hép myyyyyy hhuuuhu . GIúp toyy ikk Question hép myyyyyy hhuuuhu . GIúp toyy ikk in progress 0 Môn Toán Nho 6 years 2020-11-25T20:46:36+00:00 2020-11-25T20:46:36+00:00 2 Answers 82 views 0
Answers ( )
Đáp án:
$1)(x-3)(x-3)\\
2) (5+x).(5+x)\\
3) \left ( \dfrac{1}{2}a+2b \right ).\left ( \dfrac{1}{2}a+2b \right )\\
4)\left ( \dfrac{1}{3}+y^4 \right ).\left ( \dfrac{1}{3}+y^4 \right )\\
8) 3x.(3x+4)\\
9) (2x-5y)(2x+5y)\\
10) (2x-7)(2x+7)\\
7) (a^2-b)(a^4+a^2b+b^2)\\
8) (x-5).(x-5)\\
12)\left ( \dfrac{3}{5}x^2-\dfrac{1}{2} \right )\left ( \dfrac{3}{5}x^2+\dfrac{1}{2} \right )\\
13) (x^{16}-1)(x^{16}+1)\\
14)(2x+1).(2x+1)\\
15) (x-10).(x-10)\\
16) (y^2+7).(y^2+7)\\
10) (x+2y).(x+2y)$
Giải thích các bước giải:
$1) x^2-6x+9=x^2-2.x.3+3^2=(x-3)^2=(x-3).(x-3)\\
2) 25+10x+x^2=5^2+2.5.x+x^2=(5+x)^2=(5+x).(5+x)\\
3) \dfrac{1}{4}a^2+2ab^2+4b^4\\
=\left (\dfrac{1}{2}a \right )^2+2.\dfrac{1}{2}a.2b+(2b)^2\\
=\left ( \dfrac{1}{2}a+2b \right )^2\\ \left ( \dfrac{1}{2}a+2b \right ).\left ( \dfrac{1}{2}a+2b \right )\\
4) \dfrac{1}{9}-\dfrac{2}{3}y^4+y^8\\
=\left ( \dfrac{1}{3} \right )^2+2.\dfrac{1}{3}.y^4+(y^4)^2\\
=\left ( \dfrac{1}{3}+y^4 \right )^2\\=\left ( \dfrac{1}{3}+y^4 \right ).\left ( \dfrac{1}{3}+y^4 \right )\\
8) (3x+2)^2-4\\
=(3x+2)^2-2^2\\
=(3x+2-2)(3x+2+2)\\
=3x.(3x+4)\\
9) 4x^2-25y^2\\
=(2x)^2-(5y)^2\\
=(2x-5y)(2x+5y)\\
10) 4x^2-49\\
=(2x)^2-7^2=(2x-7)(2x+7)\\
7) a^6-b^3\\
=(a^2)^3-b^3\\
=(a^2-b)(a^4+a^2b+b^2)\\
8) x^2-10x+25\\
=x^2-2.x.5+5^2=(x-5)^2=(x-5).(x-5)\\
12) \dfrac{9}{25}x^4-\dfrac{1}{4}\\
=\left ( \dfrac{3}{5}x^2 \right )^2-\left (\dfrac{1}{2} \right )^2\\
=\left ( \dfrac{3}{5}x^2-\dfrac{1}{2} \right )\left ( \dfrac{3}{5}x^2+\dfrac{1}{2} \right )\\
13) x^{32}-1\\
=(x^{16})^2-1=(x^{16}-1)(x^{16}+1)\\
14)4x^2+4x+1\\
=(2x)^2+2.2x.1+1=(2x+1)^2=(2x+1)(2x+1)\\
15) x^2-20x+100\\
=x^2-2.x.10+10^2=(x-10)^2=(x-10)(x-10)\\
16) y^4-14y^2+49\\
=(y^2)^2-2.y^2.7+7^2=(y^2+7)^2=(y^2+7)(y^2+7)\\
10) x^2+4xy+4y^2\\
=x^2+2.x.2y+(2y)^2=(x+2y)^2=(x+2y)(x+2y)$
1, x² – 6x + 9
= x² – 2.x.3 + 3²
= (x – 3)²
= (x – 3)(x – 3)
2, 25 + 10x + x²
= 5² + 2.5.x + x²
= (5 + x)²
= (x + 5)(x + 5)
3, $\frac{1}{4}$a² + 2ab² + 4$b^{4}$
= ($\frac{1}{2}$a)² + 2.$\frac{1}{2}$a.2b² + (2b²)²
= ($\frac{1}{2}$a + 2b²)²
= ($\frac{1}{2}$a + 2b²)($\frac{1}{2}$a + 2b²)
4, $\frac{1}{9}$ – $\frac{2}{3}$$y^{4}$ + $y^{8}$
= ($\frac{1}{3}$)² – 2.$\frac{1}{3}$.$y^{4}$ + ($y^{4}$)²
= ($\frac{1}{3}$ – $y^{4}$)²
= ($\frac{1}{3}$ – $y^{4}$)($\frac{1}{3}$ – $y^{4}$)
7, $a^{6}$ – b³
= (a²)³ – b³
= (a² – b)($a^{4}$ + a²b + b²)
8, x² – 10x + 25
= x² – 2.x.5 + 5²
= (x – 5)²
= (x – 5)(x – 5)
10, x² + 4xy + 4y²
= x² + 2.x.2y + (2y)²
= (x + 2y)²
= (x + 2y)(x + 2y)
8, (3x + 2)² – 4
= (3x + 2)² – 2²
= (3x + 2 – 2)(3x + 2 + 2)
= 3x(3x + 4)
9, 4x² – 25y²
= (2x)² – (5y)²
= (2x – 5y)(2x + 5y)
10, 4x² – 49
= (2x)² – 7²
= (2x – 7)(2x + 7)
12, $\frac{9}{25}$$x^{4}$ – $\frac{1}{4}$
= ($\frac{3}{5}$x²)² – ($\frac{1}{2}$)²
= ($\frac{3}{5}$x² – $\frac{1}{2}$)($\frac{3}{5}$x² + $\frac{1}{2}$)
13, $x^{32}$ – 1
= ($x^{16}$)² – 1²
= ($x^{16}$ – 1)($x^{16}$ + 1)
= ($x^{4}$ – 1)($x^{4}$ + 1)($x^{16}$ + 1)
= (x² – 1)(x² + 1)($x^{4}$ + 1)($x^{16}$ + 1)
= (x – 1)(x + 1)(x² + 1)($x^{4}$ + 1)($x^{16}$ + 1)
14, 4x² + 4x + 1
= (2x)² + 2.2x.1 + 1²
= (2x + 1)²
= (2x + 1)(2x + 1)
15, x² – 20x + 100
= x² – 2.x.10 + 10²
= (x – 10)²
= (x – 10)(x – 10)
16, $y^{4}$ – 14y² + 49
= (y²)² – 2.y².7 + 7²
= (y² – 7)²
= (y² – 7)(y² – 7)