a. $x+5x^2=0$

$\Leftrightarrow x(1+5x)=0$

$\Leftrightarrow x=0$ hoặc $1+5x=0$

$\Leftrightarrow x=0$ hoặc $x=\frac{-1}{5}$

b.

$x+1=(x+1)^2$
$\Leftrightarrow (x+1)^2-(x+1)=0$

$\Leftrightarrow (x+1)[(x+1)-1]=0$

$\Leftrightarrow (x+1)x=0$

$\Leftrightarrow x+1=0$ hoặc $x=0$

$\Leftrightarrow x=-1$ hoặc $x=0$

c.

$x^3+x=0$

$\Leftrightarrow x(x^2+1)=0$

$\Leftrightarrow x=0$ hoặc $x^2+1=0$ 

$\Leftrightarrow x=0$ (chọn) hoặc $x^2=-1<0$ (loại)

h.

$8x^3+12x^2+6x+1=0$

$\Leftrightarrow (2x+1)^3=0$

$\Leftrightarrow 2x+1=0\Leftrightarrow x=\frac{-1}{2}$

i.

$5x(x-1)=x-1$

$\Leftrightarrow 5x(x-1)-(x-1)=0$

$\Leftrightarrow (x-1)(5x-1)=0$

$\Leftrightarrow x-1=0$ hoặc $5x-1=0$

$\Leftrightarrow x=1$ hoặc $x=\frac{1}{5}$

k.

$2(x+5)-x^2-5x=0$
$\Leftrightarrow 2(x+5)-(x^2+5x)=0$

$\Leftrightarrow 2(x+5)-x(x+5)=0$

$\Leftrightarrow (x+5)(2-x)=0$
$\Leftrightarrow x+5=0$ hoặc $2-x=0$

$\Leftrightarrow x=-5$ hoặc $x=2$

a: Xét ΔKAB và ΔKCD có

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

\(\widehat{AKB}=\widehat{CKD}\)(hai góc đối đỉnh)

Do đó: ΔKAB đồng dạng với ΔKCD

=>\(\dfrac{KA}{KC}=\dfrac{KB}{KD}\)

=>\(KA\cdot KD=KB\cdot KC\)

b: Ta có: \(\dfrac{KA}{KC}=\dfrac{KB}{KD}\)

=>\(\dfrac{KC}{KA}=\dfrac{KD}{KB}\)

=>\(\dfrac{KC}{KA}+1=\dfrac{KD}{KB}+1\)

=>\(\dfrac{KC+KA}{KA}=\dfrac{KD+KB}{KB}\)

=>\(\dfrac{AC}{KA}=\dfrac{BD}{KB}\)

=>\(\dfrac{AK}{AC}=\dfrac{BK}{BD}\left(1\right)\)

Xét ΔADC có IK//DC

nên \(\dfrac{AK}{AC}=\dfrac{IK}{DC}\left(2\right)\)

Xét ΔBDC có KQ//DC

nên \(\dfrac{KQ}{DC}=\dfrac{BK}{BD}\left(3\right)\)

Từ (1),(2),(3) suy ra IK=KQ

a) A = xy - 4y - 5x + 20

= (xy - 4y) - (5x - 20)

= y(x - 4) - 5(x - 4)

= (x - 4)(y - 5)

Thay x = 14; y = 5,5 vào A ta có:

A = (14 - 4)(5,5 - 5)

= 10 . 0,5

= 5

b) B = x² + xy - 5x - 5y

= (x² + xy) - (5x + 5y)

= x(x + y) - 5(x + y)

= (x + y)(x - 5)

Thay x = 5 1/5 = 26/5 và y = 4 4/5 24/5 vào B ta có:

(26/5 + 24/5)(26/5 - 5)

= 10 . 1/5

= 2

c) C = x³ - x²y - xy² + y³

= (x³ - x²y) - (xy² - y³)

= x²(x - y) - y²(x - y)

= (x - y)(x² - y²)

= (x - y)(x - y)(x + y)

= (x - y)²(x + y)

Thay x = 5,75 và y = 4,25 vào C ta có:

(5,75 - 4,25)².(5,75 + 4,25)

= (1,5)².10

= 2,25.10

= 22,5

cần gấp

a: \(47^2-37^2\)

\(=\left(47-37\right)\left(47+37\right)\)

\(=84\cdot10=840\)

b: \(87^2+73^2-27^2-13^2\)

\(=\left(87^2-13^2\right)+\left(73^2-27^2\right)\)

\(=\left(87-13\right)\left(87+13\right)+\left(73-27\right)\left(73+27\right)\)

\(=74\cdot100+46\cdot100\)

\(=100\left(74+46\right)=120\cdot100=12000\)

c: \(85\cdot12,7+5\cdot3\cdot12,7\)

\(=85\cdot12,7+15\cdot12,7\)

\(=12,7\left(85+15\right)=12,7\cdot100=1270\)

d: \(24^3-64\)

\(=24^3-4^3\)

\(=\left(24-4\right)\left(24^2+24\cdot4+4^2\right)\)

\(=20\cdot688=13760\)

Lời giải:

Chiều cao của cây lúc đầu: $AB+AC=9$ 

Chiều cao của cây còn lại: $AC$

Áp dụng định lý Pitago:

$AB^2=AC^2+BC^2$

$\Leftrightarrow (9-AC)^2=AC^2+3^2$
$\Leftrightarrow 81+AC^2-18AC=AC^2+9$
$\Leftrightarrow 81-18AC=9$

$\Leftrightarrow AC=4$ (m) 

Vậy chiều cao còn lại của cây là 4 m.

a: \(5x\left(x-1\right)-3x\left(x-1\right)\)

\(=\left(x-1\right)\left(5x-3x\right)\)

\(=2x\cdot\left(x-1\right)\)

b: \(x^2-9\)

\(=x^2-3^2\)

\(=\left(x-3\right)\left(x+3\right)\)

c: \(4x^2-25\)

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

\(=\left(2x-5\right)\left(2x+5\right)\)

d: \(x^6-y^6\)

\(=\left(x^3-y^3\right)\left(x^3+y^3\right)\)

\(=\left(x-y\right)\left(x+y\right)\left(x^2-xy+y^2\right)\left(x^2+xy+y^2\right)\)

e: \(9x^2+6xy+y^2\)

\(=\left(3x\right)^2+2\cdot3x\cdot y+y^2\)

\(=\left(3x+y\right)^2\)

g: \(6x-9-x^2\)

\(=-\left(x^2-6x+9\right)\)

\(=-\left(x-3\right)^2\)

f: \(x^2+4y^2+4xy\)

\(=x^2+2\cdot x\cdot2y+\left(2y\right)^2\)

\(=\left(x+2y\right)^2\)

h: \(\left(5x+1\right)^2-\left(x+1\right)^2\)

\(=\left(5x+1+x+1\right)\left(5x+1-x-1\right)\)

\(=\left(6x+2\right)\cdot4x\)

\(=4x\cdot2\cdot\left(3x+1\right)=8x\left(3x+1\right)\)

i: \(\left(x+y\right)^2-\left(x-y\right)^2\)

\(=\left(x+y-x+y\right)\left(x+y+x-y\right)\)

\(=2y\cdot2x=4xy\)

c: \(\dfrac{2x}{x+5}+\dfrac{10x}{x^2+5x}\)

\(=\dfrac{2x}{x+5}+\dfrac{10x}{x\left(x+5\right)}\)

\(=\dfrac{2x}{x+5}+\dfrac{10}{x+5}=\dfrac{2x+10}{x+5}=\dfrac{2\left(x+5\right)}{x+5}=2\)

d: \(\dfrac{x}{x^2-36}+\dfrac{x-6}{x^2+6x}+\dfrac{-36}{\left(x^2-6x\right)\left(x+6\right)}\)

\(=\dfrac{x}{\left(x-6\right)\left(x+6\right)}+\dfrac{x-6}{x\left(x+6\right)}+\dfrac{-36}{x\left(x-6\right)\left(x+6\right)}\)

\(=\dfrac{x^2+\left(x-6\right)^2-36}{x\left(x-6\right)\left(x+6\right)}\)

\(=\dfrac{x^2+x^2-12x+36-36}{x\left(x-6\right)\left(x+6\right)}=\dfrac{2x^2-12x}{x\left(x-6\right)\left(x+6\right)}\)

\(=\dfrac{2\left(x^2-6x\right)}{\left(x^2-6x\right)\left(x+6\right)}=\dfrac{2}{x+6}\)

Em cảm ơn nhìu ạ ❤️😍