1) Ta sẽ chứng minh bằng biến đổi tương đương như sau : 

Ta có : \(\left(x^{10}+y^{10}\right)\left(x^2+y^2\right)\ge\left(x^8+y^8\right)\left(x^4+y^4\right)\left(1\right)\)

\(\Leftrightarrow x^{12}+x^{10}y^2+y^{10}x^2+y^{12}\ge x^{12}+x^8y^4+y^8x^4+y^{12}\)

\(\Leftrightarrow x^{10}y^2+y^{10}x^2\ge x^8y^4+y^8x^4\)

\(\Leftrightarrow x^2y^2\left(x^8+y^8-x^6y^2-x^2y^6\right)\ge0\)

\(\Leftrightarrow x^2y^2\left[\left(x^8-x^6y^2\right)+\left(y^8-x^2y^6\right)\right]\ge0\)

\(\Leftrightarrow x^2y^2\left(x^6-y^6\right)\left(x^2-y^2\right)\ge0\)

\(\Leftrightarrow x^2y^2\left(x^3-y^3\right)\left(x^3+y^3\right)\left(x^2-y^2\right)\ge0\)

\(\Leftrightarrow x^2y^2\left(x-y\right)^2\left(x+y\right)^2\left(x^2-xy+y^2\right)\left(x^2+xy+y^2\right)\ge0\)(2)

Ta thấy : \(x^2-xy+y^2=\frac{\left(x^2-2xy+y^2\right)+x^2+y^2}{2}=\frac{\left(x-y\right)^2+x^2+y^2}{2}\ge0\)

\(x^2+xy+y^2=\frac{\left(x+y\right)^2+x^2+y^2}{2}\ge0\)  ; \(x^2y^2\left(x-y\right)^2\left(x+y\right)^2\ge0\)

Do đó (2) luôn đúng.

Vậy (1) được chứng minh. 

Câu 1: D

Câu 3: 53/144>9/170>9/230

6:ĐKXĐ: x>=0; x<>1/25

BPT=>\(\dfrac{3\sqrt{x}}{5\sqrt{x}-1}+3< =0\)

=>\(\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}\)

=>\(\dfrac{1}{25}< x< =\dfrac{25}{324}\)

7:

ĐKXĐ: x>=0

BPT \(\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1\)

=>\(\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}-1>=0\)

=>\(\dfrac{\sqrt{x}+1-2\sqrt{x}-3}{2\sqrt{x}+3}>=0\)

=>\(-\sqrt{x}-2>=0\)(vô lý)

8:

ĐKXĐ: x>=0; x<>9/4

BPT \(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)

=>\(\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)

=>\(\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)

TH1: 9căn x-14>0 và 2căn x-3<0

=>căn x>14/9 và căn x<3/2

=>14/9<căn x<3/2

=>196/81<x<9/4

TH2: 9căn x-14<0 và 2căn x-3>0

=>căn x>3/2 hoặc căn x<14/9

mà 3/2<14/9

nên trường hợp này Loại

9: 

ĐKXĐ: x>=0

\(BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}\)

=>\(\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}+\dfrac{1}{3}< =0\)

=>\(\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0\)

=>\(\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0\)(vô lý)

10: 

ĐKXĐ: x>=0; x<>1/49

\(BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0\)

=>\(\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0\)

=>\(\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0\)

=>\(\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0\)

TH1: 6căn x-1>0 và 7căn x-1>0

=>căn x>1/6 và căn x>1/7

=>căn x>1/6

=>x>1/36

TH2: 6căn x-1<0 và 7căn x-1<0

=>căn x<1/6 và căn x<1/7

=>căn x<1/7

=>0<=x<1/49

a.

\(10⋮\left(x-1\right)\)

\(\Rightarrow x-1=Ư\left(10\right)\)

\(\Rightarrow x-1=\left\{-10;-5;-2;-1;1;2;5;10\right\}\)

\(\Rightarrow x=\left\{-9;-4;-1;0;2;3;6;11\right\}\)

b.

\(\left(x+5\right)⋮\left(x-2\right)\Rightarrow\left(x-2\right)+7⋮x-2\)

\(\Rightarrow7⋮x-2\)

\(\Rightarrow x-2=Ư\left(7\right)=\left\{-7;-1;1;7\right\}\)

\(\Rightarrow x=\left\{-5;1;3;9\right\}\)

c.

\(\left(3x+8\right)⋮\left(x-1\right)\)

\(\Rightarrow\left(3x-3+11\right)⋮\left(x-1\right)\)

\(\Rightarrow3\left(x-1\right)+11⋮x-1\)

\(\Rightarrow11⋮\left(x-1\right)\)

\(\Rightarrow x-1=Ư\left(11\right)=\left\{-11;-1;1;11\right\}\)

\(\Rightarrow x=\left\{-10;0;2;12\right\}\)

Số 0 nha bn !

\(\frac{1}{2}-\left(\frac{1}{3}+\frac{3}{4}\right)\le x\le\frac{1}{24}-\left(\frac{1}{8}-\frac{1}{3}\right)\)

\(\Rightarrow\frac{1}{2}-\frac{13}{12}\le x\le\frac{1}{24}-\left(-\frac{5}{24}\right)\)

\(\Rightarrow\frac{6}{12}-\frac{13}{12}\le x\le\frac{1}{4}\)

\(\Rightarrow\frac{-7}{12}\le x\le\frac{3}{12}\)

\(\Rightarrow x\in\left\{-7;-6;-5;...;0;1;2;3\right\}\)

\(\frac{1}{2}-\left(\frac{1}{3}+\frac{3}{4}\right)\le x\le\frac{1}{24}-\left(\frac{1}{8}-\frac{1}{3}\right)\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}-\frac{3}{4}\le x\le\frac{1}{24}-\frac{1}{8}+\frac{1}{3}\)

\(\Rightarrow\frac{6-4-9}{12}\le x\le\frac{1-3+8}{24}\)

\(\Rightarrow\frac{-7}{12}\le x\le\frac{6}{24}\)

\(\Rightarrow\frac{-7}{12}\le x\le\frac{1}{4}\)

\(\Rightarrow\frac{-7}{12}\le x\le\frac{3}{12}\)

\(\Rightarrow-7\le x\le3\)

\(\Rightarrow x\in\left\{-7;-6;-5;-4;-3;....;2;3\right\}\)

a, \(A\cup B=(-4;5]\)

\(A\cap B=[-3;4)\)

\(A\backslash B=\left[4;5\right]\)

\(B\backslash A=\left(-4;-3\right)\)

b, \(A\cup B=\left(-3;7\right)\)

\(A\cap B=[1;2)\cup(3;5]\)

\(A\backslash B=\left[2;3\right]\)

\(B\backslash A=\left(-3;1\right)\cup\left(5;7\right)\)

c, \(A\cup B=\left[\dfrac{1}{2};3\right]\)

\(A\cap B=\left[1;\dfrac{3}{2}\right]\)

\(A\backslash B=[\dfrac{1}{2};1)\)

\(B\backslash A=(\dfrac{3}{2};3]\)

d, \(A\cup B=(-5;2]\cup(3;6]\)

\(A\cap B=\left\{0\right\}\cup[4;5)\)

\(A\backslash B=(0;2]\cup\left[-5;6\right]\)

\(B\backslash A=[-5;0)\cup\left(3;4\right)\)

\(a,=5^3:5^2=5\\ b,=\left(\dfrac{3}{4}\right)^{5-3}=\left(\dfrac{3}{4}\right)^2=\dfrac{9}{16}\\ c,=1728-512=1216\\ d,=x^{10}:x^8=x^2\\ e,=\left(-x\right)^{5-3}=\left(-x\right)^2=x^2\\ f,=\left(-y\right)^{5-4}=-y\)