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a) (a-b)3 - (a3 - b3)
= a3 - b3 - a3 +b3 =0
b) ( a+ b) . (a2 -ab +b2) + (a-b) .( a2+ab +b2)
=(a3 + b3) +(a3-b3) =a3 + b3+a3 -b3 =2a3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=a^2+b^2+c^2+2ab-2bc-2ac+a^2+b^2+c^2-2ab+2ac-2bc-2b^2+4bc-c^2\)
\(=2a^2\)
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Xin lỗi mk mới học lớp 5 mk cũng muốn giúp bạn lắm bạn thông cảm nhưng bài này mk chưa học
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\(\left(a+b\right)^2-\left(a-b\right)^2\)
\(=\left(a+b+a-b\right)\left(a+b-a+b\right)\)
\(=2a.2b=4ab\)
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\(\left(a^2+b^2\right)^2-4a^2b^2=a^4+b^4+2a^2b^2-4a^2b^2=a^4+b^4-2a^2b^2=\left(a^2-b^2\right)^2=5^2\)
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\(A=\left[\left(a+b\right)+\left(c+d\right)\right]^2+\left[\left(a+b\right)-\left(c+d\right)\right]^2+\left[\left(a-b\right)+\left(c-d\right)\right]^2+\left[\left(a-b\right)-\left(c-d\right)\right]^2\)
Ta có
\(\left[\left(a+b\right)+\left(c+d\right)\right]^2=\left(a+b\right)^2+2\left(a+b\right)\left(c+d\right)+\left(c+d\right)^2\)
\(\left[\left(a+b\right)-\left(c+d\right)\right]^2=\left(a+b\right)^2-2\left(a+b\right)\left(c+d\right)+\left(c+d\right)^2\)
\(\left[\left(a-b\right)+\left(c-d\right)\right]^2=\left(a-b\right)^2+2\left(a-b\right)\left(c-d\right)+\left(c-d\right)^2\)
\(\left[\left(a-b\right)-\left(c-d\right)\right]^2=\left(a-b\right)^2-2\left(a-b\right)\left(c-d\right)+\left(c-d\right)^2\)
\(A=2\left(a+b\right)^2+2\left(a-b\right)^2+2\left(c+d\right)^2+2\left(c-d\right)^2\)
\(A=2\left(a^2+2ab+b^2+a^2-2ab+b^2+c^2+2cd+d^2+c^2-2cd+d^2\right)\)
\(A=4\left(a^2+b^2+c^2+d^2\right)\)
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`Answer:`
`a)`
`A=5(x+1)^2-3(x-3)^2-4(x^2-4)`
`=>A=5(x^2+2x+1)-3(x^2-6x+9)-4x^2+16`
`=>A=5x^2+10x+5-3x^2+18x-27-4x^2+16`
`=>A=(5x^2-3x^2-4x^2)+(10x+18x)+(5-27+16)`
`=>A=-2x^2+28x-6`
`b)`
`B=5(x+1)^2-3(x-3)^2-4(x+2)(x-2)`
`=2x(3x+5)-3(3x+5)-2x(x^2-4x+4)-[(2x)^2-3^2]`
`=6x^2+10x-9x-15-2x^3+8x^2-8x-4x^2+9`
`=(6x^2-4x^2+8x^2)-2x^3+(10x-9x-8x)+(-15+9)`
Thay `x=-7` vào ta được:
`B=10(-7)^2-2(-7)^3-7(-7)-6`
`=>B=10.49-2(-343)+49-6`
`=>B=490+686+49-6`
`=>B=1219`
\(\dfrac{a^2b}{ab^2-a^2b}=\dfrac{ab.a}{ab\left(b-a\right)}=\dfrac{a}{b-a}\)