Viết mỗi biểu thức sau dưới dạng tích:\(a)27{{\rm{x}}^3} 1\)\(b)64 - 8{y^3}\)

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SGK Cánh Diều trang 21

19 tháng 7 2023 lúc 13:13

Viết biểu thức sau dưới dạng lập phương của một hiệu:

\(8{{\rm{x}}^3} - 36{{\rm{x}}^2}y + 54{\rm{x}}{y^2} - 27{y^3}\)

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Lớp 8 Toán Bài 3. Hằng đẳng thức đáng nhớ

Hà Quang Minh Giáo viên CTVVIP

10 tháng 1 lúc 21:28

\(8{{\rm{x}}^3} - 36{{\rm{x}}^2}y + 54{\rm{x}}{y^2} - 27{y^3} = {\left( {2{\rm{x}}} \right)^3} - 3.\left( {2{\rm{x}}} \right).3y + 3.2{\rm{x}}.{\left( {3y} \right)^2} - {\left( {3y} \right)^3} = {\left( {2{\rm{x}} - 3y} \right)^3}\)

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Bài 2

SGK Cánh Diều trang 23

19 tháng 7 2023 lúc 13:16

Viết mỗi biểu thức sau dưới dạng lập phương của một tổng hoặc một hiệu:a) {a^3} + 12{{rm{a}}^2} + 48{rm{a}} + 64 b) 27{{rm{x}}^3} + 54{{rm{x}}^2}y + 36{rm{x}}{y^2} + 8{y^3}c) {x^3} - 9{{rm{x}}^2} + 27{rm{x}} - 27 d) 8{{rm{a}}^3} - 12{{rm{a}}^2}b + 6{rm{a}}{b^2} - {b^3}

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Viết mỗi biểu thức sau dưới dạng lập phương của một tổng hoặc một hiệu:

a) \({a^3} + 12{{\rm{a}}^2} + 48{\rm{a}} + 64\)

b) \(27{{\rm{x}}^3} + 54{{\rm{x}}^2}y + 36{\rm{x}}{y^2} + 8{y^3}\)

c) \({x^3} - 9{{\rm{x}}^2} + 27{\rm{x}} - 27\)

d) \(8{{\rm{a}}^3} - 12{{\rm{a}}^2}b + 6{\rm{a}}{b^2} - {b^3}\)

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Lớp 8 Toán Bài 3. Hằng đẳng thức đáng nhớ

Hà Quang Minh Giáo viên CTVVIP

10 tháng 1 lúc 21:29

a) \({a^3} + 12{{\rm{a}}^2} + 48{\rm{a}} + 64 \\= {a^3} + 3{{\rm{a}}^2}.4 + 3{\rm{a}}{.4^2} + {4^3} \\= {\left( {a + 4} \right)^3}\)

b) \({x^3} - 9{{\rm{x}}^2} + 27{\rm{x}} - 27 \\= {x^3} - 3.{x^2}.3 + 3.x{.3^2} - {3^3} \\= {\left( {x - 3} \right)^3}\)

c) \(8{{\rm{a}}^3} - 12{{\rm{a}}^2}b + 6{\rm{a}}{b^2} - {b^3} \\= {\left( {2{\rm{a}}} \right)^2} - 3.{\left( {2{\rm{a}}} \right)^2}.b + 3.2{\rm{a}}.{b^2} - {b^3} \\= {\left( {2{\rm{a}} - b} \right)^3}\)

d) \(27{{\rm{x}}^3} + 54{{\rm{x}}^2}y + 36{\rm{x}}{y^2} + 8{y^3}\\= {\left( {3{\rm{x}}} \right)^3} + 3.{\left( {3{\rm{x}}} \right)^2}.2y + 3.3{\rm{x}}.{\left( {2y} \right)^2} + {\left( {2y} \right)^3} \\= {\left( {3{\rm{x}} + 2y} \right)^3}\)

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Pham Trong Bach

29 tháng 11 2017 lúc 14:35

Viết các biểu thức sau dưới dạng tích:a) x 3 + 8; b) x 3 – 64;c) 27 x 3 + 1; d) 64 m 3 – 27.

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Viết các biểu thức sau dưới dạng tích:

a) x 3 + 8; b) x 3 – 64;

c) 27 x 3 + 1; d) 64 m 3 – 27.

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Lớp 8 Toán

Cao Minh Tâm

29 tháng 11 2017 lúc 14:36

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trần khánh linh

7 tháng 10 2021 lúc 18:15

Viết biểu thức sau dưới dạng tích

a) x 2 + 2x +1

b) 9x2 +y2 - 6xy

c) x 2 – 9

d) x 3 +12x2 +48x +64

e) x 3 - 6x2 +12x -8 f

) x 3 + 8

g) 8x3 – y

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Lớp 8 Toán

Nguyễn Hoàng Minh

7 tháng 10 2021 lúc 18:19

\(a,=\left(x+1\right)^2\\ b,=\left(3x-y\right)^2\\ c,=\left(x-3\right)\left(x+3\right)\\ d,=\left(x+4\right)^3\\ e,=\left(x-2\right)^3\\ f,=\left(x+2\right)\left(x^2-2x+4\right)\\ g,=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

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Trịnh Thành Long K9A2 (...

9 tháng 7 2023 lúc 8:23

bài 4 : viết các biểu thức sau dưới dạng tích

a , 25 x mũ 2 -16

b , 16 a mũ 2 - 9 b mũ 2

c , 8 x mũ 3 + 1

d , 125 x mũ 3 + 27 y mũ 3

e , 8 x mũ 3 - 125

g , 27 x mũ 3 - y mũ 3

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Lớp 8 Toán Câu hỏi của OLM

Kiều Vũ Linh CTV

9 tháng 7 2023 lúc 8:32

a) 25x² - 16

= (5x)² - 4²

= (5x - 4)(5x + 4)

b) 16a² - 9b²

= (4a)² - (3b)²

= (4a - 3b)(4a + 3b)

c) 8x³ + 1

= (2x)³ + 1³

= (2x + 1)(4x² - 2x + 1)

d) 125x³ + 27y³

= (5x)³ + (3y)³

= (5x + 3y)(25x² - 15xy + 9y²)

e) 8x³ - 125

= (2x)³ - 5³

= (2x - 5)(4x² + 10x + 25)

g) 27x³ - y³

= (3x)³ - y³

= (3x - y)(9x² + 3xy + y²)

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Nguyễn Đức Trí

9 tháng 7 2023 lúc 8:33

a) \(25x^2-16=\left(5x-4\right)\left(5x+4\right)\)

b) \(16a^2-9b^2=\left(4a-3b\right)\left(4a+3b\right)\)

c) \(8x^3+1=\left(2x+1\right)\left(4x^2-2x+1\right)\)

d) \(125x^3+27y^3=\left(5x+3y\right)\left(25x^2-15xy+9y^2\right)\)

e) \(8x^3-125=\left(2x-5\right)\left(4x^2-10x+25\right)\)

g) \(27x^3-y^3=\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)

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Đoàn Khôi Nguyên

9 tháng 7 2023 lúc 8:39

tao phá olm đêy nè:

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Bài 2.8

SGK Kết nối tri thức với cuộc sống - Trang 36

21 tháng 7 2023 lúc 15:21

Viết các biểu thức sau dưới dạng lập phương của một tổng hoặc một hiệu.

a) \(27 + 54x + 36{x^2} + 8{x^3}\).

b) \(64{x^3} - 144{x^2}y + 108x{y^2} - 27{y^3}\).

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Lớp 8 Toán Bài 7. Lập phương của một tổng. Lập phương của một...

Hà Quang Minh Giáo viên CTVVIP

12 tháng 1 lúc 21:38

a) \(27 + 54x + 36{x^2} + 8{x^3} = {3^3} + {3.3^2}.2x + 3.3.{\left( {2x} \right)^2} + {\left( {2x} \right)^3} = {\left( {3 + 2x} \right)^3}\)

b) \(64{x^3} - 144{x^2}y + 108x{y^2} - 27{y^3} = {\left( {4x} \right)^3} - 3.{\left( {4x} \right)^2}.3y + 3.4x.{\left( {3y} \right)^2} - {\left( {3y} \right)^3} = {\left( {4x - 3y} \right)^3}\)

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Bài 1

SGK Cánh Diều trang 26

19 tháng 7 2023 lúc 13:21

Phân tích mỗi đa thức sau thành nhân tử:

\(a)4{{\rm{x}}^2} - 12{\rm{x}}y + 9{y^2}\)

\(b){x^3} + 6{{\rm{x}}^2} + 12{\rm{x}} + 8\)

\(c)8{y^3} - 12{y^2} + 6y - 1\)

\(d){\left( {2{\rm{x}} + y} \right)^2} - 4{y^2}\)

\(e)27{y^3} + 8\)

\(g)64 - 125{{\rm{x}}^3}\)

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Lớp 8 Toán Bài 4. Vận dụng hằng đẳng thức vào phân tích đa th...

Hà Quang Minh Giáo viên CTVVIP

10 tháng 1 lúc 21:33

\(a)4{{\rm{x}}^2} - 12{\rm{x}}y + 9{y^2} = {\left( {2{\rm{x}}} \right)^2} - 2.2{\rm{x}}.3y + {\left( {3y} \right)^2} = {\left( {2{\rm{x}} - 3y} \right)^2}\)

\(b){x^3} + 9{{\rm{x}}^2} + 27{\rm{x}} + 27 = {x^3} + 3.{x^2}.3 + 3.x{.3^2} + {3^3} = {\left( {x + 3} \right)^3}\)

\(c)8{y^3} - 12{y^2} + 6y - 1 = {\left( {2y} \right)^3} - 3.{\left( {2y} \right)^2}.1 + 3.2y{.1^2} - {1^3} = {\left( {2y - 1} \right)^3}\)

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Hà Quang Minh Giáo viên CTVVIP

10 tháng 1 lúc 21:33

\(\begin{array}{l}d) {\left( {2{\rm{x}} + y} \right)^2} - 4{y^2}\\ = {\left( {2{\rm{x}} + y} \right)^2} - {\left( {2y} \right)^2}\\ = \left( {2{\rm{x}} + y + 2y} \right)\left( {2{\rm{x}} + y - 2y} \right) = \left( {2{\rm{x}} + 3y} \right)\left( {2{\rm{x}} - y} \right)\end{array}\)

\(e) 27{y^3} + 8 = {\left( {3y} \right)^3} + {2^3} = \left( {3y + 2} \right)\left( {9{y^2} - 6y + 4} \right)\)

\(g) 64 - 125{{\rm{x}}^3} = {4^3} - {\left( {5{\rm{x}}} \right)^3} = \left( {4 - 5{\rm{x}}} \right)\left( {16 + 20{\rm{x}} + 25{{\rm{x}}^2}} \right)\)

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Luong Thi Van Anh

30 tháng 10 2023 lúc 21:37

Viết mỗi biểu thức sau dưới dạng tích a x^3+(2y)^3 b (2x)^3-y^3

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Lớp 8 Toán

Nguyễn Lê Phước Thịnh CTV

30 tháng 10 2023 lúc 21:38

a: \(x^3+\left(2y\right)^3=\left(x+2y\right)\left[x^2-x\cdot2y+\left(2y\right)^2\right]\)

\(=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)

b: \(\left(2x\right)^3-y^3\)

\(=\left(2x-y\right)\left[\left(2x\right)^2+2x\cdot y+y^2\right]\)

\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

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Bài 3

SGK Cánh Diều trang 23

19 tháng 7 2023 lúc 13:17

Viết mỗi biểu thức sau dưới dạng tích:

a) \(25{{\rm{x}}^2} - 16\) b) \(16{{\rm{a}}^2} - 9{b^2}\) c) \(8{{\rm{x}}^3} + 1\)

d) \(125{{\rm{x}}^3} + 27{y^3}\) e) \(8{{\rm{x}}^3} - 125\) g) \(27{{\rm{x}}^3} - {y^3}\)

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Lớp 8 Toán Bài 3. Hằng đẳng thức đáng nhớ

Hà Quang Minh Giáo viên CTVVIP

10 tháng 1 lúc 21:30

a) \(25{{\rm{x}}^2} - 16 = {\left( {5{\rm{x}}} \right)^2} - {4^2} = \left( {5{\rm{x}} + 4} \right)\left( {5{\rm{x}} - 4} \right)\)

b) \(8{{\rm{x}}^3} + 1 = {\left( {2{\rm{x}}} \right)^3} + {1^3} = \left( {2{\rm{x}} + 1} \right)\left( {4{{\rm{x}}^2} - 2{\rm{x}} + 1} \right)\)

c) \(8{{\rm{x}}^3} - 125 = {\left( {2{\rm{x}}} \right)^3} - {5^3} = \left( {2{\rm{x}} - 5} \right)\left( {4{{\rm{x}}^2} + 10{\rm{x + }}25} \right)\)

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Hà Quang Minh Giáo viên CTVVIP

10 tháng 1 lúc 21:30

d) \(27{{\rm{x}}^3} - {y^3} = {\left( {3x} \right)^3} - {y^3} = \left( {3{\rm{x}} - y} \right)\left( {9{{\rm{x}}^2} + 3{\rm{x}}y + {y^2}} \right)\)

e) \(16{{\rm{a}}^2} - 9{b^2} = {\left( {4{\rm{a}}} \right)^2} - {\left( {3b} \right)^2} = \left( {4{\rm{a}} - 3b} \right)\left( {4{\rm{a}} + 3b} \right)\)

g) \(125{{\rm{x}}^3} + 27{y^3} = {\left( {5{\rm{x}}} \right)^3} + {\left( {3y} \right)^3} = \left( {5{\rm{x}} + 3y} \right)\left( {25{{\rm{x}}^2} - 15{\rm{x}}y + 9{y^2}} \right)\)

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