a 2 x + 3 a + 9 = a 2   ⇔   a x a + 3 = a 2 - 9

Ta có \(A\left(1\right)=B\left(-2\right)\Leftrightarrow12+2a+a^2=8-\left|2a+3\right|\left(-2\right)+a^2\)

\(\Leftrightarrow4+2a=2\left|2a+3\right|\)

đk a >= -2 

\(\left[{}\begin{matrix}4a+6=4+2a\\4a+6=-2a-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=-1\left(tm\right)\\a=-\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)

2: \(8xy-24xy+16x\)

\(=8x\cdot y-8x\cdot3y+8x\cdot2\)

\(=8x\left(y-3y+2\right)=8x\left(-2y+2\right)\)

\(=-16y\left(y-1\right)\)

3: \(xy-x=x\cdot y-x\cdot1=x\left(y-1\right)\)

11: \(2mx-4m2xy+6mx\)

\(=2mx-2my\cdot4y+2mx\cdot3\)

\(=2mx\left(1-4y+3\right)\)

\(=2mx\left(4-4y\right)=8mx\left(1-y\right)\)

12: \(7x^2y^5-14x^3y^4-21y^3\)

\(=7y^3\cdot x^2y^2-7y^3\cdot2x^3y-7y^3\cdot3\)

\(=7y^3\left(x^2y^2-2x^3y-3\right)\)

13: \(2\left(x-y\right)-a\left(x-y\right)\)

\(=2\cdot\left(x-y\right)-a\cdot\left(x-y\right)\)

\(=\left(x-y\right)\left(2-a\right)\)

Đáp án A

Gọi số khối của X lần lượt là A₁, A₂, A3

Ta có: 

A 1   +   A 2   +   A 3   = 75 A 2 - A 1   = 1 0 , 79 . A 1   +   0 , 1 .   A 2   +   0 , 11   A 3 1 =   24 , 32

⇒ A 1 = 24 A 2 = 25 A 3 = 26

\(a,=\left(xy-1-x-y\right)\left(xy-1+x+y\right)\\ b,Sửa:a^3+2a^2+2a+1\\ =a^3+a^2+a^2+a+a+1=\left(a+1\right)\left(a^2+a+1\right)\\ c,=1-4a^2-a\left(a^2-4\right)=1-4a^2-a^3+4a\\ =\left(1-a\right)\left(1+a+a^2\right)+4a\left(1-a\right)\\ =\left(1-a\right)\left(1+5a+a^2\right)\\ d,=\left(a^2-a^2b^2\right)+\left(b^2-b\right)+\left(ab-a\right)\\ =a^2\left(1-b\right)\left(1+b\right)+b\left(b-1\right)+a\left(b-1\right)\\ =\left(b-1\right)\left(-a^2-ab+b+a\right)\\ =\left(b-1\right)\left(b-1\right)\left(a+b\right)\left(1-a\right)\)

\(e,=x^2y+xy^2-yz\left(y+z\right)+x^2z-xz^2\\ =\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\\ =x^2\left(y+z\right)+x\left(y-z\right)\left(y+z\right)-yz\left(y+z\right)\\ =\left(y+z\right)\left(x^2+xy-xz-yz\right)\\ =\left(y+z\right)\left(x+y\right)\left(x-z\right)\)

\(f,=xyz-xy-yz-xz+x+y+z-1\\ =xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(x-1\right)\\ =\left(z-1\right)\left(xy-y-x+1\right)=\left(z-1\right)\left(x-1\right)\left(y-1\right)\)

$P\le \sqrt{a^2+2\sqrt{a}ab+2b^2}+\sqrt{b^2+2\sqrt{2}bc+2c^2}+\sqrt{c^2+2\sqrt{2}ca+2a^2}$

$P\le \sqrt{{\left(a+\sqrt{2}b\right)}^2}+\sqrt{{\left(b+\sqrt{2}c\right)}^2}+\sqrt{{\left(c+\sqrt{2}a\right)}^2}$

$P\le \left(1+\sqrt{2}\right)\left(a+b+c\right)=1+\sqrt{2}$

Dấu "=" xảy ra khi $\left(a;b;c\right)=\left(0;0;1\right)$ và các hoán vị