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Câu 1 :
a) \(x^3-5x^2-14x\)
\(=x^3-7x^2+2x^2-14x\)
\(=x^2\left(x-7\right)+2x\left(x-7\right)\)
\(=\left(x-7\right)\left(x^2+2x\right)\)
\(=x\left(x-7\right)\left(x+2\right)\)
b) \(a^4+a^2+1\)
\(=\left(a^2\right)^2+2a^2+1-a^2\)
\(=\left(a^2+1\right)-a^2\)
\(=\left(a^2-a+1\right)\left(a^2+a+1\right)\)
c) \(x^4+64\)
\(=\left(x^2\right)^2+2\cdot x^2\cdot8+8^2-2\cdot x^2\cdot8\)
\(=\left(x^2+8\right)^2-\left(4x\right)^2\)
\(=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)
Câu 2 :
a) \(\left(a-b\right)^2=a^2-2ab+b^2\)
Ta có : \(\left(a+b\right)^2=a^2+2ab+b^2\)
\(\Rightarrow a^2+b^2=\left(a+b\right)^2-2ab=7^2-2\cdot14=25\)
\(\Rightarrow\left(a-b\right)^2=25-2\cdot12=1\)
b) tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
a2 + b2 =7
a+b=3
(a+b)2=9 =>a2 +b2 +2ab=9 <=>ab=1
=> a2 +b2 =7
a+b=3
ab=1
A=a4 +b4 = (a2 +b2 )2 -2a2b2
= 7-2.1=47
Ta có :
\(a^2+b^2=\left(a+b\right)^2-2ab=7\)
=> \(9-2ab=7\Rightarrow2ab=2\Rightarrow ab=1\)
Lại có :
\(a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2=7^2-2\cdot1=47\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(a-b\right)^2=a^2+b^2-2ab\\ \Rightarrow49=a^2+b^2-120\Rightarrow a^2+b^2=169\)
\(\left(a+b\right)^2=a^2+b^2+2ab=169+120=289\\ \Rightarrow a+b=17\)
\(a^2-b^2=\left(a-b\right)\left(a+b\right)=7\cdot17=119\)
\(a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2=169^2-2\cdot60^2\\ =28561-7200=21361\)
\(2\left(x^2+y^2\right)=\left(x-y\right)^2\\ \Rightarrow2x^2+2y^2=x^2-2xy+y^2\\ \Rightarrow x^2+2xy+y^2=0\\ \Rightarrow\left(x+y\right)^2=0\Rightarrow x+y=0\Rightarrow x=-y\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(a+b\right)^2-4ab\)
\(=a^2+2ab+b^2-4ab\)
\(=a^2-2ab+b^2\)
\(=\left(a-b\right)^2=VT\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
+ Chứng minh (a + b)2 = (a – b)2 + 4ab
Ta có:
VP = (a – b)2 + 4ab = a2 – 2ab + b2 + 4ab
= a2 + (4ab – 2ab) + b2
= a2 + 2ab + b2
= (a + b)2 = VT (đpcm)
+ Chứng minh (a – b)2 = (a + b)2 – 4ab
Ta có:
VP = (a + b)2 – 4ab = a2 + 2ab + b2 – 4ab
= a2 + (2ab – 4ab) + b2
= a2 – 2ab + b2
= (a – b)2 = VT (đpcm)
+ Áp dụng, tính:
a) (a – b)2 = (a + b)2 – 4ab = 72 – 4.12 = 49 – 48 = 1
b) (a + b)2 = (a – b)2 + 4ab = 202 + 4.3 = 400 + 12 = 412.
Ta có: \(ab=\dfrac{\left(a+b\right)^2-\left(a^2+b^2\right)}{2}=1\).
\(a^4+b^4=\left(a^2+b^2\right)-2a^2b^2=7^2-2=47\).
Sai một chút rồi bạn!
Cái chỗ \(a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2\) mới đúng bạn ạ!