b: 

1: \(\Leftrightarrow2x\left(x+2\right)=0\)

=>x=0 hoặc x=-2

d) \(4x^2-9-x\left(2x-3\right)=0\)

\(\Leftrightarrow4x^2-9-2x^2+3x=0\)

\(\Leftrightarrow2x^2+3x-9=0\)

\(\Delta=3^2-4.2.\left(-9\right)=9+72=81\)

Vậy pt có 2 nghiệm phân biệt

\(x_1=\frac{-3+\sqrt{81}}{4}=\frac{-3}{2}\);\(x_1=\frac{-3-\sqrt{81}}{4}=-3\)

e) \(x^3+5x^2+9x=-45\)

\(\Leftrightarrow x^3+5x^2+9x+45=0\)

\(\Leftrightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)

\(\Leftrightarrow\left(x^2+9\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm3i\\x=-5\end{cases}}\)

f) \(x^3-6x^2-x+30=0\)

\(\Leftrightarrow\left(x^3-x^2-6x\right)-\left(5x^2-5x-30\right)=0\)

\(\Leftrightarrow x\left(x^2-x-6\right)-5\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-2x+3x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left[x\left(x-2\right)+3\left(x-2\right)\right]=0\)

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

\(\Leftrightarrow x\in\left\{5;-3;2\right\}\)

\(\left(x^2+2x\right)^2-6x^2+12x+9=0\Leftrightarrow x^4+4x^3+4x^2-6x^2+12x+9=0\\ \Leftrightarrow x^4+4x^3-2x^2+12x+9=0\Leftrightarrow x^2+4x-2+\frac{12}{x}+\frac{9}{x^2}=0\\ \Leftrightarrow\left(x^2+\frac{9}{x^2}\right)+4\left(x+\frac{3}{x}\right)-2=0\)

Đặt \(k=x+\frac{3}{x}\Rightarrow x^2+\frac{9}{x^2}=k^2-6\)

Ta đc \(k^2-6+4k-2=0\Leftrightarrow k^2+4k-8=0\)

\(\left(x^2+2x\right)^2\)\(-6x^2\)\(+12x+9\)=0

⇔\(\left(x^2\right)^2\)\(+2.2x.x^2\)+\(2x^2\)-6x²+12x+9=0

⇔ x⁴+ 4x³+2x²-6x²+12x+9=0

⇔ x²+4x³-4x² +12x=-9

⇔x²+ 4x(x-x+3)=-9

⇔x²+12x=-9

⇔x(x+12)=-9

⇔ {x=-9 hoặc x+12=-9}

⇔ {x=-9 hoặc x=-21}

S={-9;-21}

câu hỏi là gì vậy bạn

\(\left(x^2+2x\right)^2-6x^2-12x+9=0\)

\(\Leftrightarrow\left(x^2+2x\right)^2-6\left(x^2+2x\right)+9=0\)

\(\Leftrightarrow\left(x^2-2x-3\right)^2=0\)

\(\Leftrightarrow x^2-2x-3=0\)

\(\Leftrightarrow x_1=3\) \(x_2=-1\)

Vừa đăng lên thì biết cách giải.Đề là giải pt nha :D

`Answer:`

a. \(x^3+6x^2+12=19\)

\(\Leftrightarrow x^3+6x^2+12x-19=0\)

\(\Leftrightarrow x^3-x^2+7x^2-7x+19x-19=0\)

\(\Leftrightarrow x^2.\left(x-1\right)+7x\left(x-1\right)+19\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+7x+19\right)=0\)

Ta có \(x^2+7x+19=x^2+2x.3,5+12,25+6,75=\left(x+3,5\right)^2+6,75>0\)

\(\Rightarrow x-1=0\Leftrightarrow x=1\)

b. \(5\left(x+9\right)^2.\left(x-4\right)^3-10\left(x+9\right)^3.\left(x-4\right)^2=0\)

\(\Leftrightarrow5\left(x+9\right)^2.\left(x-4\right)^2.[x-4-2\left(x+9\right)]=0\)

\(\Leftrightarrow\left(x+9\right)^2.\left(x-4\right)^2.\left(x-4-2x-18\right)=0\)

\(\Leftrightarrow\left(x+9\right)^2.\left(x-4\right)^2.\left(-x-22\right)=0\)

\(\Leftrightarrow\left(x+9\right)^2=0\) hoặc \(\left(x-4\right)^2=0\) hoặc \(-x-22=0\)

\(\Leftrightarrow x+9=0\) hoặc \(x-4=0\) hoặc \(-x=22\)

\(\Leftrightarrow x=-9\) hoặc \(x=4\) hoặc \(x=-22\)

c. \(\left(2x+3\right)^2+\left(x-2\right)^2-2\left(2x+3\right)\left(x-2\right)\)

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

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

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

e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

f: Ta có: \(x^3-6x^2+12x-19=0\)

\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)

\(\Leftrightarrow\left(x-2\right)^3=11\)

hay \(x=\sqrt[3]{11}+2\)

a) \(\sqrt[]{x^2-4x+4}=x+3\)

\(\Leftrightarrow\sqrt[]{\left(x-2\right)^2}=x+3\)

\(\Leftrightarrow\left|x-2\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-\left(x+3\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\x-2=-x-3\end{matrix}\right.\)

\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)

\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)

\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)

\(\Leftrightarrow\left|3x-1\right|=2x^2-5\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)

Giải pt (1)

\(\Delta=9+32=41>0\)

Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)

Giải pt (2)

\(\Delta=9+48=57>0\)

Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)

Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)

a) Ta có: \(3x^2+2x-1=0\)

\(\Leftrightarrow3x^2+3x-x-1=0\)

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-1;\dfrac{1}{3}\right\}\)

b) Ta có: \(x^2-5x+6=0\)

\(\Leftrightarrow x^2-2x-3x+6=0\)

\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Vậy: S={2;3}

c) Ta có: \(x^2-3x+2=0\)

\(\Leftrightarrow x^2-x-2x+2=0\)

\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Vậy: S={1;2}

d) Ta có: \(2x^2-6x+1=0\)

\(\Leftrightarrow2\left(x^2-3x+\dfrac{1}{3}\right)=0\)

mà \(2\ne0\)

nên \(x^2-3x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{23}{12}=0\)

\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{23}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{69}}{6}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{69}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{69}}{6}\\x=\dfrac{9-\sqrt{69}}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{9+\sqrt{69}}{6};\dfrac{9-\sqrt{69}}{6}\right\}\)

e) Ta có: \(4x^2-12x+5=0\)

\(\Leftrightarrow4x^2-10x-2x+5=0\)

\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)

cho vào máy tính là ra hết