Tìm x, biết : 1. (x3+27)+(x+3)(x+9)=0 2. x2+4x+4=25
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a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
a: \(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
a) (2x−1)2−25=0(2x−1)2−25=0
(2x−1)2=0+25=25(2x−1)2=0+25=25
(2x−1)2=52=(−5)2(2x−1)2=52=(−5)2
⇒[2x−1=52x−1=−5⇒[2x=62x=−4⇒[x=3x=−2⇒[2x−1=52x−1=−5⇒[2x=62x=−4⇒[x=3x=−2
b) 8x3−50x=08x3−50x=0
2x(4x2
a: Ta có: \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
a) x = -1. b) x = 4 hoặc x = 5.
c) x = ± 2 . d) x = 1 hoặc x = 2.
a,x3+3x2+3x+1
b,x2+6x+9
c,-x3+9x2-27x+27
d,x2+4x+4
k,10x-25-x2
f,(x+y)2-9x2
g,8x3+42x2y+16xy2+6xy+y3
a) \(x^3+3x^2+3x+1=x^2+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=\left(x-1\right)^3\)
b) \(x^2+6x+9=x^2+2\cdot3\cdot x+3^2=\left(x+3\right)^2\)
c) \(-x^3+9x^2-27x+27\)
\(=-\left(x^3-9x^2+27x-27\right)\)
\(=-\left(x^3-3\cdot3\cdot x^2+3\cdot3^2\cdot x-3^3\right)=-\left(x-3\right)^3\)
d) \(x^2+4x+4=x^2+2\cdot2\cdot x+2^2=\left(x+2\right)^2\)
k) \(10x-25-x^2=-x^2+10x-25=-\left(x^2-10x+25\right)\)
\(=-\left(x^2-2\cdot5\cdot x+5^2\right)=-\left(x-5\right)^2\)
f) \(\left(x+y\right)^2-9x^2=\left(x-y\right)^2-\left(3x\right)^2=\left[\left(x-y\right)-3x\right]\left[\left(x-y\right)+3x\right]\)
\(=\left(x-y-3x\right)\left(x-y+3x\right)=\left(-2x-y\right)\left(4x-y\right)\)
Bài 3
a) x² + 10x + 25
= x² + 2.x.5 + 5²
= (x + 5)²
b) 8x - 16 - x²
= -(x² - 8x + 16)
= -(x² - 2.x.4 + 4²)
= -(x - 4)²
c) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
d) (x + y)² - 9x²
= (x + y)² - (3x)²
= (x + y - 3x)(x + y + 3x)
= (y - 2x)(4x + y)
e) (x + 5)² - (2x - 1)²
= (x + 5 - 2x + 1)(x + 5 + 2x - 1)
= (6 - x)(3x + 4)
Bài 4
a) x² - 9 = 0
x² = 9
x = 3 hoặc x = -3
b) (x - 4)² - 36 = 0
(x - 4 - 6)(x - 4 + 6) = 0
(x - 10)(x + 2) = 0
x - 10 = 0 hoặc x + 2 = 0
*) x - 10 = 0
x = 10
*) x + 2 = 0
x = -2
Vậy x = -2; x = 10
c) x² - 10x = -25
x² - 10x + 25 = 0
(x - 5)² = 0
x - 5 = 0
x = 5
d) x² + 5x + 6 = 0
x² + 2x + 3x + 6 = 0
(x² + 2x) + (3x + 6) = 0
x(x + 2) + 3(x + 2) = 0
(x + 2)(x + 3) = 0
x + 2 = 0 hoặc x + 3 = 0
*) x + 2 = 0
x = -2
*) x + 3 = 0
x = -3
Vậy x = -3; x = -2
1) \(\left(x^3+27\right)+\left(x+3\right)\left(x+9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x+9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x+18\right)=0\)
\(\Leftrightarrow x+3=0\)(vì \(x^2-2x+18=\left(x-1\right)^2+17>0\))
\(\Leftrightarrow x=-3\).
2. \(x^2+4x+4=25\)
\(\Leftrightarrow\left(x+2\right)^2=5^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=5\\x+2=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-7\end{cases}}\)