`(24x^5-18x^4+30x^3) \div 6x^3`

`= 24x^5 \div 6x^3 -18x^4 \div 6x^3 + 30x^3 \div 6x^3`

`= 4x^2-3x+5`

a) 3x² .(2x² - 3yz + x³)= 6x⁴ - 6x²yz +3x⁵

b)(24x⁵ - 12x⁴ + 6x² ).6x² = 144x⁷ - 72x⁶ +36x⁴

a) 3x² . (2x² - 3yz + x³)

= 3x² . 2x² + 3x² . (- 3yz) + 3x² . x³

= 6x⁴ + (-9x²yz) + 3x⁵

= 6x⁴ - 9x²yz + 3x⁵

b) (24x⁵ - 12x⁴ + 6x²) . 6x²

= 24x⁵ . 6x² - 12x⁴ . 6x² + 6x² . 6x²

= 144x⁷ - 72x⁶ + 36x⁴

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

\(6x\left(4x-5\right)-24x^2=24x^2-30x-24x^2=-30x\)

ý B

Lượng giác hóa nghĩa là sử dụng kiến thức 11 thoải mái đúng ko nhỉ?

\(\Leftrightarrow6x+7+\sqrt[3]{6x+7}=\left(2x+2\right)^3+2x+2\)

Hàm \(f\left(t\right)=t^3+t\) có \(f'\left(t\right)=3t^2+1>0\Rightarrow f\left(t\right)\) đồng biến trên R

\(\Rightarrow\left(1\right)\Leftrightarrow2x+2=\sqrt[3]{6x+7}\Leftrightarrow\left(6x+7\right)-1=3\sqrt[3]{6x+7}\)

Đặt \(\sqrt[3]{6x+7}=t\Rightarrow t^3-3t-1=0\)

Xét hàm \(f\left(t\right)=t^3-3t-1\) bậc 3 nên có tối đa 3 nghiệm

\(f\left(-2\right).f\left(-1\right)=\left(-3\right).1< 0\) ; \(f\left(-1\right).f\left(0\right)=-1< 0\) ; \(f\left(0\right).f\left(2\right)=-1.1< 0\)

\(\Rightarrow\) Cả 3 nghiệm của t đều thuộc \(\left[-2;2\right]\)

\(\Rightarrow\dfrac{t}{2}\in\left[-1;1\right]\Rightarrow\) đặt \(\dfrac{t}{2}=cosu\) hay \(t=2cosu\)

Pt trở thành:

\(8cos^3u-6cosu-1=0\Leftrightarrow4cos^3u-3cosu=\dfrac{1}{2}\)

\(\Leftrightarrow cos3u=\dfrac{1}{2}\Rightarrow3u=\pm\dfrac{\pi}{3}+k2\pi\)

\(\Rightarrow u=\pm\dfrac{\pi}{9}+\dfrac{k2\pi}{3}\)

\(\Rightarrow t=2cosu=\left\{2cos\dfrac{\pi}{9};2cos\dfrac{5\pi}{9};2cos\dfrac{7\pi}{9}\right\}\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{6x+7}=2cos\dfrac{\pi}{9}\\\sqrt[3]{6x+7}=2cos\dfrac{5\pi}{9}\\\sqrt[3]{6x+7}=2cos\dfrac{7\pi}{9}\end{matrix}\right.\) \(\Rightarrow x=...\)

\(B=\dfrac{\sqrt{24x^3}}{\sqrt{6x}}=\dfrac{2x\sqrt{6x}}{\sqrt{6x}}=2x\)

\(B=\dfrac{\sqrt{24x^3}}{\sqrt{6x}}\)

\(=\sqrt{\dfrac{24x^3}{6x}}\)

\(=\sqrt{4x^2}=2x\)