Những câu hỏi liên quan
Pé Ken
Xem chi tiết
Đinh Thùy Linh
26 tháng 6 2016 lúc 5:47

1) \(a^3+2a^2-13a+10=a^3-a^2+3a^2-3a-10a+10=\)

\(=a^2\left(a-1\right)+3a\left(a-1\right)-10\left(a-1\right)=\left(a-1\right)\left(a^2+3a-10\right)\)

\(=\left(a-1\right)\left(a^2-2a+5a-10\right)=\left(a-1\right)\left[a\left(a-2\right)+5\left(a-2\right)\right]=\)

\(=\left(a-1\right)\left(a-2\right)\left(a+5\right)\)

b) \(\left(a^2+4b^2-5\right)^2-16\left(ab+1\right)^2=\left(a^2+4b^2-5+4ab+4\right)\left(a^2+4b^2-5-4ab-4\right)\)

\(=\left(a^2+4ab+4b^2-1\right)\left(a^2-4ab+4b^2-9\right)=\left[\left(a+2b\right)^2-1\right]\left[\left(a-2b\right)^2-9\right]=\)

\(=\left(a+2b+1\right)\left(a+2b-1\right)\left(a-2b+3\right)\left(a-2b-3\right)\)

2) \(6a-5b=1\Rightarrow5b=6a-1\Rightarrow25b^2=36a^2-12a+1\)

\(\Rightarrow4a^2+25b^2=40a^2-12a+1=40\left(a^2-2\cdot a\cdot\frac{3}{20}+\left(\frac{3}{20}\right)^2\right)+1-\frac{9}{10}\)

\(=40\left(a-\frac{3}{20}\right)^2+\frac{1}{10}\)

Vậy GTNN của \(4a^2+25b^2\)= 1/10. Xảy ra khi a = 3/20 và b = -1/50.

Bình luận (0)
Buddy
Xem chi tiết
Vui lòng để tên hiển thị
22 tháng 7 2023 lúc 8:52

`a^2 + ab + 2a + 2b = a(a+2) + b(a+2) = (a+b)(a+2)`

Bình luận (0)
Thu Thủy vũ
Xem chi tiết
kudo shinichi
23 tháng 9 2018 lúc 19:05

\(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)

\(=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\)

\(=\left[a^2-\left(b^2-2bc+c^2\right)\right].\left[\left(b^2+2bc+c^2\right)-a^2\right]\)

\(=\left[a^2-\left(b-c\right)^2\right].\left[\left(b+c\right)^2-a^2\right]\)

\(=\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(b+c+a\right)\)

\(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)

\(=\left[\left(a-b\right)^2-3^2\right].\left[\left(a+b\right)^2-1\right]\)

\(=\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)

Tham khảo nhé~

Bình luận (0)
Phan Tiến Nhật
Xem chi tiết
Kiệt Nguyễn
29 tháng 7 2019 lúc 10:52

\(\left(a^2+4b^2-5\right)^2-16\left(ab+1\right)^2\)

\(=\left(a^2+4b^2-5\right)^2-4^2\left(ab+1\right)^2\)

\(=\left(a^2+4b^2-5\right)^2-\left[4\left(ab+1\right)\right]^2\)

\(=\left(a^2+4b^2-5\right)^2-\left[4ab+4\right]^2\)

\(=\left(a^2+4b^2-5-4ab-4\right)\left(a^2+4b^2-5+4ab+4\right)\)

\(=\left(a^2+4b^2-4ab-9\right)\left(a^2+4b^2+4ab-1\right)\)

Bình luận (0)
Edogawa Conan
29 tháng 7 2019 lúc 10:55

\(\left(a^2+4b^2-5\right)^2-16\left(ab+1\right)^2\)

\(\left(a^2+4b^2-5\right)^2-\left[4\left(ab+1\right)\right]^2\)

\(\left(a^2+4b^2-5\right)^2-\left(4ab+4\right)^2\)

\(\left(a^2+4b^2-5-4ab-4\right)\left(a^2+4b^2-5+4ab+4\right)\)

\(\left(a^2+4b^2-4ab-9\right)\left(a^2+4b^2+4ab-1\right)\)

\(\left[\left(a-2b\right)^2-3^2\right]\left[\left(a+2b\right)^2-1^2\right]\)

\(\left(a-2b-3\right)\left(a-2b+3\right)\left(a+2b-1\right)\left(a+2b+1\right)\)

Bình luận (0)

\(\left(a^2+4b^2-5\right)^2-16\left(ab+1\right)^2\)

\(=\left(a^2-4b^2-5\right)^2-\left[4\left(ab+1\right)\right]^2\)

\(=\left(a^2-4b^2-5-4ab-4\right)\left(a^2-4b^2-5+4ab+4\right)\)

\(=\left(a^2-4b^2-4ab-9\right)\left(a^2-4b^2+4ab-1\right)\)

Bình luận (0)
Cô gái thất thường (Ánh...
Xem chi tiết
Ƹ̴Ӂ̴Ʒ ♐  ๖ۣۜMihikito ๖ۣ...
12 tháng 10 2018 lúc 21:00

1. \(4x^2-17xy+13y^2=4x^2-4xy-13xy+13y^2=4x\left(x-y\right)-13y\left(x-y\right)=\left(x-y\right)\left(4x-13y\right)\)

2. \(2x\left(x-5\right)-x\left(3+2x\right)=26\Leftrightarrow2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)

3. \(A=\left(2a-3b\right)^2+2\left(2a-3b\right)\left(3a-2b\right)+\left(2b-3a\right)^2\)

\(\Leftrightarrow\left(2a-3b\right)^2-2\left(2a-3b\right)\left(2b-3a\right)+\left(2b-3a\right)^2=\left(2a-3b-2b+3a\right)^2=\left(5a-5b\right)^2\)

\(=25\left(a-b\right)^2=25\cdot100=2500\)

Bình luận (0)
Hoàng Ngọc Tuyết Nhung
Xem chi tiết
Bùi Minh Anh
Xem chi tiết
alibaba nguyễn
9 tháng 8 2017 lúc 16:08

\(\left(a-b\right)\left(c-a\right)\left(c-b\right)\left(ab+bc+ca\right)\)

Bình luận (0)
Trà My
9 tháng 8 2017 lúc 16:32

\(=a^2b^2\left(a-b\right)+b^2c^2\left(b-a+a-c\right)+c^2a^2\left(c-a\right)\)

\(=a^2b^2\left(a-b\right)+b^2c^2\left(b-a+a-c\right)+c^2a^2\left(c-a\right)\)

\(=a^2b^2\left(a-b\right)+b^2c^2\left(b-a\right)+b^2c^2\left(a-c\right)+c^2a^2\left(c-a\right)\)

\(=b^2\left(a-b\right)\left(a^2-c^2\right)+c^2\left(c-a\right)\left(a^2-b^2\right)\)

\(=b^2\left(a-b\right)\left(a-c\right)\left(a+c\right)+c^2\left(c-a\right)\left(a-b\right)\left(a+b\right)\)

\(=\left(a-b\right)\left(c-a\right)\left[-b^2\left(a+c\right)+c^2\left(a+b\right)\right]\)

\(=\left(a-b\right)\left(c-a\right)\left(-ab^2-b^2c+ac^2+bc^2\right)\)

\(=\left(a-b\right)\left(c-a\right)\left[a\left(c^2-b^2\right)+bc\left(c-b\right)\right]\)

\(=\left(a-b\right)\left(c-a\right)\left[a\left(c-b\right)\left(c+b\right)+bc\left(c-b\right)\right]\)

\(=\left(a-b\right)\left(c-a\right)\left(c-b\right)\left(ab+bc+ca\right)\)

Bình luận (0)
Thu Thủy vũ
Xem chi tiết
Sắc màu
25 tháng 9 2018 lúc 21:25

a) \(\left(4x^2-25\right)^2-9\left(2x-5\right)^2\)

\(=\left(4x^2-25\right)^2-\left(6x-15\right)^2\)

\(=\left(4x^2-25-6x+15\right)\left(4x^2-25+6x-15\right)\)

\(=\left(4x^2-6x-10\right)\left(4x^2+6x-40\right)\)

\(=\left(4x^2+4x-10x-10\right)\left(4x^2+16x-10x-40\right)\)

\(=\left[4x\left(x+1\right)-10\left(x+1\right)\right]\left[4x\left(x+4\right)-10\left(x+4\right)\right]\)

\(=\left(4x-10\right)\left(x+1\right)\left(4x-10\right)\left(x+4\right)\)

\(=\left(4x-10\right)^2\left(x+1\right)\left(x+4\right)\)

\(=4\left(2x-5\right)^2\left(x+1\right)\left(x+4\right)\)

b) \(a^6-a^4+2a^3+2a^2\)

\(=a^2\left(a^4-a^2+2a+2\right)\)

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

\(=a^2\left[a^3\left(a+1\right)-a^2\left(a+1\right)+2\left(a+1\right)\right]\)

\(=a^2\left(a+1\right)\left(a^3-a^2+2\right)\)

Bình luận (0)
Diệu Anh Hoàng
Xem chi tiết
Đường Quỳnh Giang
18 tháng 9 2018 lúc 23:51

\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)

\(=\left(a+b+c\right)\left(ab+bc\right)+\left(a+b+c\right)ac-abc\)

\(=\left(ab+b^2+bc\right)\left(a+c\right)+\left(a+c\right)ac+abc-abc\)

\(=\left(a+c\right)\left(ab+b^2+bc+ac\right)\)

\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

Bình luận (0)