Ta có MC là \(60\)

\(\frac{-15}{20}=\frac{-45}{60}\)

\(\frac{9}{10}=\frac{54}{60}\)

\(\frac{26}{-30}=\frac{-52}{60}\)

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\(\frac01\)

\(a)\) \(A\left(x\right)+B\left(x\right)=\) \(-2x^3-2x^2+6x-x+5+\left(x^3-2x+1\right)\)

\(=-2x^3-2x^2+5x+5+x^3-2x+1\)

\(=-x^3-2x^2+3x+6\)

\(b)\) \(A\left(x\right)-B\left(x\right)=\) \(-2x^3-2x^2+6x-x+5-\left(x^3-2x+1\right)\)

\(=-2x^3-2x^2+5x+5-x^3+2x-1\)

\(=-3x^3-2x^2+7x+4\)

Vì \(n:2;3;4;5;7\) đều dư 1 thì \(\left(n-1\right)\) ⋮\(2;3;4;5;7\)

⇒ \(\left(n-1\right)\in BCNN\left(2;3;4;5;7\right)\)

Ta có: \(BCNN\left(2;3;4;5;7\right)=420\)

⇒ \(n-1=420\)

\(n\) \(=421\)

Vậy số tự nhiên \(n\) nhỏ nhất là \(421\)

\(a)\)

\(3x\left(x-1\right)-1+x=0\\ 3x\left(x-1\right)-\left(x-1\right)=0\\ \left(x-1\right)\left(3x-1\right)=0\)

\(TH1:\) \(x-1=0\\ x=1\)

\(TH2:\) \(3x-1=0\\ 3x=1\\ x=\dfrac{1}{3}\)

\(Vậy\) \(x\in\left\{0;\dfrac{1}{3}\right\}\)

\(b)\)

\(x^2-9x=0\\ x\left(x-9\right)=0\)

\(TH1:\) \(x=0\)

\(TH2:\) \(x-1=0\\ x=1\)

\(Vậy\) \(x\in\left\{0;1\right\}\)

\(Mom\) \(bought\) \(many\) \(books,\) \(for\) \(she\) \(adores\) \(reading.\)

\(a)\) \(x^2+25-10x=x^2-10x+25=x^2-2x\times5+5^2=\left(x-5\right)^2\)

\(b)\) \(-8y^3+x^3=x^3-8y^3=\left(x-2y\right)\left[x^2+x\times2y+\left(2y\right)^2\right]=\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)

\(a)\) \(^{\left(2x+1\right)^2=\left(2x\right)^2+2\times2x\times1+1^2=4x^2+4x+1}\)

\(b)\) \(\left(a-\dfrac{b}{2}\right)^3=a^3-3a^2\times\dfrac{b}{2}+3a\left(\dfrac{b}{2}\right)^2-\left(\dfrac{b}{2}\right)^3=a^3-\dfrac{3}{2}a^2b+\dfrac{3}{4}ab^2-\dfrac{1}{8}b^3\)