\(40:\left\{\left[11+\left(26-3^3\right)\right].2\right\}\)

\(=40:\left\{\left[11+\left(26-27\right)\right].2\right\}\)

\(=40:\left\{\left[11-1\right].2\right\}\)

\(=40:\left\{10.2\right\}\)

\(=40:20\)

\(=2\)

\(\left(2x+1\right)^4-3\left(2x+1\right)^2+2\)

\(=\left[\left(2x+1\right)^2\right]^2-2.\left(2x+1\right)^2.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{1}{4}\)

\(=\left[\left(2x+1\right)^2-\dfrac{9}{4}\right]^2-\left(\dfrac{1}{2}\right)^2\)

\(=\left[\left(2x+1\right)^2-\dfrac{9}{4}+\dfrac{1}{2}\right].\left[\left(2x+1\right)^2-\dfrac{9}{4}-\dfrac{1}{2}\right]\)

mình ko chắc là có đúng ko nếu có sai thì mong bạn nào đó giúp mình

\(53.39+47.40-53.21-47.42\)

\(=2067+1880-1113-1034\)

\(=3947-1113-1034\)

\(=2834-1034\)

\(=1800\)

\(8x^3-12x^2+6x-1=0\)

\(\left(2x\right)^3-3.\left(2x\right)^2.1+3.2x.1^2-1^3=0\)

\(\left(2x-1\right)^3=0\)

\(=>2x-1=0\)

\(2x=1\)

\(x=\dfrac{1}{2}\)

e)

\(x^4+2x^2-8\)

\(=x^4+4x^2-2x^2-8\)

\(=\left(x^4-2x^2\right)+\left(4x^2-8\right)\)

\(=x^2\left(x^2-2\right)+4\left(x^2-2\right)\)

\(=\left(x^2+4\right).\left(x^2-2\right)\)

f)

\(x^2-2xy-3y^2\)

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

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

\(=x\left(x-y\right)+3y\left(x-y\right)\)

\(=\left(x+3y\right).\left(x-y\right)\)

1.

\(\left(a+3b\right)^2-9b^2\)

\(=\left(a+3b\right)^2-\left(3b\right)^2\)

\(=\left(a+3b\right).\left(a-3b\right)\)

\(=a\left(a+6\right)\)

2.

\(4a^2-\left(a+b\right)^2\)

\(=\left(2a\right)^2-\left(a+b\right)^2\)

\(=\left(2a-a+b\right).\left(2a+a+b\right)\)

\(=\left(a-b\right).\left(3a+b\right)\)

3.

\(x^3+3x^2+3x+1\)

\(=x^3+3.x^2.1+3.x.1^2+1^3\)

\(=\left(x+1\right)^3\)