\(\cdot16\left(2x+3\right)^2-9\left(5x-2\right)^2\) (Bài này ra số to nhg mà đó là cách duy nhất thoi ;=;)
\(=16\left(4x^2+12x+9\right)-9\left(25x^2-20x+4\right)\)
\(=64x^2+192x+144-225x^2+180x-36\)
\(=-161x^2+372x+108\)
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\(\cdot\left(x+2\right)^3-x^2\left(x-6\right)=4\)
\(\Rightarrow x^3+6x^2+12x+8-x^3+6x^2=4\)
\(\Rightarrow12x=-4\)
Vậy \(x=\dfrac{-1}{3}\)
câu 1
\(=4^2\left(2x+3\right)^2-3^2\left(5x-2\right)^2\)
\(=\left[4\left(2x+3\right)\right]^2-\left[3\left(5x-2\right)\right]^2\)
\(=\left(8x+12\right)^2-\left(15x-6\right)^2\)
\(=\left(8x+12-15x+6\right)\left(8x+12+15x-6\right)\)
\(=\left(-7x+18\right)\left(23x+6\right)\)
câu 2
\(x^3+6x^2+12x+8-x^3+6x^2=4\)
\(x^3-x^3+6x^2+6x^2+12x+8=4\)
\(12x^2+12x+8-4=0\)
\(12x^2+12x+4=0\)
câu 2 bn xem lại đề phải (x-2)^3 chứ nhỉ
