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Thanh Tu Nguyen
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Lê Song Phương
5 tháng 10 2023 lúc 20:48

\(C=c\left[b\left(a+d\right)\left(b-c\right)+a\left(b+d\right)\left(c-a\right)\right]+ab\left(c+d\right)\left(a-b\right)\)

\(C=c\left[\left(ab+bd\right)\left(b-c\right)+\left(ab+ad\right)\left(c-a\right)\right]+ab\left(c+d\right)\left(a-b\right)\)

\(C=c\left[ab^2-abc+b^2d-bcd+abc-a^2b+acd-a^2d\right]+ab\left(c+d\right)\left(a-b\right)\)

\(C=c\left[\left(ab^2-a^2b\right)+\left(b^2d-a^2d\right)+\left(acd-bcd\right)\right]+ab\left(c+d\right)\left(a-b\right)\)

\(C=c\left[ab\left(b-a\right)+d\left(a+b\right)\left(b-a\right)+cd\left(a-b\right)\right]+ab\left(c+d\right)\left(a-b\right)\)

\(C=c\left(a-b\right)\left(-ab-da-db+cd\right)+ab\left(c+d\right)\left(a-b\right)\)

\(C=\left(a-b\right)\left(-abc-acd-bcd+c^2d+abc+abd\right)\)

\(C=\left(a-b\right)\left(-acd-bcd+abd+c^2d\right)\)

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

\(C=c\left(a-b\right)\left[\left(c^2-ac\right)-\left(bc-ab\right)\right]\)

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

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

 

chi lê
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Hà Võ
14 tháng 9 2016 lúc 21:53

= a2b + ab- b2c + bc+ a2c - ac2

Nguyễn Anh Tú
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Ngu Ngu Ngu
12 tháng 5 2017 lúc 11:15

Ta có:

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

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

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

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

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

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

Vũ Thu An
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Nguyễn Huệ Lam
13 tháng 7 2017 lúc 9:45

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

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

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

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

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

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

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

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

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

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

Ichigo Minako
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Thảo
13 tháng 8 2018 lúc 22:37

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

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

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

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

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

Diệu Anh Hoàng
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Đường Quỳnh Giang
18 tháng 9 2018 lúc 23:51

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

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

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

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

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

Diệu Anh Hoàng
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Đường Quỳnh Giang
18 tháng 9 2018 lúc 23:51

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

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

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

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

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

Nguyễn Bá Hùng
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Mất nick đau lòng con qu...
1 tháng 1 2019 lúc 17:17

\(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\)

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

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

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

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

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

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

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

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

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

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

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

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

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

...

Nguyệt
4 tháng 1 2019 lúc 23:54

cách này ngắn hơn nè:

\(ab.\left(a+b\right)+bc.\left(b+c\right)+ac.\left(a+c\right)+2abc\)

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

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

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

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

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

Trương Tùng Dương
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︵✰ßล∂ ß๏у®
25 tháng 6 2019 lúc 8:32

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

\(=ab^2+ac^2+abc+bc^2+ba^2+abc+ac^2+bc^2+abc\)

\(=c^2\left(b+a\right)+\left(b^2+3\text{a}b+a^2\right)c+ab^2+a^2b\)

\(=bc^2+ac^2+b^2c+3\text{a}bc+a^2c+ab^2+a^2b\)

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