Phân tích thành nhân tử:
a) \(B=\left(4x+1\right)\left(12x-1\right)\left(12x-1\right)\left(x+1\right)-4\)
b) \(C=\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
Phân tích các đa thức sau thành nhân tử:
\(A=4x^2+6x\). \(B=\left(2x+3\right)^2-x\left(2x+3\right)\). \(C=\left(9x^2-1\right)-\left(3x-1\right)^2\).
\(D=x^3-16x\). \(E=4x^2-25y^2\). \(G=\left(2x+3\right)^2-\left(2x-3\right)^2\).
\(A=4x^2+6x=2x\left(2x+3\right)\)
\(B=\left(2x+3\right)^2-x\left(2x+3\right)=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\)
\(C=\left(9x^2-1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1-3x+1\right)=2\left(3x+1\right)\)
\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)
\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)
\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)
\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)
Phân tích các đa thức sau thành nhân tử:
a) \({x^3} + 4x\)
b) \(6ab - 9a{b^2}\)
c) \(2a\left( {x - 1} \right) + 3b\left( {1 - x} \right)\)
d) \({\left( {x - y} \right)^2} - x\left( {y - x} \right)\)
`a, x^3 + 4x = x(x^2+4)`
`b, 6ab - 9ab^2 = 3ab(2-b)`
`c, 2a(x-1) + 3b(1-x)`
`= (2a-3b)(x-1)`
`d, (x-y)^2 - x(y-x)`
`= (x-y+x)(x-y)`
`= (2x-y)(x-y)`
Phân tích thành nhân tử:
a) \(B=\left(4x+1\right)\left(12x-1\right)\left(12x-1\right)\left(x+1\right)-4\)
b) \(C=\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
c) \(D=\left(x^2+2x\right)^2+9x^2+18x+20\)
d) \(E=\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
e) \(F=\left(x^2+8x+7\right)\left(x+3\right)\left(x+5\right)+15\)
vào câu hỏi tương tự mà lm tương tự như thế nha
giải pt:
a,\(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
b,\(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
PHân tích các đa thức sau thành nhân tử
a) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
b) \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
Đặt \(A=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(A=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+10=y\)
\(\Rightarrow\)\(A=y.\left(y+2\right)-24\)
\(A=y^2+2y+1-25\)
\(A=\left(y+1\right)^2-5^2\)
\(A=\left(y+1-5\right)\left(y+1+5\right)\)
\(A=\left(y-4\right)\left(y+6\right)\)
\(\Rightarrow A=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(A=\left[\left(x^2+x\right)+\left(6x+6\right)\right].\left(x^2+7x+16\right)\)
\(A=\left[x.\left(x+1\right)+6.\left(x+1\right)\right].\left(x^2+7x+16\right)\)
\(A=\left(x+1\right).\left(x+6\right).\left(x^2+7x+16\right)\)
Đặt \(B=\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(B=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
Đặt \(12x^2+11x-1=a\)
\(\Rightarrow B=a.\left(a+3\right)-4\)
\(B=a^2+3a-4\)
\(B=\left(a^2-a\right)+\left(4a-4\right)\)
\(B=a.\left(a-1\right)+4.\left(a-1\right)\)
\(B=\left(a-1\right)\left(a+4\right)\)
\(\Rightarrow B=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)
phân tích đa thức thành nhân tử:
a. \(ax^2-a^2x-x+a\)
b. \(18x^3-12x^2+2x\)
c. \(x^3-5x^2-4x+20\)
d. \(\left(x+7\right)\left(x+15\right)+15\)
\(a.\) \(ax^2-a^2x-x+a\)
\(=\left(ax^2-a^2x\right)-\left(x-a\right)\)
\(=ax\left(x-a\right)-\left(x-a\right)\)
\(=\left(ax-1\right)\left(x-a\right)\)
\(b.\) \(18x^3-12x^2+2x\)
\(=2x\left(9x^2-6x+1\right)\)
\(=2x\left(3x-1\right)^2\)
\(c.\) \(x^3-5x^2-4x+20\)
\(=\left(x^3-5x^2\right)-\left(4x-20\right)\)
\(=x^2\left(x-5\right)-4\left(x-5\right)\)
\(=\left(x^2-4\right)\left(x-5\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-5\right)\)
\(d.\) \(\left(x+7\right)\left(x+15\right)+15\)
\(=x^2+15x+7x+105+15\)
\(=x^2+22x+120\)
\(=\left(x+10\right)\left(x+12\right)\)
Phân tích các đa thức sau thành nhân tử:
a) \(4{x^2} - 1\)
b) \({\left( {x + 2} \right)^2} - 9\)
c) \({\left( {a + b} \right)^2} - {\left( {a - 2b} \right)^2}\)
a) \(4x^2-1=\left(2x+1\right)\left(2x-1\right)\)
b) \(\left(x+2\right)^2-9=\left(x-1\right)\left(x+5\right)\)
c) \(\left(a+b\right)^2-\left(a-2b\right)^2\)
\(=\left(a+b-a+2b\right)\left(a+b+a-2b\right)\)
\(=3b\left(2a-b\right)\)
`a, 4x^2-1 = (2x+1)(2x-1)`
`b, (x+2)^2-9 = (x+2-3)(x+2+3) = (x-1)(x+5)`
`c, (a+b)^2-(a-2b)^2 = (a+b+a-2b)(a+b-a+2b) = (2a-b)(3b)`
Phân tích đa thức sau thành nhân tử:
a) \(\left(x^2+8x+7\right).\left(x+3\right).\left(x+5\right)+15\)
b) \(\left(4x+1\right).\left(12x-1\right).\left(3x+2\right).\left(x+1\right)-4\)
c) \(\left(x^2+2x\right)^2+9x^2+18x+20\)
a) Ta có: \(\left(x^2+8x+7\right)\left(x+3\right)\left(x+5\right)+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
\(=\left(x^2+8x\right)^2+22\left(x^2+8x\right)+105+15\)
\(=\left(x^2+8x\right)^2+22\left(x^2+8x\right)+120\)
\(=\left(x^2+8x\right)^2+12\left(x^2+8x\right)+10\left(x^2+8x\right)+120\)
\(=\left(x^2+8x\right)\left(x^2+8x+12\right)+10\left(x^2+8x+12\right)\)
\(=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)
\(=\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)
b) Ta có: \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
\(=\left(12x^2+11x\right)^2+\left(12x^2+11x\right)-2-4\)
\(=\left(12x^2+11x\right)^2+\left(12x^2+11x\right)-6\)
\(=\left(12x^2+11x\right)^2+3\left(12x^2+11x\right)-2\left(12x^2+11x\right)-6\)
\(=\left(12x^2+11x\right)\left(12x^2+11x+3\right)-2\left(12x^2+11x+3\right)\)
\(=\left(12x^2+11x+3\right)\left(12x^2+11x-2\right)\)
c) Ta có: \(\left(x^2+2x\right)^2+9x^2+18x+20\)
\(=\left(x^2+2x\right)^2+9\left(x^2+2x\right)+20\)
\(=\left(x^2+2x\right)^2+5\left(x^2+2x\right)+4\left(x^2+2x\right)+20\)
\(=\left(x^2+2x\right)\left(x^2+2x+5\right)+4\left(x^2+2x+5\right)\)
\(=\left(x^2+2x+5\right)\left(x^2+2x+4\right)\)
