Toán 8 Tìm X

Câu 1
a) $x^2$ + 2x = x(x+2) = 0 <=> x = 0; x = -2
b) $x^3 - 5x^2 + 6x$ = x($x^2$ - 5x + 6) = 0
<=> x = 0; $x^2$ - 5x + 6 = 0 (tính $\Delta$ sẽ ra nghiệm 2 và 3)
Câu 2
A = -5($x^2$ - $\frac{4}{5}$x) +3
=-5($x^2$ - 2$\frac{2}{5}$x + $(\frac{2}{5})^2$ ) +5$(\frac{2}{5})^2$ + 3
-5(x - $\frac{2}{5})^2$ + $\frac {19}{5}$ $\leq{\frac {19}{5}}$
Vậy GTLN là $\frac{19}{5}$ tại x = $\frac {2}{5}$

Câu 1 :
a ) [tex]x^2 + 5x = 0 \Leftrightarrow x(x+5) = 0 \Leftrightarrow \begin{bmatrix} x = 0 \\ x + 5 = 0 \end{bmatrix} \Leftrightarrow \begin{bmatrix} x = 0 \\ x = - 5 \end{bmatrix}[/tex]
b ) [tex]x^3 - 5x^2 + 6x = 0 \Leftrightarrow x(x^2 - 5x + 6)=0 \Leftrightarrow x(x^2 - 3x - 2x + 6)=0 \Leftrightarrow x[x(x-3)-2(x-3)]= 0 \Leftrightarrow x(x-2)(x-3)=0 \Leftrightarrow \begin{bmatrix} x=0\\ x-2=0 \\ x - 3=0 \end{bmatrix} \Leftrightarrow \begin{bmatrix} x=0\\ x=2 \\ x=3 \end{bmatrix}[/tex]
Câu 2 :
[tex]A = 4x- 5x^2 + 3 = -5(x^2-4/5x - 3/5) = -5[x^2 - 4/5x + 4/25 - 19/25] = -5[(x-2/5)^2 - 19/25] = -5(x-2/5)^2 + 19/5 \leq 19/5( voi moi x )[/tex]
Dấu " = " xảy ra
[tex]\Leftrightarrow (x-2/5)^2 = 0 \Leftrightarrow x - 2/5 = 0 \Leftrightarrow x = 2/5[/tex]
Vậy Max A là : 19/5 [tex]\Leftrightarrow x = 2/5[/tex]