Thu gọn biểu thức : (3a-1)^2+2(9a^2-1)+(3a+1)^2
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a: Sửa đề: \(A=\left(3a-1\right)\left(9a^2+3a+1\right)-\left(3a+1\right)\left(9a^2-3a+1\right)+2a+2\)
\(=27a^3-1-27a^3-1+2a+2=2a=2\cdot5=10\)
b: \(=4x^2+2x+1-20x^3+10x^2+4x\)
\(=-20x^3+14x^2+6x+1\)
c: \(=5x^2-20xy-4y^2+20xy=5x^2-4y^2\)
\(=5\cdot\dfrac{1}{25}-4\cdot\dfrac{1}{4}=\dfrac{1}{5}-1=-\dfrac{4}{5}\)
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cho mik hỏi là dấu căn bậc 2 là của mik số 4 hay là cả 4a^6 và các phần kia cug vậy
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A=\(\dfrac{1}{3a-2}\sqrt{\left(4-12a+9a^2\right)49a^2}=\dfrac{1}{3a-2}\sqrt{\left(2-3a\right)^249a^2}\)
\(A=7a\)
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a: Ta có: \(\left(3x-1\right)^2-2\left(5x-2\right)^2-2\left(x^2+x-1\right)\left(x-1\right)\)
\(=9x^2-6x+1-2\left(25x^2-20x+4\right)-2\left(x^3-x^2+x^2-x-x+1\right)\)
\(=9x^2-6x+1-50x^2+40x-8-2\left(x^3-2x+1\right)\)
\(=-41x^2+34x-7-2x^3+4x-2\)
\(=-2x^3-41x^2+38x-9\)
b: Ta có: \(\left(3a+1\right)^2+2\left(9a^2-1\right)+\left(3a-1\right)^2\)
\(=\left(3a+1+3a-1\right)^2\)
\(=36a^2\)
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a) \(=5\left|a\right|+3a=5a+3a=8a\)
b) \(=3\left|a^2\right|+3a^2=3a^2+3a^2=6a^2\)
c) \(=5.2\left|a^3\right|-3a^3=-10a^3-3a^3=-13a^3\)
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A = (2/3 - 1/5 + 3a) - 2(1/3 - 1/10 + a)
= 7/15 + 3a - 2.7/30 - 2a
= (3a - 2a) + (7/15 - 7/15)
= a
= \(\left(3a-1\right)^2\) + \(2\left(3a-1\right)\left(3a+1\right)\) + \(\left(3a-1\right)^2\)
= \(\left(3a-1+3a+1\right)^2\)
= \(\left(6a\right)^2\)
= \(36a^2\)
\(\left(3a-1\right)^2+2\left(9a^2-1\right)+\left(3a+1\right)^2\)
\(=\left(3a-1+3a+1\right)^2\)
\(=36a^2\)