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\(a.5\cdot4^2+4\cdot2^3-27:3^2=5\cdot16+4\cdot8-27:9\)

\(=80+32-3=109.\)

\(b.\left\{56-\left[\left(2\cdot15+48\cdot5\right):9\right]\right\}-5=\left\{56-\left[30+240\right]:9\right\}-5\)

\(=\left\{56-30\right\}-5=26-6=21.\)

\(c.360:\left[300:150-\left(2\cdot5-2\cdot25\right)\right]\)

\(=360:\left[300:150-\left(-40\right)\right]=360:42=\dfrac{360}{12}=\dfrac{60}{7}=8\dfrac{4}{7}.\)

\(A=\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)+x\)

\(A=x^3+1-x^3+1+x\)

\(A=2+x\) \(tại\) \(x=-0,001.\)

\(A=2-\left(0,001\right)\)

\(A=1.\)

\(7-x^1=5\)

 \(7-x=5\)

    \(-x=5-7\)

    \(-x=-2\)

       \(x=2.\)

\(Vậy...\)

\(a.x^2-25=x^2-5^2=\left(x+5\right)\left(x-5\right).\)

\(b.x^6-y^4=\left(x^3\right)^2-\left(y^2\right)^2=\left(x^3+y^2\right)\left(x^3-y^2\right).\)

\(c.9x^2-\dfrac{1}{16}y^2=\left(3x\right)^2-\left(\dfrac{1}{4}y\right)^2=\left(3x+\dfrac{1}{4}y\right)\left(3x-\dfrac{1}{4}y\right)\)

\(=\left(3x+\dfrac{1}{4}y\right)\left(3x-\dfrac{1}{4}y\right).\)

\(d.81-\left(3x+2\right)^2=9^2-\left(3x+2\right)^2=\left(11+3x\right)\left(-3x+7\right).\)

\(a.2x^3-3x^2+3x-1=2x^3-2x^2-x^2+2x+x-1\)

\(=\left(x-1\right)\left(2x^2-x+1\right).\)

\(b.\left(x^2-2xy\right)^2+2\left(x^2-2xy\right)y^2+y^4\)

\(=x^4-4x^3y+4x^2y^2+2x^2y^2-4xy^3+y^4\)

\(=x^4-4x^3y+6x^2y^2-4xy^3+y^4\)

  \(\dfrac{4}{9}:\left(x+0,4\right)=\dfrac{2}{3}\)

\(\dfrac{4}{9}:\left(x+\dfrac{4}{10}\right)=\dfrac{2}{3}\)

                    \(60=18\left(5x+2\right)\)

      \(18\left(5x+2\right)=60\)

          \(\left(5x+2\right)=60:18\)

             \(5x+2=\dfrac{10}{3}.\)

                   \(5x=\dfrac{10}{3}-2\)

                   \(5x=\dfrac{4}{3}\)

                     \(x=\dfrac{4}{3}:5\)

                     \(x=\dfrac{4}{15}.\)

\(Vậy...\)

\(-\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=1\dfrac{2}{3}\)

\(-\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{5}{3}\)

       \(-\left(x+\dfrac{1}{2}\right)=\dfrac{5}{3}+\dfrac{3}{4}\)

       \(-\left(x+\dfrac{1}{2}\right)=\dfrac{29}{12}.\)

              \(2x+\dfrac{1}{2}=-\dfrac{29}{12}\)

               \(2x+1=-\dfrac{29}{12}\cdot2\)

               \(2x+1=-\dfrac{29}{6}\)

                     \(2x=-\dfrac{29}{6}-1\)

                    \(2x=-\dfrac{35}{6}.\)

                     \(x=-\dfrac{35}{6}:2\)

                    \(x=-\dfrac{35}{12}.\)

                    \(x=-2\dfrac{11}{12}.\)

\(Vậy...\)

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