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a: \(A=25a^2+50a+25+10\left(a^2-2a-3\right)+a^2-6a+9\)
\(=26a^2+46a+34+10a^2-20a-30\)
\(=36a^2+26a+4\)
b: \(B=\dfrac{1}{4}\left(x^2-2x+1\right)+x^2-1+x^2+2x+1\)
\(=\dfrac{1}{4}x^2-\dfrac{1}{2}x+\dfrac{1}{4}+2x^2+2x\)
\(=\dfrac{9}{4}x^2+\dfrac{3}{2}x+\dfrac{1}{4}\)
a) Điều kiện : \(a\ne-b;b\ne1;a\ne-1\)
\(P=\frac{a^2\left(1+a\right)-b^2\left(1-b\right)-a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1-b\right)\left(1+a\right)}\)
\(P=\frac{a^3+a^2+b^3-b^2-a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1-b\right)\left(1+a\right)}\)
\(P=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a+b\right)\left(a-b\right)-a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1-b\right)\left(1+a\right)}\)
\(P=\frac{\left(a+b\right)\left(a^2-ab+b^2+a-b-a^2b^2\right)}{\left(a+b\right)\left(1-b\right)\left(1+a\right)}\)
\(P=\frac{a^2+b^2-a^2b^2+a-b-ab}{\left(1-b\right)\left(1+a\right)}\)
\(P=\frac{a^2\left(1-b^2\right)-\left(1-b^2\right)+a\left(1-b\right)+\left(1-b\right)}{\left(1-b\right)\left(1+a\right)}\)
\(P=\frac{\left(1-b\right)\left(a^2+a^2b-1-b+a+1\right)}{\left(1-b\right)\left(1+a\right)}\)
\(P=\frac{a^2+a^2b+a-b}{1+a}\)
\(P=\frac{a\left(a+1\right)+b\left(a-1\right)\left(a+1\right)}{1+a}\)
\(P=\frac{\left(a+1\right)\left(a+ab-b\right)}{1+a}\)
P = a + ab - b
b)
P = 3
<=> a + ab - b = 3
<=> a(b+1) - (b+1) +1 - 3 = 0
<=> (b+1)(a-1) = 2
Ta có bảng sau với a, b nguyên
| b+1 | 1 | 2 | -1 | -2 |
| a-1 | 2 | 1 | -2 | -1 |
| b | 0 | 1 | -2 | -3 |
| a | 3 | 2 | -1 | 0 |
| so với đk | loại | loại |
Vậy (a;b) \(\in\){ (3; 0) ; (0; -3)}
\(a,\left(-4xy-5\right)\left(5-4xy\right)=\left(4xy+5\right)\left(4xy-5\right).\)
\(=\left(4xy\right)^2-5^2=16x^2y^2-25\)
\(b,\left(a^2b+ab^2\right)\left(ab^2-a^2b\right)=\left(ab^2+a^2b\right)\left(ab^2-a^2b\right)\)
\(=\left(ab^2\right)^2-\left(a^2b\right)^2=a^2b^4-a^4b^2\)
\(c,\left(3x-4\right)^2+2\left(3x-4\right)\left(4-x\right)+\left(4-x\right)^2\)
\(=\left[\left(3x-4\right)+\left(4-x\right)\right]^2\)
\(=\left(3x-4+4-x\right)^2=\left(2x\right)^2=4x^2\)
\(d,\left(a^2+ab+b^2\right)\left(a^2-ab+b^2\right)-\left(a^4+b^4\right)\)
\(=\left[\left(a^2+b^2\right)+ab\right]\left[\left(a^2+b^2\right)-ab\right]-\left(a^4+b^4\right)\)
\(=\left(a^2+b^2\right)^2-\left(ab\right)^2-a^4-b^4\)
\(=a^4+2a^2b^2+b^4-a^2b^2-a^4-b^4=a^2b^2\)
a: Ta có: \(\left(3x-1\right)^2-2\left(5x-2\right)^2-2\left(x^2+x-1\right)\left(x-1\right)\)
\(=9x^2-6x+1-2\left(25x^2-20x+4\right)-2\left(x^3-x^2+x^2-x-x+1\right)\)
\(=9x^2-6x+1-50x^2+40x-8-2\left(x^3-2x+1\right)\)
\(=-41x^2+34x-7-2x^3+4x-2\)
\(=-2x^3-41x^2+38x-9\)
b: Ta có: \(\left(3a+1\right)^2+2\left(9a^2-1\right)+\left(3a-1\right)^2\)
\(=\left(3a+1+3a-1\right)^2\)
\(=36a^2\)
\(M=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left(a^2+b^2+2ab-2ab\right)+6a^2b^2\left(a+b\right)\)
\(M=a^2+2ab+b^2-3ab+3ab-6a^2b^2+6a^2b^2\)
\(M=\left(a+b\right)^2=1\)
\(A=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\left(a^2+b^2-c^2-a^2+b^2-c^2\right)-4a^2b^2\)
\(=2a^2.2b^2-4a^2b^2=0\)
\(C=\left(2-6x\right)^2+\left(2-5x\right)^2+2\left(6x-2\right)\left(2-5x\right)\)
\(=\left[\left(2-6x\right)+\left(2-5x\right)\right]^2\)
\(=\left[4-11x\right]^2\)
\(=16-88x+121x^2\)
chúc bn học tốt
\(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\)
\(=1-3ab+3ab\left[1-2ab\right]+6a^2b^2\)
\(=1-3ab+3ab-6a^2b^2+6a^2b^2\)
=1