a)(Sửa đề) \(4(x^2-6x+9)-16(4x^2+28x+49)=0\)

\(⇔(2x-6)^2-(8x+28)^2=0\)

\(⇔(-6x-34)(10x+22)=0\)

\(⇔\left[\begin{array}{} -6x-34=0\\ 10x+22=0 \end{array}\right.\)

\(⇔\left[\begin{array}{} x=-\dfrac{17}{3}\\ x=-\dfrac{11}{5} \end{array}\right.\)

b)(Sửa đề 1) \((2x-16)^2-(x-4)^2=0\)

\(⇔(3x-20)(x-12)=0\)

\(⇔\left[\begin{array}{} 3x-20=0\\ x-12=0 \end{array}\right.\)\(⇔\left[\begin{array}{} x=\frac{20}{3}\\ x=12 \end{array}\right.\)

(Sửa đề 2) \((x^2-16)^2-(x-4)^2=0\)

\(⇔(x^2-x-12)(x^2+x-20)=0\)

\(⇔(x-4)^2(x+3)(x+5)=0\)

\(⇔\left[\begin{array}{} (x-4)^2=0\\\ x+3=0\\ x+5=0 \end{array}\right.\)\(⇔\left[\begin{array}{} x=4\\\ x=-3\\ x=-5 \end{array}\right.\)

Chọn A

A

1) \(x^2+6x+9\)

\(=\left(x+3\right)^2\)

2) \(10x-25-x^2\)

\(=-25+10x-x^2\)

\(=-\left(5-x\right)^2\)

3) \(8x^3-\dfrac{1}{8}\)

\(=\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3\)

\(=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

4) \(\dfrac{1}{25}x^2-64y^2\)

\(=\left(\dfrac{1}{5}x\right)^2-\left(8y\right)^2\)

\(=\left(\dfrac{1}{5}x+8y\right)\left(\dfrac{1}{5}x-8y\right)\)

\(x^2+6x+9=\left(x+3\right)^2\)

\(10x-25-x^2=-\left(x-5\right)^2\)

\(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

\(a,=x^2-4-x^2+2x+4=2x\\ b,=\left(x-5y\right)^2:\left(5y-x\right)=\left(5y-x\right)^2:\left(5y-x\right)=5y-x\\ c,Sửa:\left(28x-9x^2+x^3-30\right):\left(x-3\right)\\ =\left(x^3-3x^2-6x^2+18x+10x-30\right):\left(x-3\right)\\ =\left(x-3\right)\left(x^2-6x+10\right)\left(x-3\right)=x^2-6x+10\)

3: \(\Leftrightarrow a-15=0\)

hay a=15

√(x²-6x+11) + √(x²-6x+13) + √(x²-4x+5) = 3+√2 (1)

Có: \(\sqrt{x^2-6x+11}=\sqrt{\left(x-3\right)^2+2}\ge\sqrt{2}\)

(Dấu = xảy ra khi x = 3)

\(\sqrt{x^2-6x+13}=\sqrt{\left(x-3\right)^2+4}\ge\sqrt{4}=2\)

(Dấu = xảy ra khi x = 3)

\(\sqrt{x^2-4x+5}=\sqrt{\left(x-2\right)^2+1}\ge1\)

(Dấu = xảy ra khi x = 2)

Nhận xét PT (1):

\(VT\ge3+\sqrt{2}\)

\(VP=3+\sqrt{2}\)

Nên: √(x²-6x+11) + √(x²-6x+13) + √(x²-4x+5) = 3+√2 khi: x = 3 và x = 2

=> PT vô nghiệm