Đáp án `+` Giải thích các bước giải `!`
Bài `1:`
`a)`
`25x^3-9x = 0`
`<=> x(25x^2-9) = 0`
`<=> x(5x-3)(5x+3) = 0`
`⇔` \(\left[ \begin{array}{l}x=0\\5x-3=0\\5x+3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\5x=3\\5x=-3\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{array} \right.\)
Vậy `S= {0; 3/5; -(3)/(5)}`
`b)`
`(x+4)^2-(x+1)(x-1) = 16`
`<=> x^2+8x+16-x^2+1 = 16`
`<=> 8x+17 = 16`
`<=> 8x = -1`
`<=> x = -(1)/(8)`
Vậy `S= {-(1)/(8)}`
`c)`
`(2x-1)^2+(x+3)^2-5(x+7)(x-7) = 0`
`<=> (4x^2-4x+1)+(x^2+6x+9)-5(x^2-49) = 0`
`<=> 4x^2-4x+1+x^2+6x+9-5x^2+245 = 0`
`<=> (4x^2+x^2-5x^2)+(-4x+6x)+(1+9+245) = 0`
`<=> 2x+255 = 0`
`<=> 2x = -255`
`<=> x = -(255)/2`
Vậy `S= {-(255)/2}`
Bài `2:`
`a)`
`x^2+x+1`
`= (x^2+x+1/4)+3/4`
`= (x+1/2)^2+3/4`
Vì `(x+1/2)^2 >= 0` `AA x`
`=>` `(x+1/2)^2+3/4 >= 3/4 > 0` (luôn dương)
`b)`
`9x^2-6x+2`
`= (9x^2-6x+1)+1`
`= (3x-1)^2+1`
Vì `(3x-1)^2 >= 0` `AA x`
`=>` `(3x-1)^2+1 >= 1 > 0` (luôn dương)
Bài `3:`
`a)`
`A= 4x^2-12x+10`
`= (4x^2-12x+9)+1`
`= (2x-3)^2+1`
Vì `(2x-3)^2 >= 0` `AA x`
`=> (2x-3)^2+1 >= 1`
Dấu `\text{"="}` xảy ra:
`<=> (2x-3)^2 = 0`
`<=> 2x-3 = 0`
`<=> 2x = 3`
`<=> x = 3/2`
Vậy $Min_A$ `= 1 <=> x=3/2`
`b)`
`x(x-3)`
`= x^2-3x`
`= (x^2-3x+9/4)-9/4`
`= (x-3/2)^2-9/4`
Vì `(x-3/2)^2 >= 0` `AA x`
`=>` `(x-3/2)^2-9/4 >= -(9)/4`
Dấu `\text{"="}` xảy ra:
`<=> (x-3/2)^2 = 0`
`<=> x-3/2 = 0`
`<=> x = 3/2`
Vậy $Min_B$ `= -(9)/(4) <=> x=3/2`
Bài `4:`
`a)`
`(a+b)^2 = (a-b)^2+4ab` `(1)`
Ta có:
`(a+b)^2`
`= a^2+2ab+b^2`
`= a^2+2ab-4ab+b^2+4ab`
`= (a^2-2ab+b^2)+4ab`
`= (a-b)^2+4ab` `= (1)`
`b)`
`(a-b)^2 = (a+b)^2-4ab` `(2)`
Ta có:
`(a-b)^2`
`= a^2-2ab+b^2`
`= a^2-2ab+4ab+b^2-4ab`
`= (a^2+2ab+b^2)-4ab`
`= (a+b)^2-4ab` `= (2)`
