Lời giải:

Đặt \((a,b,c)=(m+4,n+5,p+6)\Rightarrow m,n,p\geq 0\)

Điều kiện đb trở thành:

\(a^2+b^2+c^2=90\Leftrightarrow m^2+n^2+p^2+8m+10n+12p=13\)

Vì \(m,n,p\geq 0\) nên:

\(13=m^2+n^2+p^2+8m+10n+12p\leq (m+n+p)^2+12(m+n+p)\)

\(\Leftrightarrow (m+n+p+13)(m+n+p-1)\geq 0\)

\(\Rightarrow m+n+p\geq 1\)

\(\Rightarrow a+b+c=m+n+p+15\geq 16\)

Ta có đpcm

Dấu bằng xảy ra khi \((a,b,c)=(4,5,7)\)

\(Taco:\)

\(Đặt:S=a^2+b^2+c^2\)

\(.Với:a=4;b=5;c=6\Rightarrow S=76< 90\)

\(Taco:4+5+6=15\)

\(mà:a=4;b=5;c=6.S< 90\Rightarrow\)ít nhất a>4 hoặc: b>5 hoặc: c>6

Vì: a²;b²,c² E N=> a,b,c E N

=> \(a+b+c\inℕ\Rightarrow a+b+c>15\Rightarrow a+b+c\ge16\left(đpcm\right)\)

Hoi nham ti ti nx to lam lai cho trua nhe

cái này đặt a=x+4,b=y+5,c=z+6 nha bạn mình nghĩ cả tối đó

Từ giả thiết ta suy ra 

(a-4)(a-9)+(b-5)(b-8)+(c-6)(c-7)\(\le\)0

⇔a²+b²+c²−13(a+b+c)+118≤0⇔a²+b²+c²−13(a+b+c)+118≤0

⇔a+b+c≥16

Dấu "=" xảy ra khi a=4,b=5,c=6

\(\left\{{}\begin{matrix}a\ge4\\b\ge5\end{matrix}\right.\) \(\Rightarrow a^2+b^2\ge16+25=41\Rightarrow c^2=90-\left(a^2+b^2\right)\le49\Rightarrow c\le7\)

Tương tự: \(b=\sqrt{90-\left(a^2+c^2\right)}\le\sqrt{90-\left(4^2+6^2\right)}=\sqrt{38}\)

\(a\le\sqrt{90-\left(5^2+6^2\right)}=\sqrt{29}\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a-4\right)\left(a-9\right)\le0\\\left(b-5\right)\left(b-8\right)\le0\\\left(c-6\right)\left(c-7\right)\le0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}13a\ge a^2+36\\13b\ge b^2+40\\13c\ge c^2+42\end{matrix}\right.\)

\(\Rightarrow13\left(a+b+c\right)\ge a^2+b^2+c^2+118=208\)

\(\Rightarrow a+b+c\ge16\)

\(P_{min}=16\) khi \(\left(a;b;c\right)=\left(4;5;7\right)\)

a>=4,b>=5,c>=6

=>a+b+c>=4+5+6>=15

hay P>=15

@Akai Haruma làm hộ mình

3

Ta có: \(a^2+b^2+c^2\ge ab+bc+ca\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca\ge3\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+2a\left(b+c\right)+\left(b+c\right)^2\ge3\left(ab+bc+ca\right)\)

\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Rightarrow\text{Đ}PCM\)

2b)

Ta có: \(x^2+y^2-4x-2y+5=0\Leftrightarrow x^2+y^2-4x-2y+4+1=0\Leftrightarrow\left(x-2\right)^2+\left(y-1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y-1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}}\)

c) \(x^4-11x^2+4x-21=0\Leftrightarrow x^4-10x^2+25-x^2+4x-4=0\)

\(\Leftrightarrow\left(x^2-5\right)^2-\left(x-2\right)^2=0\Leftrightarrow\left(x^2-x-5+2\right)\left(x^2+x-5-2\right)=0\)

đến đây tự làm

Bài 2:

b/   \(x^2+y^2-4x+2y+5=0\)

\(\Rightarrow x^2+y^2-4x+2y+4+1=0\)

\(\Rightarrow\left(x^2-4x+4\right)\left(y^2+2y+1\right)=0\)

\(\Rightarrow\left(x-2\right)^2\left(y+1\right)^2=0\)

\(\orbr{\begin{cases}x-2=0\Rightarrow x=2\\y+1=0\Rightarrow y=1\end{cases}}\)

Bài 1 :

(a^2+b^2)(x^2+y^2)=(ax+by)^2 
<=> a^2x^2 + a^2y^2 + b^2x^2 + b^2y^2 = a^2x^2 + 2abxy + b^2y^2 
<=> a^2y^2 + b^2x^2 = 2abxy 
<=> a^2y^2 + b^2x^2 - 2abxy = 0 
<=> (ay - bx)^2 = 0 
=> ay - bx = 0 
=> ay = bx 
=> a/x = b/y ( x,y khác 0)