Lời giải:

$P=4a^2+b^2+c^2+4ab+4ac+2bc=(2a+b+c)^2=(-1)^2=1$

=>4a^2-5ab+b^2=0

=>(a-b)(4a-b)=0

=>a=b hoặc b=4a(loại)

=>P=b^2/3b^2=1/3

a: \(=-4x^2+20x+2x-10=-4x^2+22x-10\)

b: =x^2-9

c: =x^3+27

d: \(=-2x^2-6x+x+3=-2x^2-5x+3\)

e: =8a^3+1

f: =(3-x)(x+1)(x+2)

=(3-x)(x^2+3x+2)

=3x^2+9x+6-x^3-3x^2-2x

=-x^3+7x+6

Ta có:

\(4a^2+b^2=5ab\Leftrightarrow4a^2+b^2-4ab-ab=0\)

\(\Leftrightarrow4a\left(a-b\right)-b\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(4a-b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\4a-b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=b\left(ktm\right)\\4a=b\left(tm\right)\end{matrix}\right.\)

\(\Rightarrow4a=b\)

\(\Rightarrow\dfrac{5ab}{3a^2+2b^2}=\dfrac{5a.4a}{3a^2+2.\left(4a\right)^2}=\dfrac{20a^2}{3a^2+32a^2}\)

\(=\dfrac{20a^2}{35a^2}=\dfrac{4}{7}\)

\(4a^2+b^2=5ab\)

\(\Rightarrow4a\left(a-b\right)-b\left(a-b\right)=0\)

\(\Rightarrow\left(a-b\right)\left(4a-b\right)=0\)

\(\Rightarrow b=4a\left(do.a\ne b\right)\)

\(\dfrac{5ab}{3a^2+2b^2}=\dfrac{20a^2}{3a^2+32a^2}=\dfrac{4}{7}\)

từng câu 1 thôi:v

a) x2-xy+5y-25
 = x(2-y)+ 5(y-2)
 = x(2-y)-5(2-y)
 = (x-5)(2-y)

h: \(=\left(a-1-b\right)\left(a-1+b\right)\)

\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)

\(2a+b=2\Rightarrow b=2-2a\)

\(\Rightarrow P=3a^2+b\left(2a+b\right)=3a^2+2b=3a^2+2\left(2-2a\right)=3a^2-4a+4=3\left(a-\dfrac{2}{3}\right)^2+\dfrac{8}{3}\ge\dfrac{8}{3}\)

\(p_{min}=\dfrac{8}{3}\) khi \(a=\dfrac{2}{3}\)

A

A

A