rút gọn :
(a - 1)3 - 4m(m + 1)(m - 1) + 3(a - 1)(a2 + 2 + 1)
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\(N=a^3+b^3+3ab\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\)
=1
\(M=\left(a^2+b^2+2-a^2-b^2+2\right)\left[\left(a^2+b^2+2\right)^2+\left(a^2+b^2+2\right)\left(a^2+b^2-2\right)+\left(a^2+b^2-2\right)^2\right]-12\left(a^2+b^2\right)^2\\ M=4\left(a^4+b^4+4+4a^2+4b^2+2a^2b^2+\left(a^2+b^2\right)^2-4+a^4+b^4+4-4a^2-4b^2+2a^2b^2\right)-12\left(a^4+2a^2b^2+b^4\right)\\ M=4\left(3a^4+3b^4+4+6a^2b^2\right)-12\left(a^4+2a^2b^2+b^4\right)\\ M=4\left(3a^4+3b^4+4+6a^2b^2-3a^4-6a^2b^2-3b^4\right)\\ M=4\cdot4=164\)
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\(A=1+\frac{3}{m-1}\cdot\sqrt{m^2-4m+4}\)
\(=1+\frac{3}{m-1}\cdot\sqrt{\left(x-2\right)^2}\)
\(=1+\frac{3}{m-1}\cdot\left|x-2\right|\)
\(=\frac{m-1}{m-1}+\frac{3\left(m-2\right)}{m-1}\)
\(=\frac{4m-7}{m-1}\)
\(B=\sqrt{1-10x+25x^2}-4x\)
\(=\sqrt{\left(5x-1\right)^2}-4x\)
\(=\left|5x-1\right|-4x\)
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1:
\(=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{2}{3\sqrt{x}-6}\right):\dfrac{2\sqrt{x}+3}{3\sqrt{x}}\)
\(=\dfrac{3+2\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{3\sqrt{x}}{2\sqrt{x}+3}=\dfrac{1}{\sqrt{x}-2}\)
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b: Ta có: \(N=a^3+b^3+3ab\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\)
\(=1-3ab+3ab\)
=1
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Tại a = -9 ta được:
= 3√-(-9) - |3 + 2(-9)|
= 3√32 - |3 - 18|
= 3.3 - |-15| = 9 - 15 = -6
Tại a = √2 ta được:
= |1 - 5√2| - 4√2
= (5√2 - 1) - 4√2
= √2 - 1
Tại x = -√3 ta được:
= 4(-√3) - |3(-√3) + 1|
= -4√3 - |-3√3 + 1|
= -4√3 - (3√3 - 1)
= -7√3 + 1
Có lẽ bạn bị nhầm đề giữa:
(a - 1)3 - 4m(m + 1)(m - 1) + 3(a - 1)(a2 + 2 + 1)
và \(\left(a-1\right)^3-4m\left(m+1\right)\left(m-1\right)+3\left(a-1\right)\left(a^2+a+1\right)\) nên mk sẽ làm 2 cái luôn ^-^