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a) 2x3 + 8x2 - 8x
= 2x(x2 + 4x - 4)
= 2x(x2 + 4x + 4 - 8)
= 2x[(x + 2)2 - 8]
= \(2x\left(x+2-\sqrt{8}\right)\left(x+2+\sqrt{8}\right)\)
b) a2 - b2 + 4a + 4b
= (a - b)(a + b) + 4(a + b)
= (a + b)(a - b + 4)
c) x2 - 2x - 3
= x2 + x - 3x - 3
= x(x + 1) - 3(x + 1)
= (x + 1)(x - 3)
d) x2 - 4x - 3
= x2 - 4x + 4 - 7
= (x + 2)2 - 7
= \(\left(x+2-\sqrt{7}\right)\left(x+2+\sqrt{7}\right)\)
1/Tự chép lại đb nha :v
=a2 - 9b2+2ab+3a2-8b2-12ab+6ab-3b2-2a2+ab
= 2a2-3ab-20b2
= (2a2+5ab) - (8ab+20b2)
= a(2a+5b) - 4b(2a+5b)
=(2a+5b)(a-4b)
câu 2 tương tự nhé :)
1,
b,a2+4ab+4b2
=a2+2.a.2b+(2b)2
=(a+2b)2
c,5x.(x-2y)+2(x-2y)2
=5x(x-2y)+2.(x-2y).(x-2y)
=(x-2y).[5x+2.(x-2y)]
=(x-2y).(5x+2x-4y)
=(x-2y).(7x-4y)
nhớ t*** mình nha mỏi tay quá!!!
còn câu a sử dụng hằng đẳng thức hiệu hai bình phương nha
c) \(\left(2x+5\right)^2-\left(x-9\right)^2\)
\(=\left(2x+5+x-9\right)\left(2x+5-x+9\right)\)
\(=\left(3x-4\right)\left(x+14\right)\)
d) \(\left(3x+1\right)^2-4\left(x-2\right)^2\)
\(=\left(3x+2\right)^2-\left[2\left(x-2\right)\right]^2\)
\(=\left[3x+2+2\left(x-2\right)\right]\left[3x+2-2\left(x-2\right)\right]\)
\(=\left(3x+2+2x-4\right)\left(3x+2-2x+4\right)\)
\(=\left(5x-2\right)\left(x+6\right)\)
e) \(9\left(2x+3\right)^2-4\left(x+1\right)^2\)
\(=\left[3\left(2x+3\right)\right]^2-\left[2\left(x+1\right)\right]^2\)
\(=\left[3\left(2x+3\right)+2\left(x+1\right)\right]\left[3\left(2x+3\right)-2\left(x+1\right)\right]\)
\(=\left(6x+9+2x+2\right)\left(6x+9-2x-2\right)\)
\(=\left(8x+11\right)\left(4x+7\right)\)
f) \(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(2bc\right)^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(2bc+b^2+c^2-a^2\right)\left(2bc-b^2-c^2+a^2\right)\)
\(=\left[\left(b+c\right)^2-a^2\right]\left[a^2-\left(b-c\right)^2\right]\)
\(=\left(a+b+c\right)\left(b+c-a\right)\left(a-b+c\right)\left(a+b-c\right)\)
Ta có : \(\left(2x+5\right)^2-\left(x-9\right)^2\)
\(=\left(4x^2+20x+25\right)-\left(x^2-18x+81\right)\)
\(=4x^2+20x+25-x^2+18x-81\)
\(=3x^2+38x-56\)
\(=3x^2+42x-4x-56\)
\(=3x\left(x+14\right)-\left(4x+56\right)\)
\(=3x\left(x+14\right)-4\left(x+14\right)\)
\(=\left(x+14\right)\left(3x-4\right)\)
a) \(y^2+2y^2-3y=y.y+y.2y-y.3\)
\(=y\left(y+2y-3\right)\)
b) \(2x^4-x^3+2x^2+1=2x^2.x^2-x^2.x+2x.x^2+1\)
\(=x^2\left(2x^2-x+2x\right)+1=x^2.x\left(2x-1+2\right)+1\)
k mình nha
a. y2+2y2-3y
=y(y+2y-3)
b. 2x4-x3+2x2+1
=2x2(x2+1)-(x3-1)
=2x2(x+1)(x-1)-(x-1)(x2+x+1)
=(x-1)[2x2(x+1)-(x2+x+1)]
=(x-1)(2x3+2x2-x2-x-1)
=(x-1)(2x3+x2-x-1)
Em sửa lại tên đi nhé!
\(\left(x^2-1\right)^2-x\left(x^2-1\right)-2x^2\)
= \(\left(x^2-1\right)^2-2.\left(x^2-1\right).\frac{x}{2}+\frac{x^2}{4}-\frac{x^2}{4}-2x^2\)
= \(\left(x^2-1-\frac{x}{2}\right)^2-\frac{9}{4}x^2\)
\(=\left(x^2-1-\frac{x}{2}-\frac{3}{2}x\right)\left(x^2-1-\frac{x}{2}+\frac{3}{2}x\right)\)
= \(\left(x^2-2x-1\right)\left(x^2-x-1\right)\)
Phân tích tiếp được đấy:
\(x^2-2x-1=\left(x-1\right)^2-2=\left(x-1-\sqrt{2}\right)\left(x-1+\sqrt{2}\right)\)
\(x^2-x-1=\left(x-\frac{1}{2}\right)^2-\frac{5}{4}=\left(x-\frac{1}{2}-\frac{\sqrt{5}}{2}\right)\left(x-\frac{1}{2}+\frac{\sqrt{5}}{2}\right)\)
Thay vào nhé!
câu a sai đề ko bạn