phan tích đa thức thanh nhan tử:
\(\text{a}^3x-\text{a}b+b-x\)
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phan tich đa thức thanh nhan tử
\(4\text{a}^2b^2-\left(\text{a}^2+b^2-c^2\right)^2\)
Ta có: 4a2b2 -(a2+b2 -c2)2 =[2ab-(a2+b2-c2)][2ab+a2+b2-c2]
=[c2-(a-b)2][(a+b)2-c2]
=(-a+b+c)(a-b+c)(a+b+c)(a+b-c)
Có sai sót nào mong bỏ qua :)
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ta có \(4a^2b^4-\left(a^2+b^2-c^2\right)^2\)
= \(\left(2ab\right)^2-\left(a^2+b^2-c^2\right)^2\)
Theo hằng đẳng thức thứ 3 ta có:
\(\left(2ab+a^2+b^2-c^2\right)\left(2ab-a^2-b^2+c^2\right)\)
=\(\left[\left(a^2+2.a.b+b^2\right)-c^2\right]\left[-\left(a^2-2.a.b+b^2\right)+c^2\right]\)
=\(\left[\left(a+b\right)^2-c^2\right]\left[-\left(a-b\right)^2+c\right]\)
=\(\left(a+b+c\right)\left(a+b-c\right)\left(a+b+c\right)\left(a+b-c\right)\)
=\(2\left(a+b+c\right)2\left(a+b-c\right)\)
=\(\left(2a+2b+2c\right)\left(2a+2b-2c\right)\)
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phan tích nhan tử thanh nhan tử:a)3x^2-12y^2b)5xy^2-10xyt+5xt^2c)x^3+3x^2+3x+1-27x^3d)text{a}^3x-text{a}b+b-xe)3x^2left(text{a}+b+cright)+36xyleft(text{a}+b+cright)+108y^2left(text{a}+b+cright)f)text{a}bleft(text{a}-bright)+bcleft(b-cright)+ctext{a}left(c-text{a}right)g)left(text{a}+b+cright)^3-text{a}^3-b^3-c^3h)4text{a}^2b^2-left(text{a}^2+b^2-c^2right)^2
Đọc tiếp
phan tích nhan tử thanh nhan tử:
a)\(3x^2-12y^2\)
b)\(5xy^2-10xyt+5xt^2\)
c)\(x^3+3x^2+3x+1-27x^3\)
d)\(\text{a}^3x-\text{a}b+b-x\)
e)\(3x^2\left(\text{a}+b+c\right)+36xy\left(\text{a}+b+c\right)+108y^2\left(\text{a}+b+c\right)\)
f)\(\text{a}b\left(\text{a}-b\right)+bc\left(b-c\right)+c\text{a}\left(c-\text{a}\right)\)
g)\(\left(\text{a}+b+c\right)^3-\text{a}^3-b^3-c^3\)
h)\(4\text{a}^2b^2-\left(\text{a}^2+b^2-c^2\right)^2\)
phan tích da thuc thanh nhan tu
\(\text{a}b\left(\text{a}-b\right)+bc\left(b-c\right)+c\text{a}\left(c-\text{a}\right)\)
\(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)
\(=ab\left(a-b\right)+bc\left[\left(b-a\right)-\left(c-a\right)\right]+ca\left(c-a\right)\)
\(=ab\left(a-b\right)-bc\left(a-b\right)-bc\left(c-a\right)+ca\left(c-a\right)\)
\(=\left(a-b\right)\left(ab-bc\right)-\left(c-a\right)\left(bc-ca\right)\)
\(=b\left(a-b\right)\left(a-c\right)-c\left(c-a\right)\left(b-a\right)\)
\(=b\left(a-b\right)\left(a-c\right)-c\left(a-c\right)\left(a-b\right)\)
\(=\left(a-c\right)\left(a-b\right)\left(b-c\right)\)
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phan tích đa thức thanh nhan tử
\(\left(x^2+x+1\right)\left(x^2+3x+1\right)+x^2\)
\(\left(x^2+x+1\right)\left(x^2+3x+1\right)+x^2\)
\(=x^4+x^3+x^2+3x^3+3x^2+3x+x^2+x+1+x^2\)
\(=x^4+4x^3+6x^2+4x+1\)
\(=\left(x+1\right)^4\)
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phan tích đa thức thanh nhan tử
\(\left(x^2-x-x\right)^2+4x^2+4x\)
phan tich da thuc thanh nhan tu
\(\left(\text{a}+b+c\right)^3-\text{a}^3-b^3-c^3\)
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3-a^3-b^3-c^3\)
\(=a^3+b^3+3a^2b+3ab^2+3c\left(a^2+2ab+b^2\right)+3ac^2+3bc^2-a^3-b^3\)
\(=3a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^2\)
\(=3\left(a^2b+ab^2+a^2c+ac^2+2abc+b^2c+bc^2\right)\)
\(=3\left(a^2b+ab^2+a^2c+ac^2+abc+abc+b^2c+bc^2\right)\)
\(=3\left[ab\left(a+b\right)+c^2\left(a+b\right)+ac\left(a+b\right)+bc\left(a+b\right)\right]\)
\(=3\left(a+b\right)\left(ab+c^2+ac+bc\right)\)
\(=3\left(a+b\right)\left[c\left(a+c\right)+b\left(a+c\right)\right]\)
\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
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phan tích đa thức thanh nhan tử
\(\left(x+2y-3\right)^2-4\left(x+2y-3\right)+4\)
(x + 2y - 3)2 - 4(x + 2y - 3) + 4
= (x + 2y - 3)2 - 2. 2. (x + 2y - 3) + 22 (hằng đẳng thức số 2, bình phương của một hiệu)
= ( x + 2y - 3 - 2)2
= ( x + 2y - 5)2
. x^3 + 2x^2 + x – 9x giúp mình với Phân tích đa thức thành nhân tử. chỉ tao coi tụi mày
X×134-x×24-x×10=1300
a)\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
phan tích đa thúc thanh nhan tử
\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(=x\left(x+3\right)\left(x+1\right)\left(x+2\right)+1\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)(1)
Đặt \(x^2+3x+1=t\)thay vào (1) ta được :
\(\left(t-1\right)\left(t+1\right)+1\)
\(=t^2-1+1\)
\(=t^2\)Thay \(t=x^2+3x+1\)ta được:
\(\left(x^2+3x+1\right)^2\)
\(=\left(x^2+2.\frac{3}{2}x+\frac{9}{4}-\frac{9}{4}+1\right)^2\)
\(=\left[\left(x+\frac{3}{2}\right)-\frac{5}{4}\right]^2\)
\(=\left(x+\frac{3}{2}-\frac{\sqrt{5}}{2}\right)^2\left(x+\frac{3}{2}+\frac{\sqrt{5}}{2}\right)^2\)
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