tìm x
36 x2-8x + 4/9 =0
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\(a,25x^2-1=15\)\(< =>x^2=\dfrac{16}{25}< =>x=\pm\dfrac{4}{5}\)
\(b,\left(x-4\right)^2-\left(5x+2\right)^2=0\)\(< =>\left(-4x-6\right)\left(6x-2\right)=0\)
\(< =>\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(c,\left(x-1\right)\left(x-9\right)=0< =>\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\)
a: Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x^2-3\right)=0\)
\(\Leftrightarrow x^3-27-x^3+3x=0\)
\(\Leftrightarrow x=9\)
b: Ta có: \(8x^4+x=0\)
\(\Leftrightarrow x\left(8x^3+1\right)=0\)
\(\Leftrightarrow x\left(2x+1\right)\left(4x^2-2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(a,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\\ c,\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a) x = -1. b) x = 4 hoặc x = 5.
c) x = ± 2 . d) x = 1 hoặc x = 2.
Lời giải:
a. $(x^2-9)(5x+15)=0$
$\Rightarrow x^2-9=0$ hoặc $5x+15=0$
Nếu $x^2-9=0$
$\Rightarrow x^2=9=3^2=(-3)^2$
$\Rightarrow x=3$ hoặc $-3$
Nếu $5x+15=0$
$\Rightarrow x=-3$
b.
$x^2-8x=0$
$\Rightarrow x(x-8)=0$
$\Rightarrow x=0$ hoặc $x-8=0$
$\Rightarrow x=0$ hoặc $x=8$
c.
$5+12(x-1)^2=53$
$12(x-1)^2=53-5=48$
$(x-1)^2=48:12=4=2^2=(-2)^2$
$\Rightarrow x-1=2$ hoặc $x-2=-2$
$\Rightarrow x=3$ hoặc $x=0$
d.
$(x-5)^2=36=6^2=(-6)^2$
$\Rightarrow x-5=6$ hoặc $x-5=-6$
$\Rightarrow x=11$ hoặc $x=-1$
e.
$(3x-5)^3=64=4^3$
$\Rightarrow 3x-5=4$
$\Rightarrow 3x=9$
$\Rightarrow x=3$
f.
$4^{2x}+2^{4x+3}=144$
$2^{4x}+2^{4x}.8=144$
$2^{4x}(1+8)=144$
$2^{4x}.9=144$
$2^{4x}=144:9=16=2^4$
$\Rightarrow 4x=4\Rightarrow x=1$
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
\(\Leftrightarrow x^2\left(x-1\right)^2-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)^2\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.\)
e: ta có: \(4x^2+4x-6=2\)
\(\Leftrightarrow4x^2+4x-8=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
f: Ta có: \(2x^2+7x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)