\(x^3-6x^2+12x+19=0\)
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B
4
T
19 tháng 2 2020
a, x^2 - x - 20 = 0
=> x^2 - 5x + 4x - 20 = 0
=> x(x - 5) + 4(x - 5) = 0
=> (x + 4)(x - 5) = 0
=> x + 4 = 0 hoặc x - 5 = 0
=> x = -4 hoặc x = 5
b, x^3 - 6x^2 + 12x + 19 = 0
=> x^3 + x^2 - 7x^2 - 7x + 19x + 19 = 0
=> x^2(x + 1) - 7x(x + 1) + 19(x + 1) = 0
=> (x^2 - 7x + 19)(x + 1) = 0
x^2 - 7x + 19 > 0
=> x + 1 = 0
=> x = -1
BA
19 tháng 2 2020
\(a,x^2-x-20=0\)
\(x^2-5x+4x-20=0\)
\(\left(x-5\right)\left(x-4\right)=0\)
\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=4\end{cases}}}\)
\(b,x^3-6x^2+12x+19=0\)
\(\left(x^3+x^2\right)-\left(7x^2+7x\right)+\left(19x+19\right)=0\)
\(\left(x+1\right)\left(x^2-7x+19\right)=0\)
Vì \(\left(x^2-7x+19\right)>0\forall x\)
\(x+1=0\)
\(x=-1\)
18 tháng 11 2017
\(=\left(x^3+x^2\right)-\left(7x^2+7x\right)+\left(19x+19\right)=\left(x+1\right)\left(x^2-7x+19\right)=0\)
Ta thấy: \(x^2-7x+19=x^2-2\times\frac{7}{2}x+\frac{7}{2}^2+\frac{27}{4}=\left(x-\frac{7}{2}\right)^2+\frac{27}{4}\ge\frac{27}{4}\)lớn hơn 0
\(\Rightarrow x+1=0\Rightarrow x=-1\)
18 tháng 11 2017
\(x^3-6x^2+12x+19=0\)
\(\Leftrightarrow\left(x^3+x^2\right)-\left(7x^2+7x\right)+\left(19x+19\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-7x+19\right)=0\)
Mà \(x^2-7x+19>0\)với \(\forall x\)
\(\Rightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy \(x=-1\)
26 tháng 9 2021
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 \(x=\sqrt[3]{11}+2\)
YN
22 tháng 3 2022
`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. \(\left(2x+3\right)^2+\left(x-2\right)^2-2\left(2x+3\right)\left(x-2\right)\)
\(=\left(2x+3\right)^2-2\left(2x+3\right)\left(x-2\right)+\left(x-2\right)^2\)
\(=\left(2x+3-x+2\right)^2\)
\(=\left(x+5\right)^2\)
DT
0
DT
1
22 tháng 7 2017
a) \(x^3-6x^2+12x-9=0\)
\(\Leftrightarrow x^3-6x^2+12x-8-1=0\)
\(\Leftrightarrow\left(x-2\right)^3=1\)
\(\Leftrightarrow x-2=1\Leftrightarrow x=3\)
b) \(8x^3+12x^2+6x-26=0\)
\(\Leftrightarrow8x^3+12x^2+6x+1-27=0\)
\(\Leftrightarrow\left(2x+1\right)^3=27\)
\(\Leftrightarrow2x+1=3\Leftrightarrow x=1\)
\(\Leftrightarrow x^3-6x^2+12x-8=-27\)
\(\Leftrightarrow\left(x-2\right)^3=\left(-3\right)^3\)
\(\Leftrightarrow x-2=-3\)
\(\Leftrightarrow x=-1\)
x3+6x^2+12x−19=0
⇔(x^3+6x^2+12x+8)−27=0
⇔(x+2)^3=3
⇔x+2=3
⇒x=1
Vậy...