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\(\left(a-x-y\right)^3-\left(a+x-y\right)^3\)
\(=\left[\left(a-x-y\right)-\left(a+x-y\right)\right]\left[\left(a-x-y\right)^2+\left(a-x-y\right)\left(a+x-y\right)+\left(a+x-y\right)^2\right]\)
\(=-2x.\left[a^2+x^2+y^2-2ax+2xy-2ay+\left(a-y\right)^2-x^2+a^2+x^2+y^2+2ax-2xy-2ay\right]\)
\(=-2x\left[a^2+x^2+y^2-2ax+2xy-2ay+a^2-2ay+y^2-x^2+a^2+x^2+y^2+2ax-2xy-2ay\right]\)
\(=-2x\left(3a^2+x^2+3y^2-4ay\right)\)
1. (A+B)2 = A2+2AB+B2
2. (A – B)2= A2 – 2AB+ B2
3. A2 – B2= (A-B)(A+B)
4. (A+B)3= A3+3A2B +3AB2+B3
5. (A – B)3 = A3- 3A2B+ 3AB2- B3
6. A3 + B3= (A+B)(A2- AB +B2)
7. A3- B3= (A- B)(A2+ AB+ B2)
8. (A+B+C)2= A2+ B2+C2+2 AB+ 2AC+ 2BC
* CHÚ Ý;
a/ a+b= -(-a-b) ; b/ (a+b)2= (-a-b)2 ; c/ (a-b)2= (b-a)2 ; d/ (a+b)3= -(-a-b)3 e/ (a-b)3=-(-a+b)3
(a+b)^2=a^2+2ab+b^2
(a-b)^2=a^2-2ab+b^2
a^2-b^2=(a+b)(a-b)
(a+b)^3=a^3+3a^2b+3ab^2+b^3
(a-b)^3=a^3-3a^2b+3ab^2-b^3
a^3+b^3=(a+b)(a^2-ab+b^2)
a^3-b^3=(a-b)(a^2+ab+b^2)
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\(A=\left(3x-2\right)^2-\left(x+3\right)^2\)
\(=\left(3x-2+x+3\right)\left(3x-2-x-3\right)\)
\(=\left(4x+1\right)\left(2x-5\right)\)
\(B=\left(x+2y+3z\right)^2-\left(x-2y-3z\right)^2\)
\(=\left(x+2y+3z-x+2y+3z\right)\left(x+2y+3z+x-2y-3z\right)\)
\(=2x\left(4y+6z\right)\)
\(=4x\left(2y+3z\right)\)
Ta có :
\(4x^2+12x+10>0\)
\(\Leftrightarrow\)\(\left(4x^2+12x+9\right)+1>0\)
\(\Leftrightarrow\)\(\left[\left(2x\right)^2+2.2x.3+3^2\right]+1>0\)
\(\Leftrightarrow\)\(\left(2x+3\right)^2+1\ge1>0\)
Vậy \(4x^2+12x+10\) luôn dương với mọi giá trị x
Chúc bạn học tốt ~
\(x^2+xy+y^2+1=\left(x^2+xy+\frac{y^2}{4}\right)+\frac{3y^2}{4}+1=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+1>0\)
B = -x2 + 24x - 405
= -(x2 - 24x + 405)
= -(x2 - 2.x.12 + 122) - 261
= -(x - 12)2 - 261