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

7) \(\dfrac{-5}{17}+\dfrac{3}{17}\le\dfrac{x}{17}\le\dfrac{13}{17}+\dfrac{-11}{17}\)

\(\Rightarrow\dfrac{-2}{17}\le\dfrac{x}{17}\le\dfrac{2}{17}\)

\(\Rightarrow-2\le x\le2\)

\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)

8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)

\(\Rightarrow\dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{6}{12}-\dfrac{2}{12}\right)\)

\(\Rightarrow\dfrac{2}{3}\cdot\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}\cdot\dfrac{4}{12}\)

\(\Rightarrow\dfrac{22}{36}\le\dfrac{x}{18}\le\dfrac{28}{36}\)

\(\Rightarrow\dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\)

\(\Rightarrow x\in\left\{11;12;13;14\right\}\)

8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{3}{6}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}.\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}.\dfrac{2}{6}\\ \dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\\ \Rightarrow11\le x\le14\\ \Rightarrow x\in\left\{11;12;13;14\right\}\)

7) \(\dfrac{-5}{17}+\dfrac{3}{17}\le\dfrac{x}{17}\le\dfrac{13}{17}+\dfrac{-11}{17}\\ \dfrac{-2}{17}\le\dfrac{x}{17}\le\dfrac{2}{17}\\ \Rightarrow-2\le x\le2\\ \Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)

a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< =x< =\dfrac{-13}{5}:\dfrac{21}{15}\)

=>-10<=x<=-13/7

hay \(x\in\left\{-10;-9;...;-2\right\}\)

b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< =x< =-\dfrac{2}{3}\cdot\dfrac{-11}{12}\)

=>-13/9<=x<=11/18

hay \(x\in\left\{-1;0\right\}\)

a) Ta có: \(\dfrac{1}{7}+x=-\dfrac{2}{3}\)

\(\Leftrightarrow x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14}{21}-\dfrac{3}{21}\)

hay \(x=-\dfrac{17}{21}\)

Vậy: \(x=-\dfrac{17}{21}\)

b) Ta có: \(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)

\(\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{-5}{6}=\dfrac{-2}{3}\cdot\dfrac{6}{-5}=\dfrac{-12}{-15}=\dfrac{4}{5}\)

Vậy: \(x=\dfrac{4}{5}\)

c) Ta có: \(\left(\dfrac{3}{5}-2x\right)\cdot\dfrac{5}{8}=1\)

\(\Leftrightarrow\left(\dfrac{3}{5}-2x\right)=1:\dfrac{5}{8}=\dfrac{8}{5}\)

\(\Leftrightarrow-2x=\dfrac{8}{5}-\dfrac{3}{5}=1\)

hay \(x=-\dfrac{1}{2}\)

Vậy: \(x=-\dfrac{1}{2}\)

d) Ta có: \(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)

\(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29}{60}-\dfrac{45}{60}=\dfrac{-16}{60}=\dfrac{-4}{15}\)

hay \(x=\dfrac{-4}{15}:\dfrac{2}{5}=\dfrac{-4}{15}\cdot\dfrac{5}{2}=\dfrac{-20}{30}=-\dfrac{2}{3}\)

Vậy: \(x=-\dfrac{2}{3}\)

e) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)

hay \(x=-\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{-1}{4}\cdot\dfrac{20}{7}=\dfrac{-20}{28}=\dfrac{-5}{7}\)

Vậy: \(x=-\dfrac{5}{7}\)

f) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow-x+\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=0\)

\(\Leftrightarrow-x+\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}=0\)

\(\Leftrightarrow-x-\dfrac{9}{60}=0\)

\(\Leftrightarrow-x=\dfrac{9}{60}=\dfrac{3}{20}\)

hay \(x=-\dfrac{3}{20}\)

Vậy: \(x=-\dfrac{3}{20}\)

g) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)

\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{-1}{2}+4=\dfrac{-1}{2}+\dfrac{8}{2}=\dfrac{7}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{2}\\x+\dfrac{1}{3}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{21}{6}-\dfrac{2}{6}=\dfrac{19}{6}\\x=-\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{-21}{6}-\dfrac{2}{6}=\dfrac{-23}{6}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{19}{6};-\dfrac{23}{6}\right\}\)

a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3

=>x=-1/3+3/4=-4/12+9/12=5/12

b: =>x(1/2-5/6)=7/2

=>-1/3x=7/2

hay x=-21/2

c: (4-x)(3x+5)=0

=>4-x=0 hoặc 3x+5=0

=>x=4 hoặc x=-5/3

d: x/16=50/32

=>x/16=25/16

hay x=25

e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4

=>2x=-7/4+3=5/4

hay x=5/8

1. 

ĐK: \(x\ne3;x\ne-2\)

\(\dfrac{5}{x-3}+\dfrac{3}{x+2}\le\dfrac{3+2x}{x^2-x-6}\)

\(\Leftrightarrow\dfrac{5\left(x+2\right)+3\left(x-3\right)}{x^2-x-6}\le\dfrac{3+2x}{x^2-x-6}\)

\(\Leftrightarrow\dfrac{8x+1-3-2x}{x^2-x-6}\le0\)

\(\Leftrightarrow\dfrac{6x-2}{x^2-x-6}\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x-2\ge0\\x^2-x-6< 0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}6x-2\le0\\x^2-x-6>0\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}6x-2\ge0\\x^2-x-6< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\-2< x< 3\end{matrix}\right.\Leftrightarrow\dfrac{1}{3}\le x< 3\)

TH2: \(\left\{{}\begin{matrix}6x-2\le0\\x^2-x-6>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\\left[{}\begin{matrix}x>3\\x< -2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x< -2\)

Vậy ...

2.

ĐK: \(x\ne\pm2\)

\(\dfrac{1}{x^2-4}+\dfrac{2}{x+2}>-\dfrac{3}{x-2}\)

\(\Leftrightarrow\dfrac{1}{x^2-4}+\dfrac{2\left(x-2\right)+3\left(x+2\right)}{x^2-4}>0\)

\(\Leftrightarrow\dfrac{5x+3}{x^2-4}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}5x+3>0\\x^2-4>0\end{matrix}\right.\\\left\{{}\begin{matrix}5x+3< 0\\x^2-4< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{3}{5}< x< 2\\x< -2\end{matrix}\right.\)

Vậy ...

8.

\(\sqrt{x^2+6x+9}-2x+1>0\)

\(\Leftrightarrow\sqrt{x^2+6x+9}>2x-1\)

\(\Leftrightarrow\sqrt{x^2+6x+9}>2x-1\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-1< 0\\x^2+6x+9\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1\ge0\\x^2+6x+9>\left(2x-1\right)^2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{1}{2}\\\dfrac{1}{2}\le x< 4\end{matrix}\right.\)

Vậy ...

Chia nhỏ ra

a: =>1/2x=7/2-2/3=21/6-4/6=17/6

=>x=17/3

b: =>2/3:x=-7-1/3=-22/3

=>x=2/3:(-22/3)=-1/11

c: =>1/3x+2/5x-2/5=0

=>11/15x=2/5

hay x=6/11

d: =>2x-3=0 hoặc 6-2x=0

=>x=3/2 hoặc x=3

a) \(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30.75\right).x-8=\left(\dfrac{3}{5}+0.415\right)\)

\(=\left(\dfrac{1}{12}+3\dfrac{1}{6}-\dfrac{123}{4}\right).x-8=\left(\dfrac{3}{5}+\dfrac{83}{200}\right)\)

\(=\dfrac{-55}{2}.x-8=\dfrac{203}{200}\)\(=\dfrac{-55}{2}.x=\dfrac{203}{200}+8=\dfrac{1803}{200}\)

\(x=\dfrac{1803}{200}:\dfrac{-55}{2}=\dfrac{-1803}{5500}\)

a, \(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,75\right).x-8=\dfrac{3}{5}+0,415\)

\(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,75\right).x-8=\dfrac{203}{200}\)

\(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,75\right).x=\dfrac{203}{200}+8\)

\(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,75\right).x=\dfrac{1803}{200}\)

\(\left(\dfrac{13}{4}-30,75\right).x=\dfrac{1803}{200}\)

\(\dfrac{-55}{2}.x=\dfrac{1803}{200}\)

\(x=\dfrac{1803}{200}:\dfrac{-55}{2}\)

\(x=\dfrac{-1803}{5500}\)

Nếu là tìm số nguyên thì hình như đề sai rồi bạn
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b, \(4\dfrac{1}{3}.\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\le x\le\dfrac{2}{3}.\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)

Cho \(A=4\dfrac{1}{3}.\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\)

\(A=4\dfrac{1}{3}.\dfrac{-1}{3}\)

\(A=\dfrac{13}{3}.\dfrac{-1}{3}\)

\(A=\dfrac{-13}{9}\)

Cho \(B=\dfrac{2}{3}.\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)

\(B=\dfrac{2}{3}.\left(\dfrac{-1}{6}-\dfrac{3}{4}\right)\)

\(B=\dfrac{2}{3}.\dfrac{-11}{12}\)

\(B=\dfrac{-11}{18}\)

Ta có: \(A\le x\le B\)

\(\dfrac{-13}{9}\le x\le\dfrac{-11}{18}\)

\(\Rightarrow x=-1\)