\(M=x^2-2x+2014\)

\(M=x^2-2\cdot x\cdot1+1^2+2013\)

\(M=\left(x-1\right)^2+2013\ge2013\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy M_min = 2013 khi và chỉ khi x = 1

\(x\left(x-3\right)-3x+9=0\)

\(x\left(x-3\right)-3\left(x-3\right)=0\)

\(\left(x-3\right)\left(x-3\right)=0\)

\(\left(x-3\right)^2=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

Vậy x = 3

(x • (x - 3) -  3x) +  9  = 0 

(x - 3)2  = 0 

(x-3)² = 0 

:   x-3  = 0 

x = 3

học tốt

Giơ hai tay

trước

a)\(x^2-y^2+7x+7y\)

\(=\left(x-y\right).\left(x+y\right)+7.\left(x+y\right)\)

\(=\left(x+y\right).\left(x-y+7\right)\)

\(b,x^2+5x+4\)

\(=x^2+x+4x+4\)

\(=x.\left(x+1\right)+4.\left(x+1\right)\)

\(=\left(x+4\right).\left(x+1\right)\)

\(c,x^3-9x^2\)

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

\(d,x^3+x^2+2x\)

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

\(e,3x^2+3y^2-6xy-1^2\)

\(=\left(3x-3y\right)^2-1^2\)

\(=\left(3x-3y-1\right).\left(3x-3y+1\right)\)

cho sửa câu cuối cái nha :>

\(=3.\left(x-y\right)^2-1^{^2}\)

\(=3x^2-3y^2-1^2\)

đến đây tịt r :< sory 

\(a+b=5\Leftrightarrow a^2+2ab+b^2=25\)

\(\Leftrightarrow a^2+2.4+b^2=25\Leftrightarrow a^2+b^2=17\)

\(\Leftrightarrow\left(a^2+b^2\right)^2=289\Leftrightarrow a^4+2\left(ab\right)^2+b^4=289\)

\(\Leftrightarrow a^4+b^4+2.4^2=289\Leftrightarrow a^4+b^4=257\)

tích đi

=31

có thằng Phạm Huy đấy , bảo nó ở bên 

#Girl2k6#

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 \(x-\frac{1}{2}=0\)

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