\(A=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)\\ A=\left(x^2-5x+4\right)\left(x^2-5x+6\right)\\ A=\left(x^2-5x+5-1\right)\left(x^2-5x+5+1\right)\\ A=\left(x^2-5x+5\right)^2-1\ge-1\)

đẳng thức xảy ra khi :

\(x^2-5x+5=0\\ x^2-2.\dfrac{5}{2}x+\dfrac{25}{4}=\dfrac{25}{4}-5\\ \left(x-\dfrac{5}{2}\right)^2=\dfrac{5}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=\sqrt{\dfrac{5}{4}}\\x-\dfrac{5}{2}=-\sqrt{\dfrac{5}{4}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{\sqrt{5}+5}{2}\\x=-\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{5-\sqrt{5}}{2}\end{matrix}\right.\)

vậy GTNN của A =-1 tại \(\left[{}\begin{matrix}x=\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{\sqrt{5}+5}{2}\\x=-\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{5-\sqrt{5}}{2}\end{matrix}\right.\)

a, \(A_{\left(x\right)}=2x^2+2xy+y^2-2x+2y+2\)

\(=\left(x^2+y^2+1+2xy+2x+2y\right)+\left(x^2-4x+4\right)-3\)

\(=\left(x+y+1\right)^2+\left(x-2\right)^2-3\ge-3\) hay \(A_{\left(x\right)}\ge-3\)

Dấu ''='' xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y+1\right)^2=0\\\left(x-2\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+y+1=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=-3\\x=2\end{matrix}\right.\)

Vậy \(minA_{\left(x\right)}=-3\) khi x=-3; y=2

b, \(B_{\left(x\right)}=x^2-4xy+5y^2+10x-22y+28\)

\(=\left(x^2+4y^2+25-4xy+10x-20y\right)+\left(y^2-2y+1\right)+2\)

\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\Leftrightarrow B_{\left(x\right)}\ge2\)

Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\left(x-2y+5\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-2y+5=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

Vậy \(minB_{\left(x\right)}=2\Leftrightarrow x=-3;y=1\)

c, \(C_{\left(x\right)}=x^2-10xy+26y^2+14x-76y+59\)

\(=\left(x^2+25y^2+49-10xy+14x-70y\right)+\left(y^2-6y+9\right)+1\)

\(=\left(x-5y+7\right)^2+\left(y-3\right)^2+1\ge1\Leftrightarrow C_{\left(x\right)}\ge1\)

Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\left(x-5y+7\right)^2=0\\\left(y-3\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-5y+7=0\\y-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\)

Vậy \(minC_{\left(x\right)}=1\Leftrightarrow x=8;y=3\)

d, \(D_{\left(x\right)}=4x^2-4xy+2y^2-20x-4y+174\)

\(=\left(4x^2+y^2+25-4xy-20x+10y\right)+\left(y-14y+49\right)+74\)

\(=\left(2x-y-5\right)^2+\left(y-7\right)^2+74\ge74\Leftrightarrow D_{\left(x\right)}\ge74\)

Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\left(2x-y-5\right)^2=0\\\left(y-7\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x-y-5=0\\y-7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=6\\y=7\end{matrix}\right.\)

Vậy \(minD_{\left(x\right)}=74\Leftrightarrow x=6;y=7\)

e, \(E_{\left(x\right)}=x^2-2x+y^2+4y+5\)

\(=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=\left(x-1\right)^2+\left(y+2\right)^2\ge0\)

Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Vậy \(minE_{\left(x\right)}=0\Leftrightarrow x=1;y=-2\)

bạn ơi! Sao cái chỗ A(x) =(x+y+1)²+(x-2)²-3 mà chuyển sang lại là -3 v

a) y(x²-y²)(x²+y²)-y(x⁴-y⁴)=y[(x²)²-(y²)²] - y(x⁴-y⁴)=y(x⁴-y⁴)-y(x⁴-y⁴)=0

vậy giá trị biểu thức không phụ thuộc vào biến (đpcm)

b) \(\left(\frac{1}{3}+2x\right)\left(4x^2-\frac{2}{3}x+\frac{1}{9}\right)-\left(8x^3-\frac{1}{27}\right)\)

\(=\left[\left(2x\right)^3+\left(\frac{1}{3}\right)^3\right]-\left(8x^3-\frac{1}{27}\right)=8x^3+\frac{1}{27}-8x^3+\frac{1}{27}=\frac{1}{54}\)

vậy giá trị biểu thức không phụ thuộc vào biến (đpcm)

c) (x - 1)^3 - (x - 1)(x^2 + x + 1) - 3(1 - x)x

= (x - 1)(x^2 + x + 1) - (x - 1)(x^2 + x + 1) - 3x(1 - x)

= x^3 - 3x^2 + 3x - 1 - x^3 + 1 - 3x + 3x^2

= 0 (đpcm)

Bài 2 thay 2 vào x rồi giải bình thường tìm k 

1.a) Không tồn tại\(\)

   b) 1997 tại x=4

   c) 4 tại x=1;y=2

   d) 164 tại x=8

2.a) x>3 và x<-1

   b) Không tốn tại x

M = ( x - y )³ - ( x - y )² 

   = 7³ - 7² = 294

N = x³ + x²  - y² + y² + xy - 3x²y +3xy² - 3xy - 95

  = ( x - y )³ + ( x - y )² - 95

  = 7³ + 7² - 95 = 297

Mình không chép lại đề nhé !

Bạn chép sai đề rồi , câu b ( x - y + 1 ) mới đúng nha

\(C=\left(x^3+y^3\right)+3xy\left(x^2+y^2+2xy\left(x+y\right)\right)\)

\(C=\left(x^3+y^3+3x^2y+3xy^2-3x^2y-3xy^2\right)+3xy\left(x^2+y^2+2xy\right)\) (vì x+y=1)

\(C=\left(x+y\right)^3-3x^2y-3xy^2+3xy\left(x+y\right)^2\)

\(C=1^3-3xy\left(x+y\right)+3xy.1^2\) (vì x+y=1)

\(C=1-3xy+3xy\)(vì x+y=1)

\(C=1\)

\(D=2\left(\left(x+y\right)^3-3xy\left(x+y\right)\right)-3\left(\left(x+y\right)^2-2xy\right)\)

\(D=2\left(1^3-3xy\right)-3\left(1^2-2xy\right)\)(vì x+y=1)

\(D=2-6xy-3+6xy\)

\(D=-1\)

\(F=a^2\left(a+1\right)-b^2\left(b-1\right)+ab-3ab\left(1-1\right)\)(vì a-b=1)

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

\(F=a^3+a^2-b^3+b^2+ab\)

\(F=\left(a^3-b^3\right)+a^2+b^2+ab\)

\(F=\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a^2+ab+b^2\right)\)

\(F=\left(a^2+ab+b^2\right)+\left(a^2+ab+b^2\right)\)(vì a-b=1)

\(F=2\left(a^2+ab+b^2\right)\)

\(F=2\left(a^2-2ab+b^2+3ab\right)\)

\(F=2\left(\left(a-b\right)^2+3ab\right)\)

\(F=2\left(1+3ab\right)\)

\(F=2+6ab\)

ta có x+y+z=0 

=> \(\left(x+y+z\right)^2=0\)

\(< =>x^2+y^2+z^2+2xy+2xz+2yx=0\)

\(< =>x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)

\(< =>x^2+y^2+z^2+2.0=0\)(vì xy+xz+yz=0)

\(< =>x^2+y^2+z^2=0\)

\(< =>\hept{\begin{cases}x^2=0\\y^2=0\\z^2=0\end{cases}< =>x=y=z=0}\)

thay x=y=z=0 vào 

\(K=\left(x-1\right)^{2014}+y^{2015}+\left(z+1\right)^{2016}\)

\(K=\left(0-1\right)^{2014}+0^{2015}+\left(0+1\right)^{2016}\)

\(K=1+0+1=2\)

\(\)

thanks nhìu