\(\Leftrightarrow4x^2-12x-4x^2+9=-3\)

=>-12x=-12

hay x=1

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

\(4x^2-12x-4x^2+9+3=0\)

\(12-12x=0\\ \Rightarrow1-x=0\\ \Rightarrow x=1\)

=>(x-2)(3x+5)=0

=>x=2 hoặc x=-5/3

3x(x-2)+5x-10=0

3x²-6x+5x-10=0

3x²-x-10=0

\(\Rightarrow\)\(\left\{{}\begin{matrix}x=2\\x=\dfrac{-5}{3}\end{matrix}\right.\)

\(a,\left(x-2\right)^2-x\left(x+2\right)=20\\ \Leftrightarrow x^2-4x+4-x^2-2x=20\\ \Leftrightarrow-6x+4=20\\ \Leftrightarrow-6x=16\\ \Leftrightarrow x=-\dfrac{8}{3}\)

\(\Leftrightarrow x^2-4x+4-x^2-2x=20\)

=>-6x=16

hay x=-8/3

b: Ta có: \(\left(4x^4-3x^3\right):\left(-x^3\right)+\left(15x^2+6x\right):3x=0\)

\(\Leftrightarrow-4x+3+5x+2=0\)

\(\Leftrightarrow x=-5\)

ChọnC

C.

ĐKXĐ: x≥0

Ta có: (\(\sqrt{5+\sqrt{x}}\))\(^2\)= 4\(^2\) ⇔ l\(5+\sqrt{x}\)l = 16

TH1) \(5+\sqrt{x}=16\)⇔\(\sqrt{x}\)= 11⇔x = 121 (TM)

TH2) 5+\(\sqrt{x}\) = -16⇔ \(\sqrt{x}\) = -21(vô nghiệm)

Vậy x = 121=> Chọn C

\(\left(x+2\right)-2=0\)

\(\Rightarrow x+2-2=0\)

\(\Rightarrow x=0\)

\(\left(x+3\right)+1=7\)

\(\Rightarrow x+3+1=7\)

\(\Rightarrow x+4=7\)

\(\Rightarrow x=3\)

\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)

\(\Rightarrow3x=12\)

\(\Rightarrow x=4\)

\(\left(5x+4\right)-1=13\)

\(\Rightarrow5x+4-1=13\)

\(\Rightarrow5x+3=13\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=2\)

\(\left(4x-8\right)-3=5\)

\(\Rightarrow4x-8-3=5\)

\(\Rightarrow4x-11=5\)

\(\Rightarrow4x=16\)

\(\Rightarrow x=4\)

\(8-\left(2x+4\right)=2\)

\(\Rightarrow8-2x-4=2\)

\(\Rightarrow4-2x=2\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

\(7+\left(5x+2\right)=14\)

\(\Rightarrow7+5x+2=14\)

\(\Rightarrow9+5x=14\)

\(\Rightarrow5x=5\)

\(\Rightarrow x=1\)

\(5-\left(3x-11\right)=1\)

\(\Rightarrow5-3x+11=1\)

\(\Rightarrow16-3x=1\)

\(\Rightarrow3x=15\)

\(\Rightarrow x=5\)

\(\Leftrightarrow x^2-12x+36-x^2+10x=40\)

=>-2x=4

hay x=-2

\(45:\frac{2}{\frac{4}{7}}-1,25=45:\left(2.\frac{7}{4}\right)-1,25=45:\frac{7}{2}-\frac{5}{4}=45.\frac{2}{7}-\frac{5}{4}=\frac{90}{7}-\frac{5}{4}=\frac{325}{28}\)

a: \(x^2-4x-5=\left(x-5\right)\left(x+1\right)\)

b: \(x^2-3x+2=\left(x-2\right)\left(x-1\right)\)

d: \(2x^2-3x+1=\left(x-1\right)\left(2x-1\right)\)

k: \(4x^2-9=\left(2x-3\right)\left(2x+3\right)\)