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https://hoc24.vn/hoi-dap/question/177629.html (câu 2)
https://hoc24.vn/hoi-dap/question/940816.html?pos=2486212 (câu 1)Bài 1:
\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)
\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)
Bài 2:
\(2x^2-6x+xy-3y=2x.\left(x-3\right)+y.\left(x-3\right)=\left(x-3\right).\left(2x+y\right)\)
Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9
Bài 4:
\(\left(\frac{1}{x+1}+\frac{2x}{1-x^2}\right).\left(\frac{1}{x-1}\right)\) \(\frac{1}{\left(x+1\right).\left(x-1\right)}+\frac{2x}{\left(1-x^2\right).\left(x-1\right)}=\frac{x-1}{\left(x+1\right).\left(x-1\right)^2}+\frac{-2x}{\left(x-1\right)^2.\left(x+1\right)}\)
= \(\frac{x-1-2x}{\left(x+1\right).\left(x-1\right)^2}=\frac{-\left(x+1\right)}{\left(x+1\right).\left(x-1\right)^2}=\frac{-1}{\left(x-1\right)^2}\)
a)\(\left(x-2\right)\left(x+3\right)-\left(2x-1\right)^2\)
\(=x^2+x-6-\left(4x^2-4x+1\right)\)
\(=x^2+x-6-4x^2+4x-1\)
\(=5x-3x^2-7\)
b)\(\left(2a+3\right)^2-a\left(5a-2\right)\)
\(=4a^2+12a+9-5a^2+2a\)
\(=14a-a^2+9\)
c)\(3a\left(a-1\right)\left(a+2\right)-\left(3a+1\right)\left(1-3a\right)\)
\(=\left(3a^2-3a\right)\left(a+2\right)+9a^2-1\)
\(=3a^3+3a^2-6a+9a^2-1\)
\(=3a^3+12a^2-6a-1\)
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: \(=5x^2-20xy-4y^2+20xy=5x^2-4y^2\)
\(=5\cdot\dfrac{1}{25}-4\cdot\dfrac{1}{4}=\dfrac{1}{5}-1=-\dfrac{4}{5}\)