\(\left(8x+5\right)^2\left(4x+3\right)\left(2x+1\right)=9\)

\(\Leftrightarrow\left(64x^2+8x+25\right)\left(8x^2+10x+3\right)-9=0\)

Đặt \(a=8x^2+10x+3\)

\(\left(8a+1\right)a-9=0\)

\(\Leftrightarrow8a^2+a-9=0\)

\(\Leftrightarrow\left(a-1\right)\left(8a+9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}a=1\\a=-\frac{9}{8}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}8x^2+10x+3=1\\8x^2+10x+3=-\frac{9}{8}\end{cases}}\)

Mà \(8x^2+10x+3>0\Rightarrow8x^2+10x+3>-\frac{9}{8}\)

\(\Rightarrow8x^2+10x+3=1\Rightarrow8x^2+10x+2=0\Rightarrow2\left(x+1\right)\left(4x+1\right)=0\Rightarrow\orbr{\begin{cases}x=-1\\x=-\frac{1}{4}\end{cases}}\)

MK K BIET !

XIN LOI NHA BAN !

5B=-25x² -20x+5 = 9 - (25x² +20x +4) = 9- (5x+2)² \(\le9\)

=> B\(\le\frac{9}{5}\)<=> x=-2/5

Tìm GTLN của: \(B=-5x^2-4x+1\)

Ta có 

\(B=-5x^2-4x+1\)

\(B=-5\left(x^2+\frac{4}{5}x-\frac{1}{5}\right)\)

\(B=-5\left[x^2+2x.\frac{2}{5}+\left(\frac{2}{5}\right)^2-\frac{4}{25}-\frac{5}{25}\right]\)

\(B=-5\left[\left(x+\frac{2}{5}\right)^2-\frac{9}{25}\right]\)

\(B=-5\left(x+\frac{2}{5}\right)^2+\frac{9}{5}\)

Mà \(-5\left(x+\frac{2}{5}\right)^2\le0\). Dấu "=" xảy ra khi và chỉ khi \(x=\frac{-2}{5}\)

=> \(-5\left(x+\frac{2}{5}\right)^2+\frac{9}{5}\le\frac{9}{5}\). Dấu "=" xảy ra khi và chỉ khi \(x=\frac{-2}{5}\)

Vậy B có GTLN bằng \(\frac{9}{5}\)khi \(x=\frac{-2}{5}\).

Tìm GTLN của: \(C=-2x^2+10x+3\)

Ta có

\(C=-2x^2+10x+3\)

\(C=-2\left(x^2-5x-\frac{3}{2}\right)\)

\(C=-2\left[x^2-2x.\frac{5}{2}+\left(\frac{5}{2}\right)^2-\frac{25}{4}-\frac{9}{4}\right]\)

\(C=-2\left[\left(x-\frac{5}{2}\right)^2-\frac{17}{2}\right]\)

\(C=-2\left(x-\frac{5}{2}\right)^2+17\)

Mà \(-2\left(x-\frac{5}{2}\right)^2\le0\). Dấu "=" xảy ra khi và chỉ khi \(x=\frac{5}{2}\)

=> \(-2\left(x-\frac{5}{2}\right)^2+17\le17\). Dấu "=" xảy ra khi và chỉ khi \(x=\frac{5}{2}\)

Vậy C có GTLN bằng 17 khi \(x=\frac{5}{2}\)

con GTNN duoc khong ban

Ta có: \(\left(x-2\right)\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)

Th1: x = 2

\(\Rightarrow\left(4.2-5\right)\left(2-7\right)=3.\left(-5\right)=-15\)

Th2: x = -3

\(\Rightarrow\left[4.\left(-3\right)-5\right].\left[\left(-3\right)-7\right]=-17.\left(-10\right)=170\)

vậy ..

hok tốt!!

May ban oi cau hoi nay la rut gon xong roiu moi tinh nah

(2x-1)(4x^2x+1)+(3+2x)(9-6x+4x^2)-7

= 8x^4+4x^3+2x+19

nha bạn chúc bạn học tốt nha 

cam on ban tiep nha

Đặt \(A\) , ta có :

\(A=\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)+3\left(x-1\right)\left(x^2+x+1\right)\)

\(A=\left(x-1\right)^3-4x.\left(x^2-1^2\right)+3.\left(x^3-1\right)\)

Thay \(x=2\) vào biểu thức , ta có :

\(A=\left(-2-1\right)^3-4.\left(-2\right).\left[\left(-2\right)^2-1\right]+3.\left[\left(-2\right)^3-1\right]\)

\(A=\left(-3\right)^3+8.3+3.\left(-9\right)\)

\(A=-27+24-27\)

\(A=-30\)

\(A=\left(x-1\right)^3-4x.\left(x+1\right).\left(x-1\right)+3.\left(x-1\right).\left(x^2+x+1\right)\)

\(A=\left(x-1\right)^3-4x.\left(x^2-1^2\right)+3.\left(x^3-1\right)\)

Thay x=2 vào biểu thức ta có

\(A=\left(-2-1\right)^3-4.\left(-2\right).\left[\left(-2\right)^2-1\right]+3.\left[\left(-2\right)^3-1\right]\)

\(A=\left(-3\right)^3+8.3+3.\left(-9\right)\)

\(A=-27+24-27\)

\(A=-30\)