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\(\left(a+b+c\right)^2+\left(a+b-c\right)^2-4c^2\)
\(=\left(a+b+c\right)^2+\left(a+b-c+2c\right)\left(a+b-c-2c\right)\)
\(=\left(a+b+c\right)^2+\left(a+b+c\right)\left(a+b-3c\right)\)
\(=\left(a+b+c\right)\left(a+b+c+a+b-3c\right)\)
\(=\left(a+b+c\right)\left(2a+2b-2c\right)\)
\(=2\left(a+b+c\right)\left(a+b-c\right)\)
Phân tích đa thức sau thành nhân tử
\(A=a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(A=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b^2-c^2\right)\)
\(A=\left(b-c\right)\left(a^2+bc-ab-ac\right)\)
\(A=\left(b-c\right)\left[a\left(a-b\right)-c\left(a-b\right)\right]\)
\(A=\left(b-c\right)\left(a-b\right)\left(a-c\right)\)
Auto cách khác:3
\(A=a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a+b\right)-\left(a-b\right)\left(b-c\right)\left(c+b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
\(\left(a+b\right)\left(a^2-b^2\right)+\left(b+c\right)\left(b^2-c^2\right)+\left(c+a\right)\left(c^2-a^2\right)\)
\(=\left(a+b\right)\left(a^2-b^2\right)-\left(b+c\right)\left[c^2-a^2+a^2-b^2\right]+\left(c+a\right)\left(c^2-a^2\right)\)
\(=\left(a+b\right)\left(a^2-b^2\right)-\left(b+c\right)\left(c^2-a^2\right)-\left(b+c\right)\left(a^2-b^2\right)+\left(c+a\right)\left(c^2-a^2\right)\)
\(=\left(a^2-b^2\right)\left(a+b-b-c\right)+\left(c^2-a^2\right)\left(c+a-b-c\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(a-c\right)+\left(c-a\right)\left(c+a\right)\left(a-b\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(a+b-c-a\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
Chúc bạn học tốt.
\(B=\left(a^2+b^2\right)^3+\left(c^2-a^2\right)^3-\left(b^2+c^2\right)^3\)
\(=\left(a^2+b^2+c^2-a^2\right)\left[\left(a^2+b^2\right)^2-\left(c^2-a^2\right)\left(a^2+b^2\right)+\left(c^2-a^2\right)^2\right]-\left(b^2+c^2\right)^2\)
\(=\left(b^2+c^2\right)\left[\left(a^2+b^2\right)^2-\left(c^2-a^2\right)\left(a^2+b^2\right)+\left(c^2-a^2\right)^2\right]-\left(b^2+c^2\right)^2\)
\(=\left(b^2+c^2\right)\left(a^4+2a^2b^2+b^4-a^2c^2+a^4-b^2c^2+a^2b^2-b^4-2b^2c^2-c^4\right)\)
\(=\left(b^2+c^2\right)\left(2a^4-c^4+3a^2b^2-a^2c^2-3b^2c^2\right)\)
ko chắc
\(\left(a+b+c\right)^2+\left(a+b-c\right)^2-4c\)
\(=\left(a+b\right)^2+2.c\left(a+b\right)+c^2+\left(a+b\right)^2-2.c\left(a+b\right)+c^2-4c^2\)
\(=a^2+2ab+b^2+2ca+2cb+c^2+a^2+2ab+b^2-2ca-2cb+c^2-4c^2\)
\(=2a^2+4ab+2b^2-2c^2\)
\(=2\left(a^2+2ab+b^2-c^2\right)\)
\(=2\left[\left(a+b\right)^2-c^2\right]\)
\(=2\left(a+b+c\right)\left(a+b-c\right)\)
Mn vào tcn của con này, https://olm.vn/thanhvien/kimmai123az, PTD/KM ?, nó chuyên đi copy bài của ng khác và câu hỏi tương tự