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\(A=\left(a+3\right)^2\left(a^2-3a+9\right)-\left(a-3\right)\left(a^2+3a+9\right)\)
\(A=\left(a^3+3^3\right)-\left(a^3-3^3\right)\)
\(A=a^3+3^3-a^3+3^3\)
\(A=3^3+3^3=54\)
a) (a2 - 1)3- ( a4 + a2+1)(a2-1)
= (a2 - 1)3 - (a2 - 1)3 =0
b) (a4 - 3a2+ 9)(a2+3) - (3+a2)3
= (3+a2)3 - (3+a2)3
=0
\(A=9+3a+3a^2+a^3\)
\(=\left(a+3\right)^3-18\)
\(=3,5^3-18=24,875\)
ta có : \(\frac{a^3-4a^2-a+4}{a^3-7a^2+14a-8}\)
= \(\frac{\left(a-1\right)\left(a+1\right)\left(a-4\right)}{\left(a-4\right)\left(a-2\right)\left(a-1\right)}\)
= \(\frac{a+1}{a-2}\)
nhớ nha
\(\left(a+3\right)\left(a^2-3a+9\right)-\left(a-3\right)\left(a^2+3a+9\right)\)
\(=a^3+27-\left(a^3-27\right)\)
\(=a^3+27-a^3+27\)
\(=54\)
Bài 1 :
a) (3a+4b)3+(3a-4b)3-48a2b2
=27a3+108a2b+144ab2+64b3+27a3-108a2b+144ab2-64b3-48a2b2
=54a3+288ab2-48a2b2
=2a(27a2+144b2-24ab)
b) (5x+2y)(5x-2y)+(2x-y)3+(2x+y)3
=25x2-4y2+8x3-12x2y+6xy2-y3+8x3+12x2y+6xy2+y3
=16x3+25x2-y2+12xy2
=x2(16x+25)-y2(1-12x)
Bài 2 :
\(x^2-8x+7=0\)
\(\Leftrightarrow x^2-x-7x+7=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-7=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=7\end{cases}}\)
b)\(x^3-4x^2+3x=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm\sqrt{3}\\x=1\end{cases}}\)
c)Nếu đề đổi thành =1 thì có vẻ hợp lí hơn
d)\(\left(3x-1\right)^3-3\left(3x+2\right)^2+13=0\)
\(\Leftrightarrow27x^3-27x^2+9x-1-3\left(9x^2+12x+4\right)+13=0\)
\(\Leftrightarrow27x^3-27x^2+9x-1-27x^2-36x-12+13=0\)
\(\Leftrightarrow27x^3-54x^2-27x=0\)
\(\Leftrightarrow27x\left(x^2-2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}27x=0\\x^2-2x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\-\left(x^2+2x+1\right)=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\-\left(x+1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
#H
A = (22 - 1) (22 +1)(24 +1)...(264 +1) + 1 = (24 - 1)(24 +1)...(264 +1) + 1 = (28 -1)...(264 +1) + 1 = 2128 -1 + 1 = 2128
(a4 - 3a2 + 9).(a2 + 3) - (3 + a2)3
= a6 + 27 - 27 - 9a4 - 27a2 - a6
= -9a4 - 27a2
\(\left(a^4-3a^2+9\right)\left(a^2+3\right)-\left(3+a^2\right)^3\)
\(=\left[\left(a^2\right)^3+3^3\right]-\left(27+9a^4+27a^2+a^6\right)\)
\(=a^6+27-27-9a^4-27a^2-a^6\)
\(=-9a^4-27a^2\)