Đặt biểu thức trên là A
Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\ne0\)

\(\Rightarrow x=ak,y=bk,z=ck\)

Nên \(A=\frac{\text{[}\left(ak\right)^2+\left(bk\right)^2+\left(ck\right)^2\text{]}.\left(a^2+b^2+c^2\right)}{\left(a.ak+b.bk+c.bk\right)^2}\)

\(=\frac{\left(a^2k^2+b^2k^2+c^2k^2\right).\left(a^2+b^2+c^2\right)}{\left(a^2k+b^2k+c^2k\right)^2}\)
\(=\frac{k^2\left(a^2+b^2+c^2\right).\left(a^2+b^2+c^2\right)}{\text{[}k\left(a^2+b^2+c^2\right)\text{]}^2}\)

\(=\frac{k^2.\left(a^2+b^2+c^2\right)^2}{k^2.\left(a^2+b^2+c^2\right)}\)

\(=1\)

Vậy A=1

à quên sửa dòng trên chỗ A=1 cái chỗ mẫu là \(k^2.\left(a^2+b^2+c^2\right)^2\)nhen :v

a, \(A=2x^2-8x-10=2\left(x^2-4x+4\right)-18=2\left(x-2\right)^2-18\ge-18\)

Dấu "=" xảy ra <=> x-2=0 <=> x=2

Vậy MinA = -18 khi x=2

b, \(B=x-x^2=-\left(x^2-x+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

Dấu "=" xảy ra <=> x-1/2=0 <=> x=1/2

Vậy MaxB = 1/4 khi x=1/2

a) \(A=2x^2-8x-10\)
\(=2\left(x^2-4x-5\right)\)

\(=2\left(x^2-2.x.2+2^2-2^2-5\right)\)
\(=2\left[\left(x-2\right)^2-9\right]\)
\(=2\left(x-2\right)^2-18\)

Vì \(2\left(x-2\right)^2\ge0\forall x\)

Nên \(2\left(x-2\right)^2\ge-18\)

Hay \(A\ge-18\)

Vậy gtnn của A là -18 khi \(2\left(x-2\right)^2=0\)

\(x-2=0\)

\(x=2\)

b) \(B=x-x^2\)

\(=-x^2-x\)

\(=-\left(x^2-x\right)\)

\(=-\text{[}x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\text{]}\)

\(=-\text{[}\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\text{]}\)

\(=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\)

Vì \(-\left(x-\frac{1}{2}\right)^2\le0\forall x\)
Nên \(-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x \)
Vậy gtln của B là \(\frac{1}{4}\)khi \(x-\frac{1}{2}=0\)

\(x=\frac{1}{2}\)

(x-1)(x+1) +x(x-9)=2x² -4

=>x²-1+x²-9x=2x²-4

=>2x²-2x²-9x=3

=>9x=3

=>x=\(\frac{1}{3}\)

Vậy \(x=\frac{1}{3}\)

\(\left(x-1\right)\left(x+1\right)+x\left(x-9\right)=2x^2-4.\)

\(x^2-1+x^2-9x=2x^2-4\)

\(\left(2x^2-2x^2\right)-1-9x=-4\)

\(-1-9x=-4\)

\(-9x=-3\)

\(\Rightarrow\)\(x=\frac{1}{3}\)

17 + 9 + 2005 = 2031

 Chúc bạn học tốt ^^

\(17+9+2005=2031\)

(2x³-4x²+3x+12)=(x+2).(2x²-8x+21)-30

MK ko bk đúng ko nhé

100-50=50

chúc bạn hok giỏi

100-50=50(ko đổi k đc vì mk ko có điểm hỏi dáp , k cx như k)

Sửa đề cm a²⁰¹⁸+b²⁰¹⁸=2

Ta có:\(a^3+b^3=3ab-1\)

\(\Leftrightarrow a^3+b^3+1-3ab=0\)

\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+1-3ab=0\)

\(\Leftrightarrow\left(a+b+1\right)\left[\left(a+b\right)^2-\left(a+b\right)+1\right]-3ab\left(a+b+1\right)=0\)

\(\Leftrightarrow\left(a+b+1\right)\left(a^2+2ab+b^2-a-b+1-3ab\right)=0\)

\(\Leftrightarrow\left(a+b+1\right)\left(a^2+ab+b^2-a-b+1\right)=0\)

Vì a,b > 0 => a + b + 1 > 0

=>\(a^2+ab+b^2-a-b+1=0\)

=>2a²+2ab+2b²-2a-2b+2=0

=>(a²+2ab+b²)+(a²-2a+1)+(b²-2b+1)=0

=>(a+b)²+(a-1)²+(b-1)²=0

Mà \(\hept{\begin{cases}\left(a+b\right)^2\ge0\\\left(a-1\right)^2\ge0\\\left(b-1\right)^2\ge0\end{cases}}\Rightarrow VT\ge0\)

=>\(\hept{\begin{cases}a+b=0\\a-1=0\\b-1=0\end{cases}}\)=> a=b=1

=>\(a^{2018}+b^{2018}=1+1=2\)

2bc(b + 2c) + 2ac(c - 2a) - 2ab(a + 2b) - 7abc

= 2b²c + 4bc² + 2ac² - 4a²c - 2ab(a + 2b) - 7abc

= 2b²c + abc + 4bc² + 2ac² - 4a²c - 8abc - 2ab(a + 2b)

= bc(2b + a) + 2c²(2b + a) - 4ac(a + 2b) - 2ab(a + 2b)

= (a + 2b)(bc + 2c² - 4ac - 2ab)

= (a + 2b)[c(b + 2c) - 2a(2c + b)]

= (a + 2b)(b + 2c)(c - 2a)