\frac{2}{3}.\frac{12}{12}+\frac{1}{3}\)

\(=\frac{2}{3}.1+\frac{1}{3}\)

\(=\frac{2}{3}+\frac{1}{3}\)

\(=\frac{3}{3}=1\)

Dấu "." bằng dấu x

\(\dfrac{-5}{12}\times\dfrac{2}{11}+\dfrac{-5}{11}\times\dfrac{9}{11}+\dfrac{5}{12}\\ =\dfrac{5}{12}\times\left(1-\dfrac{2}{11}\right)+\dfrac{-5}{11}\times\dfrac{9}{11}\\ =\dfrac{5}{12}\times\dfrac{9}{11}+\dfrac{-5}{11}\times\dfrac{9}{11}\\ =\dfrac{9}{11}\times\left(\dfrac{5}{12}-\dfrac{5}{11}\right)\\ =\dfrac{9}{11}\times\dfrac{-5}{132}=-\dfrac{15}{484}\)

A = \(\dfrac{-5}{12}\)x\(\dfrac{2}{11}\)+\(\dfrac{-5}{12}\)x\(\dfrac{9}{11}\)+\(\dfrac{5}{12}\)

A= \(\dfrac{-5}{12}\)(\(\dfrac{2}{11}\)+\(\dfrac{9}{11}\)-1)

A = 0

Đáp án D