RÚT GỌN BIỂU THỨC
a, A= (5a+5) ² + 10 (a - 3) (1 + a) +a ² - 6a + 9
b, B= (a-b) ³ - 3(b-a) ² . c + 3(a-b) . c ² - c ³v
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\(a,\left(2a-3\right)\left(a+1\right)+\left(a^2+6a+9\right):\left(a+3\right)\\ =2a^2-a-3+\left(a+3\right)^2:\left(a+3\right)\\ =2a^2-a-3+a+3\\ =2a^2\\ b,\left(3x-5y\right)\left(-xy\right)^2-3x^2y^2+4x^2y^3\\ =3x^3y^2-5x^2y^3-3x^2y^2+4x^2y^3\\ =3x^3y^2-3x^2y^2-x^2y^3\\ c,x\left(x-2\right)^2-\left(x+2\right)\left(x^2-2x+4\right)+4x^2\\ =x^3-4x^2+4x-x^3-8+4x^2\\ =4x-8\)
\(A=\left(a+b+c\right)^3+\left(a-b-c\right)^3-6a\left(b+c\right)^2\)
\(=\left[a+\left(b+c\right)\right]^3+\left[a-\left(b+c\right)\right]^3-6a\left(b+c\right)^2\)
\(=a^3+3a^2\left(b+c\right)+3a\left(b+c\right)^2+\left(b+c\right)^3+a^3-3a^2\left(b+c\right)+3a\left(b+c\right)^2-\left(b+c\right)^3-6a\left(b+c\right)^2\)
\(=2a^3\)
a) a ≠ 0 , a ≠ − 5
b) Ta có A = a 3 + 4 a 2 − 5 a 2 a ( a + 5 ) = a ( a − 1 ) ( a + 5 ) 2 a ( a + 5 ) = a − 1 2
c) Thay a = -1 (TMĐK) vào a ta được A = -1
d) Ta có A = 0 Û a = 1 (TMĐK)
\(B=\dfrac{\left(a+3\right)^2}{2a^2+6a}\cdot\dfrac{1-6a-18}{a^2-9}\\ a,ĐK:a\ne0;a\ne\pm3\\ b,B=\dfrac{\left(a+3\right)^2}{2a\left(a+3\right)}\cdot\dfrac{-17-6a}{\left(a-3\right)\left(a+3\right)}=\dfrac{-17-6a}{2a\left(a-3\right)}\\ c,B=0\Leftrightarrow-17-6a=0\Leftrightarrow a=-\dfrac{17}{6}\left(tm\right)\\ d,B=1\Leftrightarrow-17-6a=2a^2-6a\\ \Leftrightarrow2a^2=-17\Leftrightarrow a\in\varnothing\)
a) \(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}=\left|3a^2\right|=3a^2\)
b) \(2\sqrt{a^2}-5a=2\left|a\right|-5a=-2a-5a=-7a\)
c) \(\sqrt{16\left(1+4x+4x^2\right)}=\sqrt{\left[4\left(1+2x\right)\right]^2}=\left|4\left(1+2x\right)\right|=4\left(1+2x\right)\)