phân tích đa thức thành nhân tử :
a ) a^3 x ( b - c ) + b^3 x ( c - a ) + c^3 x ( a - b )
b ) ( a + b ) x ( a^2 - b^2 ) + ( b + c ) x ( b^2 - c^2 ) + ( c + a ) x ( c^2 - a^2 )
phân tích đa thức thành nhân tử :
a ) a^3 x ( b - c ) + b^3 x ( c - a ) + c^3 x ( a - b )
b ) ( a + b ) x ( a^2 - b^2 ) + ( b + c ) x ( b^2 - c^2 ) + ( c + a ) x ( c^2 - a^2 )
\(D=a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)
\(D=a^3\left(b-c\right)+\left[b^3\left(c-a\right)+c^3\left(a-b\right)\right]\)
\(D=a^3\left(b-c\right)\left(b^3c-ab^3+ac^3-bc^3\right)\)
\(D=a^3\left(b-c\right)\left[\left(b^3c-bc^3\right)-\left(ab^3-ac^3\right)\right]\)
\(D=a^3\left(b-c\right)\left[bc\left(b^2-c^2\right)-a\left(b^3-c^3\right)\right]\)
\(D=a^3\left(b-c\right)\left[bc\left(b-c\right)\left(b+c\right)-a\left(b-c\right)\left(b^2+bc+c^2\right)\right]\)
\(D=\left(b-c\right)\left[a^3+bc\left(b+c\right)-a\left(b^2+bc+c^2\right)\right]\)
\(D=\left(b-c\right)\left(a^3+b^2c+bc^2-ab^2-abc-ac^2\right)\)
\(D=\left(b-c\right)\left[\left(b^2c-ab^2\right)+\left(bc^2-abc\right)-\left(ac^2-a^3\right)\right]\)
\(D=\left(b-c\right)\left[b^2\left(c-a\right)+bc\left(c-a\right)-a\left(c^2-a^2\right)\right]\)
\(D=\left(b-c\right)\left[b^2\left(c-a\right)+bc\left(c-a\right)-a\left(c-a\right)\left(c+a\right)\right]\)
\(D=\left(b-c\right)\left(c-a\right)\left[b^2+bc-a\left(c+a\right)\right]\)
\(D=\left(b-c\right)\left(c-a\right)\left(b^2+bc-ac-a^2\right)\)
\(D=\left(b-c\right)\left(c-a\right)\left[\left(bc-ac\right)+\left(b^2-a^2\right)\right]\)
\(D=\left(b-c\right)\left(c-a\right)\left[c\left(b-a\right)+\left(b-a\right)\left(b+a\right)\right]\)
\(D=\left(b-c\right)\left(c-a\right)\left(b-a\right)\left(c+b+a\right)\)
\(D=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)
Chúc bạn học tốt.
a) (x + 1)(x + 2)(x + 3)(x + 4) - 24
= [(x + 1)(x + 4)].[(x + 2)(x + 3)] - 24
= (x2 + 5x + 4)(x2 + 5x + 6) - 24
= (x2 + 5x + 5 - 1)(x2 + 5x + 5 + 1) - 24
= (x2 + 5x + 5)2 - 1 - 24 = (x2 + 5x + 5)2 - 25
= (x2 + 5x)(x2 + 5x + 10)
= x(x + 5)(x2 + 5x + 10)
1/ Cho a,b,c đối 1 khác nhau thỏa mãn điều kiện (a + b + c)^2 = a^2 + b^2 + c^2 (^ là mũ)
Rút gọn biểu thức: P= (a^2)/(a^2+2bc) + (b^2)/(b^2+2ac)+(c^2)/(c^2+2ab)
2/ Phân tích đa thức thành nhân tử: (x + 1)^4 + (x^2 + x +1)^2
3/ Phân tích đa thức thành nhân tử: ab(a - b) + bc(b - c) + ca(c - a)
\(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)
\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-c\right)\)
\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-b+b-c\right)\)
\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-b\right)-ca\left(b-c\right)\)
\(=\left(a-b\right)\left(ab-ca\right)+\left(b-c\right)\left(bc-ca\right)\)
\(=\left(a-b\right)a\left(b-c\right)+\left(b-c\right)c\left(b-a\right)\)
\(=\left(a-b\right)a\left(b-c\right)-\left(b-c\right)c\left(a-b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
mình làm vội, có chỗ nào sai bạn thông cảm nha
phân tích đa thức thành nhân tử
a/(x+y)^2 -2(x+y)+1
b/(a+b+c)^3-a^3-b^3-c^3
c/ a^3+b^3+c^3 -3abc
a) \(\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)
b) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3c\left(a+b\right)\left(a+b+c\right)+c^3-a^3-b^3-c^3\)
\(=a^3+b^3+3ab\left(a+b\right)+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
c) \(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
phân tích đa thức thành nhân tử
1)(x^2+3x+1)(x^2+3x+2)-6
2)(x+1)(x+3)(x+5)(x+7)+15
3)a^3(b-c)+b^3(c-a)+c^3(a-b)
1)(x^2+3x+1)(x^2+3x+2)-6
Đặt t = x2 + 3x + 1
Khi đó PT có dạng:
t.(t + 1) - 6
= t2 + t - 6
= t2 - 2t - 3t - 6
= t.(t - 2) + 3.(t - 2)
= (t + 3).(t - 2)
= (x2 + 3x + 1 + 3).(x2 + 3x + 1 - 2)
= (x2 + 3x + 4).(x2 + 3x - 1)
\(1\hept{\begin{cases}\left(x^2+3x+2-1\right)\left(x^2+2x+2\right)-6\\\left(t-1\right)\left(t\right)-6\\t^2-t-6\end{cases}}.\) " đặt x^2+3x+2 = t
\(\hept{\begin{cases}t^2-\frac{2t.1}{2}+\frac{1}{4}-\left(\frac{24+1}{4}\right)\\\left(t-\frac{1}{2}\right)^2-\frac{25}{4}\\\left(t-\frac{1}{2}\right)^2-\frac{25}{4}\end{cases}}\)
\(\hept{\begin{cases}\left(t-\frac{1}{2}-\frac{5}{2}\right)\left(t-\frac{1}{2}+\frac{5}{2}\right)\\\left(t-\frac{7}{2}\right)\left(t+\frac{4}{2}\right)\\\left(t-\frac{7}{2}\right)\left(t+\frac{4}{2}\right)\end{cases}}\)
2) \(\hept{\begin{cases}\left\{\left(x+1\right)\left(x+7\right)\right\}\left\{\left(x+5\right)\left(x+3\right)\right\}+15\\\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\\t\left(t+8\right)+15\end{cases}}\)
\(\hept{\begin{cases}t^2+8t+15\\\left(t^2+8t+16\right)-1\\\left(t+4\right)^2-1\end{cases}}\Leftrightarrow\left(t+5\right)\left(t+4\right)\)
\(\hept{\begin{cases}a^3\left(b-c\right)+b^3\left(c-a+b-b\right)+c^3\left(a-b\right)\\a^3\left(b-c\right)-b^3\left(-c+a-b+b\right)+c^3\left(a-b\right)\\a^3\left(b-c\right)-b^3\left(a-b\right)-b^3\left(b-c\right)+c^3\left(a-b\right)\end{cases}\Leftrightarrow\hept{\begin{cases}\left(b-c\right)\left(a^3-b^3\right)-\left(a-b\right)\left(b^3-c^3\right)\\\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)-\left(a-b\right)\left(b-c\right)\left(b^2+ab+c^2\right)\\\left(a-b\right)\left(b-c\right)\left(a^2+2ab+2b^2+c^2\right)\end{cases}}}\)
1. Phân tích đa thức thành nhân tử:
a. (ab-1)^2+(a+b)^2
b. x^3+2x^2+2x+1
c. x^3-sx^2+12x-27
d. x64-2x^3+2x-1
e. x^4+2x^3+2x^2+2x+1
f. x^2-2x-4y^2-4y
g. x^4+2x^3-4x-4
h. x^2(1-x^2)-4-4x^2
i. (1+2x)(1-2x)-x(x+2)(x-2)
j. x^2+y^2-x^2.y^2+xy-x-y
2.Phân tích đa thức thành nhân tử:
a. a(b^2+c^2+bc)+b(c^2+a^2+ac)+c(a^2+b^2+ab)
b.(a+b+c)(ab+bc+ca)-abc
c. a(a+2b)^3-b(2a+b)^3
Phân tích đa thức thành nhân tử
1).a(b+c)^2(b-c)+b(c+a)^2(c-a)+c(a+b)^2(a-b)
2).x^8+x^4+1
3).x^12+1
trình bày cách lm nhaq
Phân tích đa thức thành nhân tử bằng phương pháp tách hạng tử:
a) x^2 - 6x +8
b) a^2 ( b-c ) + b^2 ( c-a ) + c^2 ( a-b )
c) x^3 - 7x - 6
a) \(=x^2-2x-4x+8\)
\(=x\left(x-2\right)-4\left(x-2\right)\)
\(=\left(x-2\right)\left(x-4\right)\)
c) \(=x^3-x-6x-6\)
\(=x\left(x^2-1\right)-6\left(x+1\right)\)
\(=x\left(x+1\right)\left(x-1\right)-6\left(x+1\right)\)
\(=x\left(x+1\right)\left(x-1-6\right)\)
\(=x\left(x+1\right)\left(x-7\right)\)
Phân tích các đa thức sau thành nhân tử
x^2((x^2+1/x^2)+6(x-1/x)+7)
(a-b)^3+(b-c)^3+(c-a)^3