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:

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

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

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

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

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

\(20x^2+24x+18=500\)    

\(20x^2+24x-482=0\) 

\(10x^2+12x-241=0\) 

\(\orbr{\begin{cases}x=\frac{-6+\sqrt{2446}}{10}\\x=\frac{-6-\sqrt{2446}}{10}\end{cases}}\)

20x² + 24x + 18 = 500

<=> 20x² + 24x + 18 - 500 = 0

<=> 20x² + 24x - 482 = 0

<=> 2( 10x² + 12x - 241 ) = 0

<=> 10x² + 12x - 241 = 0 (*)

\(\Delta'=b'^2-ac=\left(\frac{b}{2}\right)^2-ac=6^2-10\cdot\left(-241\right)=36+2410=2446\)

\(\Delta'>0\)nên (*) có hai nghiệm phân biệt :

\(\hept{\begin{cases}x_1=\frac{-b'+\sqrt{\Delta'}}{a}=\frac{-6+\sqrt{2446}}{10}\\x_2=\frac{-b'-\sqrt{\Delta'}}{a}=\frac{-6-\sqrt{2446}}{10}\end{cases}}\)

Lớp 8 sao nghiệm xấu thế nhỉ ;-;

Ta có: \(20x^2+24x+18=500\)

    \(\Leftrightarrow20x^2+24x-482=0\)

    \(\Leftrightarrow100x^2+120x-2410=0\)

    \(\Leftrightarrow\left(100x^2+120x+36\right)-2446=0\)

    \(\Leftrightarrow\left(10x+6\right)^2=2446\)

    \(\Leftrightarrow10x+6=\pm\sqrt{2446}\)

 + \(10x+6=\sqrt{2446}\)\(\Leftrightarrow x=\frac{\sqrt{2446}-6}{10}\approx4,345705208\)

 + \(10x+6=-\sqrt{2446}\)\(\Leftrightarrow x=\frac{-\sqrt{2446}-6}{10}\approx-5,545705208\)

Vậy \(S=\left\{4,345705208;-545705208\right\}\)

\(A=x^2+3x-5=x^2+3x+\frac{9}{4}-\frac{29}{4}\)

\(=\left(x+\frac{3}{2}\right)^2-\frac{29}{4}\ge-\frac{29}{4}\)

Vậy \(A_{min}=-\frac{29}{4}\Leftrightarrow x+\frac{3}{2}=0\Leftrightarrow x=-\frac{3}{2}\)

16x+20x+24x+................+40x+44x=4800

=(16+20+24+.....+40+44)x=4800

=240x=4800

x=20

 16x + 20x + 24x + .............. + 40x + 44x = 4800

(16 + 20 + 24 + ..................+ 40 + 44). x = 4800

\(\frac{\left(16+44\right).8}{2}\). x = 4800

240 . x = 4800

x = 4800 : 240 

x = 20 

đáp số: 20

16x+20x+24x+........+40x+44x=4800

(16+20+.......+40+44)x=4800

240x=4800

     x=4800:240

     x=20