Bài 1:

|x|<=4

=>-4<=x<=4

=>A=[-4;4]

-5<x-1<=8

=>-5+1<x<=8+1

=>-4<x<=9

=>B=(-4;9]

A\(\cup\) B=[-4;4]\(\cup\) (-4;9]

=[-4;9]

R\(A\(\cup\) B)

=R\[-4;9]

=(-∞;-4)\(\cup\) (9;+∞)

A\(\cap\) B=[-4;4]\(\cap\) (-4;9]

=(-4;4]

A\B=[-4;4]\(-4;9]

={-4}

B\A=(-4;9]\[-4;4]

=(4;9]

Bài 2:

\(x^2\le4\)

=>-2<=x<=2

=>A=[-2;2]

-2<x-1<3

=>-2+1<x<3+1

=>-1<x<4

=>B=(-1;4)

A\(\cup\) B=[-2;2]\(\cup\) (-1;4)

=[-2;4)

R\(A\(\cup\) B)

=R\[-2;4)

=(-∞;-2)\(\cup\) [4;+∞)

A\(\cap\) B=[-2;2]\(\cap\) (-1;4)

=(-1;2]

A\B=[-2;2]\(-1;4)

=[-2;-1)

B\A=(-1;4)\[-2;2]

=(2;4)

1+1=3

a) \(2-x\ge0\Leftrightarrow x\le2\)

b) \(2+x\ge0\Leftrightarrow x\ge-2\)

c) \(7-x\ge0\Leftrightarrow x\le7\)

b) Ta có: \(9x^4+8x^2-1=0\)

\(\Leftrightarrow9x^4+9x^2-x^2-1=0\)

\(\Leftrightarrow9x^2\left(x^2+1\right)-\left(x^2+1\right)=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(9x^2-1\right)=0\)

mà \(x^2+1>0\forall x\)

nên \(9x^2-1=0\)

\(\Leftrightarrow9x^2=1\)

\(\Leftrightarrow x^2=\dfrac{1}{9}\)

hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3}\right\}\)

Vậy: \(S=\left\{\dfrac{1}{3};-\dfrac{1}{3}\right\}\)

a) Ta có: \(x^2-20x+101=x^2-2.x.10+10^2+1=\left(x-10\right)^2+1\)

Vì \(\left(x-10\right)^2\ge0\left(\forall x\in Z\right)\)

\(\Rightarrow\left(x-10\right)^2+1>1>0\)

Vậy x²-20x+101 >0 với mọi x

b) \(4a^2+4a+2=\left(2a\right)^2+2.2a.1+1+1=\left(2a+1\right)^2+1\)

Vì \(\left(2a+1\right)^2\ge0\left(\forall a\in Z\right)\)

\(\Rightarrow\left(2a+1\right)^2+1>1>0\)

Vậy 4a²+4a+2 > 0 với mọi a

c) \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+16+8\right)+16\)

\(=\left(x^2+10x+16\right)^2+8\left(x^2+10x+16\right)+16\)

\(=\left(x^2+10x+20\right)^2\) \(\ge0\left(\forall x\right)\)

Giúp mình với !!

\(\left(x-5\right)^4=\left(x-5\right)^6\)

=> \(\left(x-5\right)^4-\left(x-5\right)^6=0\)

=> \(\left(x-5\right)^4\left[1-\left(x-5\right)^2\right]=0\)

=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\1-\left(x-5\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\\left(x-5\right)^2=1\end{cases}}\)

+) \(\left(x-5\right)^2=\left(\pm1\right)^2\Rightarrow x-5=\pm1\)

=> \(\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)

Mà với \(x\ge5\)nên loại x = 4

Vậy x = 5 và x = 6

Đặt A=.....
Dễ dàng biến đổi \(A=\frac{x^2}{y-1}+\frac{y^2}{x-1}\)
Có :\(\frac{x^2}{y-1}+4\left(y-1\right)\ge4x\)và \(\frac{y^2}{x-1}+4\left(x-1\right)\ge4y\)
Khi đó :\(A\ge4x+4y-4\left(x-1\right)-4\left(y-1\right)=8\)
Dấu = xảy ra \(\Leftrightarrow x=y=2\)
Phần dấu = tớ làm hơi tắt. bạn nên tb rõ nhé 

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

Áp dụng BĐT Côsy Schwarz \(\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}\ge\frac{\left(a_1+a_2\right)^2}{b_1+b_2}\)(Bạn có thể chứng minh được theo Bunhiacopxki - hoặc xem về BĐT Côsy Schwarz trên mạng)

cho các số dương a1=x;a2=y;b2=x-1;b2=y-1. Ta có:

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

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

Vì x+y-2 >0. Áp dụng BĐT Cô sy cho 2 số \(\left(x+y-2\right);\frac{4}{x+y-2}\)

\(A\ge4+\left\{\left(x+y-2\right)+\frac{4}{x+y-2}\right\}\ge4+2\sqrt{\left(x+y-2\right)\cdot\frac{4}{x+y-2}}=4+2\sqrt{4}=8\)

Vậy A>=8. Dấu bằng xảy ra khi x=y=2 (ĐPCM).