1.Cho x+y+z=0. CMR:

a) \(5\left(x^3+y^3+z^3\right)\left(x^2+y^2+z^2\right)=6\left(x^5+y^5+z^5\right)\)

b) \(x^7+y^7+z^7=7xyz\left(x^2y^2+y^2z^2+z^2x^2\right)\)

c) \(10\left(x^7+y^7+z^7\right)=7\left(x^2+y^2+z^2\right)\left(x^5+y^5+z^5\right)\)

d) \(2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)

2. Tìm n∈ N để biểu thức sau là số nguyên tố 

a) \(A=n^3-4n^2-4n-1\)

b) \(B=n^3-6n^2+9n-2\)

c) \(C=n^{1975}+n^{1973}+1\)

Bài 1:

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

\(b)x^2\left(y+3\right)=yz^2\) \(x,y,z\in Z_+\)

\(c)x\left(x+1\right)\left(x+7\right)\left(x+8\right)=y^2\) \(x,y\in Z_+\)

\(d)x^4+x^2-y^2+y+10=0\) \(x,y\in Z\)

\(e)x^3-y^3-2y^2-3y-1=0\) \(x,y\in Z_+\)

\(f)x^4-y^4+z^4+2x^2y^2+3x^2+4z^2+1=0\) \(x,y,z\in Z\)

thực hiện phép tính

a,\(x^3+\left[\frac{x\left(2y^3-x^3\right)}{x^3+y^3}\right]^3-\left[\frac{y\left(2x^3-y^3\right)}{x^3+y^3}\right]^3\)

b,\(\frac{\frac{x\left(x+y\right)}{x-y}+\frac{x\left(x+z\right)}{x-z}}{1+\frac{\left(y-z\right)^2}{\left(x-y\right)\left(x-z\right)}}+\frac{\frac{y\left(y+z\right)}{y-z}+\frac{y\left(y+x\right)}{y-x}}{1+\frac{\left(z-x\right)^2}{\left(y-z\right)\left(y-x\right)}}+\frac{\frac{z\left(z+x\right)}{z-x}+\frac{z\left(z+y\right)}{z-y}}{1+\frac{\left(x-y\right)^2}{\left(z-x\right)\left(z-y\right)}}\)

c,\(\left[\frac{y+z-2x}{\frac{\left(y-z\right)^3}{y^3-z^3}+\frac{\left(x-y\right)\left(x-z\right)}{y^2+yz+z^2}}+\frac{z+x-2y}{\frac{\left(z-x\right)^3}{z^3-x^3}+\frac{\left(y-z\right)\left(y-x\right)}{z^2+xz+x^2}}+\frac{x+y-2z}{\frac{\left(x-y\right)^3}{x^3-y^3}+\frac{\left(z-x\right)\left(z-y\right)}{x^2+xy+y^2}}\right]:\frac{1}{x+y+z}\)

1. a) Tìm \(n\in N\)*, \(n>2008\) sao cho \(2^{2008}+2^{2012}+2^{2013}+2^{2014}+2^{2016}+2^n\) là số chính phương

b) tìm x,y > 0 thỏa mãn \(x^2+y^2=2\left(x+y\right)\left(\sqrt{x}+\sqrt{y}-2\right)\)

2. a) \(\left\{{}\begin{matrix}a\ge0\\a+b\ge1\end{matrix}\right.\). Min \(A=\frac{8a^2+b}{4a}+b^2\)

b) \(\left\{{}\begin{matrix}a,b\ge0\\\left(a-b\right)^2=a+b+2\end{matrix}\right.\). Cmr: \(\left(1+\frac{a^3}{\left(b+1\right)^3}\right)\left(1+\frac{b^3}{\left(b+1\right)^3}\right)\le9\)

c) \(x,y>0;\left(x+\sqrt{1+x^2}\right)\left(y+\sqrt{1+y^2}\right)=2020\). Min P = x + y

d) \(x,y,z>0;\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}=6\). Min \(P=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)

e) \(\left\{{}\begin{matrix}x,y,z>0\\x+y+z+4xyz=4\end{matrix}\right.\) Cmr: \(\left(1+xy+\frac{y}{z}\right)\left(1+yz+\frac{z}{x}\right)\left(1+zx+\frac{x}{y}\right)\ge27\)

f) \(\left\{{}\begin{matrix}x,y,z\ge1\\3x^2+4y^2+5z^2=52\end{matrix}\right.\). Min P = x + y + z

g) \(x,y>0\). Min \(P=\frac{2}{\sqrt{\left(2x+y\right)^3+1}-1}+\frac{2}{\sqrt{\left(x+2y\right)^3+1}-1}+\frac{\left(2x+y\right)\left(x+2y\right)}{4}-\frac{8}{3\left(x+y\right)}\)

Rút gọn:
a) \(\dfrac{3\left(x-y\right)\left(x-z\right)^2}{6\left(x-y\right)\left(x-z\right)}\)
b) \(\dfrac{6x^2y^2}{8xy^5}\)
c) \(\dfrac{3x\left(1-x\right)}{2\left(x-1\right)}\)
d) \(\dfrac{9-\left(x+5\right)^2}{x^2+4x+4}\)
e) \(\dfrac{x^2-2x+1}{x^2-1}\)
f) \(\dfrac{8x-4}{8x^3-1}\)
g) \(\dfrac{x^2+5x+6}{x^2+4x+4}\)
k) \(\dfrac{20x^2-45}{\left(2x+3\right)^2}\)

chị QA

ta có đề bài <=> 

\(\frac{x^2}{y}-2x+y+\frac{y^2}{z}-2y+z+\frac{z^2}{x}-2z+x+\left(x+y+z\right)-\left(x-y\right)^2-\left(y-z\right)^2-\left(z-x\right)^2\)

=\(\frac{\left(x-y\right)^2}{y}-\left(x-y\right)^2+...+\left(x+y+z\right)\)

=\(\left(x-y\right)^2\left(\frac{1}{y}-1\right)+....+\left(x+y+z\right)\)

mà \(\sqrt{x}+\sqrt{y}+\sqrt{z}=1\Rightarrow x,y,z\in\left[0;1\right]\)

=> \(\frac{1}{y}-y>0\)

=> \(A\ge x+y+z\ge\frac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{3}=\frac{1}{3}\)

Bài 1: Rút gọn các biểu thức sau:

a) \(3x^2\) - 2x( 5+ 1,5x) +10

b) 7x ( 4y- x) + 4y( y-7x) - 2( \(2y^2\) - 3,5x)

c) \(\left\{2x-3\left(x-1\right)-5\left[x-4\left(3-2x\right)+10\right]\right\}.\left(-2x\right)\)

Bài 2: Tìm x, biết:

a) 3( 2x -1) - 5( x -3) + 6( 3x -4) = 24

b) \(2x^2+3\left(x^2-1\right)=5x\left(x+1\right)\)

c) \(2x\left(5-3x\right)+2x\left(3x-5\right)-3\left(x-7\right)=3\)

d) \(3x\left(x+1\right)-2x\left(x+2\right)=-1-x\)

Bài 3: Tính giá trị của các biểu thức sau:

a)\(A=x^2\left(x+y\right)-y\left(x^2+y^2\right)+2002\) Với \(x=1;y=-1\)

b) \(B=5x\left(x-4y\right)-4y\left(y-5x\right)-\dfrac{11}{20}\) Với \(x=-0,6;y=-0,75\)

Bài 4: Chứng tỏ rằng giá trị của biểu thức sau không phụ thuộc vào giá trị biến:

a) \(2\left(2x+x^2\right)-x^2\left(x+2\right)+\left(x^3-4x+3\right)\)

b) \(z\left(y-x\right)+y\left(z-x\right)+x\left(y+z\right)-2yz+100\)

c) \(2y\left(y^2+y+1\right)-2y^2\left(y+1\right)-2\left(y+10\right)\)

Bài 5: Tính giá trị của biểu thức:

a) \(A=\left(x-3\right)\left(x-7\right)-\left(2x-5\right)\left(x-1\right)\) Với \(x=0;x=1;x=-1\)

b) \(B=\left(3x+5\right)\left(2x-1\right)+\left(4x-1\right)\left(3x+2\right)\) Với \(\left|x\right|=2\)

c) \(C=\left(2x+y\right)\left(2z+y\right)+\left(x-y\right)\left(y-z\right)\) Với \(x=1;y=1;z=\left|1\right|\)

\(\hept{\begin{cases}3x^2+2y+1=2z\left(x+2\right)\\3y^2+2z+1=2x\left(y+2\right)\\3z^2+2x+1=2y\left(z+2\right)\end{cases}\Leftrightarrow\hept{\begin{cases}3x^2+2y+1=2xz+4z\\3y^2+2z+1=2xy+4x\\3z^2+2x+1=2yz+4y\end{cases}}}\)

Cộng 3 vế vào rồi chuyển vế ta được

\(2x^2+2y^2+2z^2-2xy-2yz-2zx+\left(x^2+2x+1\right)+\left(y^2+2y+1\right)+\left(z^2+2z+1\right)=0\)

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

Dễ thấy VP > 0

Dấu "=" khi x = y = z = -1