\(a,=\left(2x^3-x^2+x+4x^2-2x+2-x+1\right):\left(2x^2-x+1\right)\\ =\left[x\left(2x^2-x+1\right)+2\left(2x^2-x+1\right)-x+1\right]:\left(2x^2-x+1\right)\\ =x+2\left(\text{dư }-x+1\right)\\ b,=\left[x^2\left(2x-5\right)+3\left(2x-5\right)\right]:\left(2x-5\right)\\ =x^2+3\)

\(\left(2x^3+5x^2+6x-15\right):\left(2x-5\right)=\left[x^2\left(2x-5\right)+3\left(2x-5\right)\right]:\left(2x-5\right)=\left[\left(2x-5\right)\left(x^2+3\right)\right]:\left(2x-5\right)=x^2+3\)

-15(x+2) + 7(2x3) = -50

-15(x+2) + 7 x 6   = -50

-15(x+2) + 42       = -50

-15(x+2)               = -50 - 42

-15(x+2)               = -92

(x+2)                    = -92 : -15

(x+2)                    = 6,1333

x                          = 6,1333 + 2

x                          = 8,1333

-15(x+2)+7(2.3)=-50

-15(x+2)+7.6=-50 

-15(x+2)+42=-50

-15(x+2)=-50-42

-15(x+2)=-92

x+2=-92:-15

x+2=92/15

x=92/15-2

x=62/15

moi nguoi nho tk va ket ban voi minh nhe

sorry. mk ko biết

me too

=\(\frac{28}{15}\)x\(\frac{1}{4}\)x3+\(\frac{17}{60}\):1+\(\frac{23}{24}\)

=\(\frac{7}{15}\)x3+\(\frac{17}{60}\)+\(\frac{23}{24}\)

=\(\frac{7}{5}\)+\(\frac{17}{60}\)+\(\frac{23}{24}\)

=\(\frac{168}{120}\)+\(\frac{34}{120}\)+\(\frac{115}{120}\)

=\(\frac{317}{120}\)

\(A=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right).2}\)

\(A=\frac{11.3^{22+7}-\left(3^2\right)^{15}}{2^2.\left(3^{14}\right)^2}\)

\(A=\frac{11.3^{29}-3^{30}}{4.3^{28}}\)

\(A=\frac{11.3^{29}-3^{29}.3}{4.3^{28}}\)

\(A=\frac{3^{29}.\left(11-3\right)}{3^{28}.4}\)

\(A=\frac{3^{28}.3.8}{3^{28}.4}\)

\(A=\frac{3^{28}.3.4.2}{3^{28}.4}\)

\(A=6\)

\(A=3.2\)

Vậy : \(A=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=6\)

1: \(=x^2+1\)

3: \(=\left(x-y-z\right)^2\)