a/ x^3z+xyz-x^3-xyz^2

=x³z-xyz²-x³+xyz

=xz.(x²-xyz)-x(x²-xyz)

=(x²-xyz)(xz-x)

=x(x-yz)x(z-1)

=x²(x-yz)(z-1)

b/ x^2-axy-bxy+aby^2

=x(x-ay)-by(x-ay)

=(x-ay)(x-by)

c/ abx^2+a^2xy+aby^2+b^2xy

=ax(bx+ay)+by(ay+bx)

=(ay+bx)(ax+by)

\(\Leftrightarrow x^2+10x+25-x^2-4x+21-1=0\Leftrightarrow6x=-45\Leftrightarrow x=-\frac{45}{6}\)

=> x² +10x+25 -( x²+7x-3x-21)=1

=> x² +10x+25 - x² -7x+3x+21=1

=> 6x =1-25-21

=>6x=-45

=>x=-15/2 
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a) \(M=-\left(x^2+6x+9\right)+23=23-\left(x-3\right)^2\le23\Rightarrow MaxM=23\Leftrightarrow x=3\)

b) \(N=\left(9x^2+12x+4\right)+16=\left(3x+2\right)^2+16\ge16\Leftrightarrow MinN=16\Leftrightarrow x=-\frac{2}{3}\)

c) \(P=-\left(x^2-4x+4\right)-\left(4y^2+4y+1\right)+8=8-\left(x-2\right)^2-\left(2y+1\right)^2\le8\Rightarrow MaxP=8\Leftrightarrow x=2;y=-\frac{1}{2}\)

câu b k tìm đc GTLN chỉ tìm được GTNN thôi nha

câu b mình ghi đề sai,phải là N=-9x^2+12x+20 nha bạn

B=(x+1)(x+4)(x+2)(x+3)\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\)

đặt x^2+5x+5 =t (t>=0)

=> B=\(\left(t+1\right)\left(t-1\right)=t^2-1\) ta có: \(t^2\ge0\Rightarrow t^2-1\ge-1\Rightarrow MinB=-1\Leftrightarrow t=0\Leftrightarrow x^2+5x+5=0\Leftrightarrow\left(x^2+5x+\frac{25}{4}\right)=\frac{5}{4}\Leftrightarrow\left(x+\frac{5}{2}\right)^2=\frac{5}{4}\Rightarrow x=-\frac{5}{2}+-\frac{\sqrt{5}}{2}\)

B=(x+1)(x+2)(x+3)(x+4)T=(x+1)(x+2)(x+3)(x+4)

=(x+1)(x+4)(x+2)(x+3)=(x+1)(x+4)(x+2)(x+3)

=(x2+5x+4)(x2+5x+6)=(x2+5x+4)(x2+5x+6)

Đặt :x^2+5x+4=ax2+5x+4=a ⇒T=(a−1)(a+1)⇒T=(a−1)(a+1)

=a^2−1=(x2+5x+5)2−1≥−1=a2−1=(x2+5x+5)2−1≥−1

Vậy MinT=−1MinT=−1 khi

x2+5x+5=0⇒(x2+5x+254)−54=0x2+5x+5=0⇒(x2+5x+254)−54=0⇔(x+52)2=54⇔(x+52)2=54

\(\Rightarrow\orbr{\begin{cases}x+\frac{5}{2}=\sqrt{\frac{5}{4}}\\x+\frac{5}{2}=-\sqrt{\frac{5}{4}}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\sqrt{\frac{5}{4}-\frac{5}{2}}\\x=-\sqrt{\frac{5}{4}-\frac{5}{2}}\end{cases}}\)

4x²+4x+5

=(2x)²+2.2x.1+1+4

=(2x+1)²+4

mà (2x+1)²\(\ge\)0 => (2x+1)²+4 \(\ge\)4 => biểu thức có GTNN là 4 <=> 2x+1=0

                                                                                                          2x=-1

                                                                                                            x=-1/2

x = -1/2

**** cho tui nha mấy chế

 x² + 4x –y² + 4
= (x² +4x + 4) – y²
= (x +2)² – y²
= ( x + 2 + y)(x + 2 –y)

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

a + b +c =0 => ( a +b + c)^2 =0 => a^2 +b^2 +c^2 + 2ab +2bc + 2ac = 0

=> 1 + 2(ab + bc +ac) = 0 => 2(ab +bc +ac) = -1 ==> ab + bc +ac = -1/2

( ab + bc+ac)^2 = 1/4 => a^2.b^2 + b^2.c^2 + c^2.a^2 + 2ab^2.c +2ab.c^2 + 2 a^2.b.c = 1/4 

=> a^2 . b^2 + b^2 . c^2 + c^2 . a^2 + 2abc ( a+ b+ c) = 1/4

=> a^2 . b^2  + b^2 . c^2 + c^2 . a^2  + 2abc . 0 = 1/4

=> 2( a^2 . b^2 +  + b^2 . c^2 + c^2 . a^2 ) = 2.1/4 = 1/2 

=> 2a^2 . b^2 +  2 b^2 . c^2 + 2c^2 . a^2 = 1/2  

( a^2 + b^2 + c^2 )^2 = 1

=> a^4 + b^4 + c^4 + 2a^2.b^2 + 2b^2.c^2 + 2 c^2 . a^2 = 1

=> a^4 + b^ 4 + c^4 + 1/2 = 1 

=> a^4 + b^4 + c^4 = 1/2