Question

1.tìm GTNN của

A=x²-4x+1

B=x²-5x-2

C=-25x²-x+1

D=2x²+3x+1

E=3x²+12x-1

2.Tìm x biết

a,(4x-3)*(4x+3)-15*(x-1)*(x+1)-(x+6)-3x=1

b,(5x+1)*(5x-1)-25*(x+3)*(x-1)=4

Answer 1

Bài 2 :

a )

\(\left(4x-3\right)\left(4x+3\right)-15\left(x-1\right)\left(x+1\right)-\left(x+6\right)-3x=1\)

\(\Leftrightarrow16x^2-9-15x^2+15-x-6-3x=1\)

\(\Leftrightarrow x^2-4x-1=0\)

\(\Delta=16+4=20>0\)

\(\Rightarrow\left[{}\begin{matrix}x_1=\dfrac{4+\sqrt{20}}{2}=2+\sqrt{5}\\\dfrac{4-\sqrt{20}}{2}=2-\sqrt{5}\end{matrix}\right.\)

Vậy \(x=2-\sqrt{5}\) hoặc \(x=2+\sqrt{5}\)

b )

\(\left(5x+1\right)\left(5x-1\right)-25\left(x+3\right)\left(x-1\right)=4\)

\(\Leftrightarrow25x^2-1-25x^2-50x+75=4\)

\(\Leftrightarrow-50x+70=0\)

\(\Leftrightarrow x=\dfrac{70}{50}\)

Vậy \(x=\dfrac{70}{50}\)

Answer 2

1) A=x²-4x+4-3=(x-2)²-3

(x-2)²≥0 (Với mọi x)

=> (x-2)²-3 ≥ -3 (V...)

=> Min A=-3

Làm tương tự với những câu khác nha

Answer 3

2) a)( 4x-3)(4x+3)-15(x-1)(x+1)-x-6-3x=1

<=> 16x²-9-15x²+15-4x-6=1

<=> x²-4x=1 <=> x²-4x-1=0 <=> (x-2)²=5 =>x-2=\(\sqrt{5}\) => x=\(\sqrt{5}+2\)

làm tương tự nha

Answer 4

Gợi ý bạn câu b) x=1,4