\(6x^2+9x+12=6\left(x^2+2.\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{69}{8}=6\left(x+\dfrac{3}{4}\right)^2+\dfrac{69}{8}\ge\dfrac{69}{8}\)

Dấu "=" xảy ra khi \(x=-\dfrac{3}{4}\)

a. \(2x\left(x+5\right)-3\left(x+5\right)=\left(x+5\right)\left(2x-3\right)\)

b. \(\left(x-2\right)^2y^2=\left(x-2\right)y\left(x-2\right)y\)

a.

\(M=\left(x+1\right)^3+x\left(3-x\right)+x+x^3+9\\ =x^3+3x^2+3x+1+3x-3x^2+x+x^3+9\\ =2x^3+7x+10\)

b. \(M\left(-1\right)=2\left(-1\right)^3+7\left(-1\right)+10=1\)

c. M có thể nhận giá trị âm, thay x = -2 vào ta thấy liền. Xem lại đề câu này nhé bạn

\(Q=\left(a-b\right)^2=\left(a+b\right)^2-4ab=15^2-4.26=121\)

\(R=\left(a^2+b^2\right)^2-2a^2b^2=\left[\left(a+b\right)^2-2ab\right]^2-2\left(ab\right)^2=\left[15^2-2.26\right]^2-2.\left(26\right)^2=28577\)

\(P=\dfrac{\left(2a-1\right)\left(3a+1\right)+\left(5-a\right)\left(3a-1\right)}{\left(3a-1\right)\left(3a+1\right)}\)

\(=\dfrac{6a^2-a-1-3a^2+16a-5}{9a^2-1}=\dfrac{3a^2+15a-6}{9a^2-1}\)

\(=\dfrac{\left(30a^2+15a\right)-\left(27a^2-3\right)-9}{9a^2-1}\)

\(=\dfrac{3\left(10a^2+5a\right)-9-3\left(9a^2-1\right)}{9a^2-1}\)

\(=\dfrac{3.3-9-3\left(9a^2-1\right)}{9a^2-1}=-3\) (vì 10a² + 5a = 3)