cho đt:p=3a^2x^2+4b^2x^2-2a^2x^2-3b^2x^2+2021;tìm GTNN của p biết a,b khác 0
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a) \(\left(2a-b\right)\left(b+4a\right)+2a\left(b-3a\right)\)
\(=2ab+8a^2-b^2-4ab+2ab-6a^2\)
\(=\left(2ab+2ab-4ab\right)+\left(8a^2-6a^2\right)-b^2\)
\(=2a^2-b^2\)
b) \(\left(3a-2b\right).\left(2a-3b\right)-6a\left(a-b\right)\)
\(=6a^2-9ab-4ab+6b^2-6a^2+6ab\)
\(=\left(6a^2-6a^2\right)-\left(9ab+4ab-6ab\right)+6b^2\)
\(=-7ab+b^2\)
c) \(5b\left(2x-b\right)-\left(8b-x\right)\left(2x-b\right)\)
\(=10bx-5b^2-\left(16bx-8b^2-2x^2+bx\right)\)
\(=10bx-5b^2-16bx+8b^2+2x^2-bx\)
\(=\left(10bx-16bx-bx\right)-\left(5b^2-8b^2\right)+2x^2\)
\(=-7bx+3b^2+2x^2\)
d) \(2x\left(a+15x\right)+\left(x-6a\right)\left(5a+2x\right)\)
\(=2ax+30x^2+5ax+2x^2-30a^2-12ax\)
\(=\left(2ax+5ax-12ax\right)+\left(30x^2+2x^2\right)-30a^2\)
\(=-5ax+32x^2-30a^2\)
a: =2ab+8a^2-b^2-4ab+2ab-6a^2
=2a^2-b^2
b: =6a^2-9ab-4ab+6b^2-6a^2+6ab
=-7ab+6b^2
c: =10bx-5b^2-16bx+8b^2+2x^2-xb
=3b^2+2x^2-7xb
d: =2xa+30x^2+5ax+2x^2-30a^2-12ax
=32x^2-30a^2-5ax
1/Tự chép lại đb nha :v
=a2 - 9b2+2ab+3a2-8b2-12ab+6ab-3b2-2a2+ab
= 2a2-3ab-20b2
= (2a2+5ab) - (8ab+20b2)
= a(2a+5b) - 4b(2a+5b)
=(2a+5b)(a-4b)
câu 2 tương tự nhé :)
\(\left(2x+1\right)^2-2\left(2x+1\right)\left(3-x\right)+\left(x-3\right)^2\)
\(=\left(2x+1\right)^2+2\left(2x-1\right)\left(x-3\right)+\left(x-3\right)^2\)
\(=\left(2x+1+x-3\right)^2\)
\(=\left(3x-2\right)^2\)
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\(a^3+3a^2-6a-8\)
\(=a^3+4a^2-a^2-4a-2a-8\)
\(=\left(a^3+4a^2\right)-\left(a^2+4a\right)-\left(2a+8\right)\)
\(=a^2\left(a+4\right)-a\left(a+4\right)-2\left(a+4\right)\)
\(=\left(a+4\right)\left(a^2-a-2\right)\)
\(=\left(a+4\right)\left(a^2-2a+a-2\right)\)
\(=\left(a+4\right)\left[\left(a^2-2a\right)+\left(a-2\right)\right]\)
\(=\left(a+4\right)\left[a\left(a-2\right)+\left(a-2\right)\right]\)
\(=\left(a+4\right)\left(a-2\right)\left(a+1\right)\)
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\(2x^2-5x+2\)
\(=2x^2-4x-x+2\)
\(=\left(2x^2-4x\right)-\left(x-2\right)\)
\(=2x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(2x-1\right)\)
-----------------------------------------
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x-4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+2y-2\right)\)
-------------------------------------
\(a^2-1+4b-4b^2\)
\(=a^2-\left(1-4b+4b^2\right)\)
\(=a^2-\left(1-2b\right)^2\)
\(=\left(a-1+2b\right)\left(a+1-2b\right)\)
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\(a^4+6a^2b+9b^2-1\)
\(=\left(a^4+6a^2b+9b^2\right)-1\)
\(=\left(a^2+3b\right)^2-1\)
\(=\left(a^2+3b-1\right)\left(a^2+3b+1\right)\)
---------------------------------
\(2x^3+16y^3\)
\(=2\left(x^3+8y^3\right)\)
\(=2\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
Lần sau ghi đề tách riêng từng câu ra nhé em. Ghi dính chùm vậy khó nhìn lắm. Sẽ ít ai giải cho em
(3x^3 - 2x^2 + x + 2)(5x^2)
= 15x^5 - 10x^4 + 5x^3 + 10x^2
(3x^2 + 5x - 2)(2x^2 - 4x + 3)
= 3x^4 - 12x^3 + 9x^2 + 10x^3 - 20x^2 + 15x - 4x^2 + 8x - 6
= 6x^4 - 2x^3 - 15x^2 + 23x - 6
Áp dụng định lý Bezout ta có:
f(x) chia hết cho x-3 \(\Rightarrow f\left(3\right)=0\)
\(\Leftrightarrow2a+3b=-87\left(1\right)\)
g(x) chia hết cho x-3 \(\Rightarrow g\left(3\right)=0\)
\(\Leftrightarrow-3a+2b=-318\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\hept{\begin{cases}2a+3b=-87\\-3a+2b=-318\end{cases}\Leftrightarrow}\hept{\begin{cases}a=60\\b=-69\end{cases}}\)
Vậy ...