f(k)=k

<=> k²-3k-5=k

<=> k²-3k-5-k=0

<=> k²-4k-5=0

<=> k²-4k-4-1=0

<=> (k-2)²=1

<=> k-2=1 hoặc k-2=-1

<=> k=3 hoặc k=1

ko bit

\(Q=\frac{12x^2+20x+3}{6x^2+43x+7}=\frac{12x^2+18x+2x+3}{6x^2+42x+x+7}=\frac{6x\left(2x+3\right)+\left(2x+3\right)}{6x\left(x+7\right)+\left(x+7\right)}=\frac{\left(6x+1\right)\left(2x+3\right)}{\left(6x+1\right)\left(x+7\right)}=\frac{2x+3}{x+7}\)\(P=\frac{8x^2+36x+36}{x^2+10x+21}=\frac{4\left(2x^2+9x+9\right)}{x^2+3x+7x+21}=\frac{4\left(2x^2+3x+6x+9\right)}{x\left(x+3\right)+7\left(x+3\right)}=\frac{4\left[x\left(2x+3\right)+3\left(2x+3\right)\right]}{\left(x+3\right)\left(x+7\right)}=\frac{4\left(x+3\right)\left(2x+3\right)}{\left(x+3\right)\left(x+7\right)}=\frac{4\left(2x+3\right)}{x+7}\)

=> \(Q:P=\frac{2x+3}{x+7}:\frac{4\left(2x+3\right)}{x+7}=\frac{2x+3}{x+7}.\frac{x+7}{4\left(2x+3\right)}=\frac{1}{4}\)

=>\(Q=\frac{1}{4}P\)

\(\frac{Q\left(0\right)}{P\left(0\right)}=\frac{3.21}{7.36}=\frac{1}{4}\Rightarrow Q=\frac{1}{4}P\)

kết bạn ko 

A=a⁴-2a³+3a²-4a+5

=a⁴-2a³+a²+2a²-4a+2+3

=(a²-1)²+2(a-1)²+3 >= 3 với mọi x (do 2 cái ngoặc >= 0)

minA=3,dấu "=" xảy ra <=> a=1

bạn viết sai rồi phải là (a²-a)² chứ

Mik nè, mình nhé!