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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 \(x=\sqrt[3]{11}+2\)
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) \(x^3+5x^2+9x=-45\)
\(\Leftrightarrow x^3+5x^2+9x+45=0\)
\(\Leftrightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)
\(\Leftrightarrow\left(x^2+9\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm3i\\x=-5\end{cases}}\)
\(1.6x\left(x-10\right)-2x+20=0\)
⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)
⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)
⇔ x = 10 hoặc x = \(\dfrac{1}{3}\)
KL....
\(2.3x^2\left(x-3\right)+3\left(3-x\right)=0\)
⇔ \(3\left(x-3\right)\left(x^2-1\right)=0\)
⇔ \(x=+-1\) hoặc \(x=3\)
KL....
\(3.x^2-8x+16=2\left(x-4\right)\)
⇔ \(\left(x-4\right)^2-2\left(x-4\right)=0\)
⇔ \(\left(x-4\right)\left(x-6\right)=0\)
⇔ \(x=4\) hoặc \(x=6\)
KL.....
\(4.x^2-16+7x\left(x+4\right)=0\)
\(\text{⇔}4\left(x+4\right)\left(2x-1\right)=0\)
⇔ \(x=-4hoacx=\dfrac{1}{2}\)
KL.....
\(5.x^2-13x-14=0\)
⇔ \(x^2+x-14x-14=0\)
\(\text{⇔}\left(x+1\right)\left(x-14\right)=0\)
\(\text{⇔}x=14hoacx=-1\)
KL......
Còn lại tương tự ( dài quá ~ )
a) \(\left(x+6\right)^2-x\left(x+9\right)=0\)
\(\Leftrightarrow\)\(x^2+12x+36-x^2-9x=0\)
\(\Leftrightarrow\)\(3x+36=0\)
\(\Leftrightarrow\)\(x=-12\)
Vậy...
b) \(6x\left(2x+5\right)-\left(3x+4\right)\left(4x-3\right)=9\)
\(\Leftrightarrow\)\(12x^2+30x-12x^2-7x+12=9\)
\(\Leftrightarrow\)\(23x+12=9\)
\(\Leftrightarrow\)\(x=-\frac{3}{23}\)
Vậy
c) \(2x\left(8x+3\right)-\left(4x+1\right)=13\)
\(\Leftrightarrow\)\(16x^2+6x-4x-1=13\)
\(\Leftrightarrow\)\(16x^2+2x-14=0\)
\(\Leftrightarrow\)\(8x^2+x-7=0\)
\(\Leftrightarrow\)\(\left(x+1\right)\left(8x-7\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\\x=\frac{7}{8}\end{cases}}\)
Vậy
d) \(\left(x-4\right)^2-x\left(x+4\right)=0\)
\(\Leftrightarrow\)\(x^2-8x+16-x^2-4x=0\)
\(\Leftrightarrow\)\(-12x+16=0\)
\(\Leftrightarrow\)\(x=\frac{4}{3}\)
Vậy
e) \(\left(x-2\right)^2-\left(2x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\)\(x^2-4x+4-2x^2+x+6=0\)
\(\Leftrightarrow\)\(-x^2-3x+10=0\)
\(\Leftrightarrow\)\(\left(2-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
Vậy
`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. \(\left(2x+3\right)^2+\left(x-2\right)^2-2\left(2x+3\right)\left(x-2\right)\)
\(=\left(2x+3\right)^2-2\left(2x+3\right)\left(x-2\right)+\left(x-2\right)^2\)
\(=\left(2x+3-x+2\right)^2\)
\(=\left(x+5\right)^2\)