\(x^2=0\)

\(\Rightarrow x^2=0^2\)

\(\Rightarrow x=0\)

-----------

\(x^2=16\)

\(\Rightarrow x^2=\left(\pm4\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x^2=\left(-4\right)^2\\x^2=4^2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-4\\x=4\end{matrix}\right.\)

\(x^2=0\)

\(\Rightarrow x=0\left(tm\right)\)

Vậy: \(x=0.\)

\(---\)

\(x^2=16\)

\(\Rightarrow x^2=\left(\pm4\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=-4\left(tm\right)\end{matrix}\right.\)

Vậy: \(x\in\left\{4;-4\right\}.\)

b) 

\(x^2=16\)

\(=>x^2=4^2\) hoặc  \(x^2=\left(-4\right)^2\)

Vậy x = 4 hoặc x = -4

c) Kiểm tra lại nha

d)

\(\left(x-2\right)^5-32=0\)

\(=>\left(x-2\right)^5=0+32\)

\(=>\left(x-2\right)^5=32\)

\(=>\left(x-2\right)^5=2^5\)

\(=>x-2=2\)

\(=>x=2+2\)

\(=>x=4\)

lấy 24 cộng 1 = 25

À bạn ơi cái câu c là = 0 nha, mình ghi thiếu ạ

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

a) x = 4                

b) x = 3     

c) x = 2                

d) x = 1. 

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.

$4^{2x}+2^{4x+3}=144$
$2^{4x}+2^{4x}.8=144$

$2^{4x}(1+8)=144$

$2^{4x}.9=144$

$2^{4x}=144:9=16=2^4$

$\Rightarrow 4x=4\Rightarrow x=1$

a) x = -1.                      b) x = 4 hoặc x = 5.

c) x = ± 2 .                  d) x = 1 hoặc x = 2.

a, \(\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)

b, \(3\left(x-2\right)+13⋮x-2\Rightarrow x-2\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)

c, \(x\left(x+7\right)+2⋮x+7\Rightarrow x+7\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

a) x = 4

b) x = 3

c) x = 2

d) x = 1

e) x = 3

f) x = 2

g) x = 4

h) x = 3

Câu 17:

Xét ΔADC có OE//DC

nên \(\dfrac{OE}{DC}=\dfrac{AO}{AC}\left(1\right)\)

Xét ΔBDC có OH//DC

nên \(\dfrac{OH}{DC}=\dfrac{BO}{BD}\left(2\right)\)

Xét ΔOAB và ΔOCD có

\(\widehat{OAB}=\widehat{OCD}\)(hai góc so le trong, AB//CD)

\(\widehat{AOB}=\widehat{COD}\)(hai góc đối đỉnh)

Do đó: ΔOAB đồng dạng với ΔOCD
=>\(\dfrac{OA}{OC}=\dfrac{OB}{OD}\)

=>\(\dfrac{OC}{OA}=\dfrac{OD}{OB}\)

=>\(\dfrac{OC}{OA}+1=\dfrac{OD}{OB}+1\)

=>\(\dfrac{OC+OA}{OA}=\dfrac{OD+OB}{OB}\)

=>\(\dfrac{AC}{OA}=\dfrac{BD}{OB}\)

=>\(\dfrac{OA}{AC}=\dfrac{OB}{BD}\left(3\right)\)

Từ (1),(2),(3) suy ra \(\dfrac{OE}{DC}=\dfrac{OH}{DC}\)

=>OE=OH

Câu 15:

a: \(3x\left(x-1\right)+x-1=0\)

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

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

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

b: \(x^2-6x=0\)

=>\(x\cdot x-x\cdot6=0\)

=>x(x-6)=0

=>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)