1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

a) \(\sqrt{2x-1}=\sqrt{5}\)

ĐK : \(x\ge\frac{1}{2}\)

Bình phương hai vế

pt <=> \(2x-1=25\)

    <=> \(2x=26\)

    <=> \(x=13\left(tm\right)\)

Vậy S = { 13 }

b) \(\sqrt{4-5x}=12\)

ĐK : \(x\le\frac{4}{5}\)

Bình phương hai vế

pt <=> \(4-5x=144\)

    <=> \(-5x=140\)

    <=> \(x=-28\left(tm\right)\)

Vậy S = { -28 }

c) \(\sqrt{x^2+6x+9}=3x-1\)< chắc hẳn là như này :]> 

<=> \(\sqrt{\left(x+3\right)^2}=3x-1\)

<=> \(\left|x+3\right|=3x-1\)

<=> \(\orbr{\begin{cases}x+3=3x-1\left(x\ge-3\right)\\-3-x=3x-1\left(x< -3\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}x=2\left(tm\right)\\x=-\frac{1}{2}\left(ktm\right)\end{cases}}\)

Vậy S = { 2 }

d) \(2\sqrt{x}\le\sqrt{10}\)

ĐK : \(x\ge0\)

Bình phương hai vế

bpt <=> \(4x\le10\)

      <=> \(x\le\frac{10}{4}\)

Kết hợp với ĐK => Nghiệm của bất phương trình là \(0\le x\le\frac{10}{4}\)

a) \(ĐKXĐ:x\ge\frac{1}{2}\)

 \(\sqrt{2x-1}=\sqrt{5}\)\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow2x-1=5\)\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)( thỏa mãn ĐKXĐ )

Vậy nghiệm của phương trình là \(x=3\)

b) \(ĐKXĐ:x\le\frac{4}{5}\)

\(\sqrt{4-5x}=12\)\(\Leftrightarrow4-5x=144\)( bình phương 2 vế )

\(\Leftrightarrow5x=-140\)\(\Leftrightarrow x=-28\)( thỏa mãn ĐKXĐ )

Vậy nghiệm của phương trình là \(x=-28\)

c) \(ĐKXĐ:x\ge\frac{1}{3}\)

\(\sqrt{x^2+6x+9}=3x-1\)

\(\Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-1\)

\(\Leftrightarrow\left|x+3\right|=3x-1\)

+) TH1: Nếu \(x+3< 0\)\(\Leftrightarrow x< -3\)

thì \(\left|x+3\right|=-\left(x+3\right)=-x-3\)

\(\Rightarrow-x-3=3x-1\)\(\Leftrightarrow4x=-2\)

\(\Leftrightarrow x=\frac{-1}{2}\)(  không thỏa mãn ĐKXĐ )

+) TH2: \(x+3\ge0\)\(\Rightarrow x\ge-3\)

thì \(\left|x+3\right|=x+3\)

\(\Rightarrow x+3=3x-1\)\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)( thỏa mãn ĐKXĐ )

Vậy nghiệm của phương trình là \(x=2\)

d) \(ĐKXĐ:x\ge0\)

Vì \(2\sqrt{x}\ge0\); \(\sqrt{10}>0\)

Bình phương 2 vế ta có:

\(4x\le10\)\(\Leftrightarrow x\le\frac{5}{2}\)

Kết hợp với ĐKXĐ \(\Rightarrow0\le x\le\frac{5}{2}\)

Vậy nghiệm của bất phương trình là \(0\le x\le\frac{5}{2}\)

a: 7x+35=0

=>7x=-35

=>x=-5

b: \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)

=>8-x-8(x-7)=1

=>8-x-8x+56=1

=>-9x+64=1

=>-9x=-63

hay x=7(loại)

a, \(7x=-35\Leftrightarrow x=-5\)

b, đk : x khác 7 

\(8-x-8x+56=1\Leftrightarrow-9x=-63\Leftrightarrow x=7\left(ktm\right)\)

vậy pt vô nghiệm 

2, thiếu đề 

1.

\(a,7x+35=0\\ \Rightarrow7x=-35\\ \Rightarrow x=-5\\ b,ĐKXĐ:x\ne7\\ \dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\\ \Leftrightarrow\dfrac{8-x}{x-7}-\dfrac{8\left(x-7\right)}{x-7}-\dfrac{1}{x-7}=0\\ \Leftrightarrow\dfrac{8-x-8x+56-1}{x-7}=0\\ \Rightarrow-9x+63=0\\ \Leftrightarrow-9x=-63\\ \Leftrightarrow x=7\left(ktm\right)\)

2.đề thiếu

1) Ta có: \(4x+8=3x-1\)

\(\Leftrightarrow4x-3x=-1-8\)

\(\Leftrightarrow x=-9\)

2) Ta có: \(10-5\left(x+3\right)>3\left(x-1\right)\)

\(\Leftrightarrow10-5x-15-3x+3>0\)

\(\Leftrightarrow-8x>2\)

hay \(x< \dfrac{-1}{4}\)

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

Bài 1: 

a) Ta có: \(2\left(3-4x\right)=10-\left(2x-5\right)\)

\(\Leftrightarrow6-8x-10+2x-5=0\)

\(\Leftrightarrow-6x+11=0\)

\(\Leftrightarrow-6x=-11\)

hay \(x=\dfrac{11}{6}\)

b) Ta có: \(3\left(2-4x\right)=11-\left(3x-1\right)\)

\(\Leftrightarrow6-12x-11+3x-1=0\)

\(\Leftrightarrow-9x-6=0\)

\(\Leftrightarrow-9x=6\)

hay \(x=-\dfrac{2}{3}\)

a) \(3x-2\sqrt{x-1}=4\) (ĐK: x ≥ 1)

\(\Rightarrow3x-2\sqrt{x-1}-4=0\)

\(\Rightarrow3x-6-2\sqrt{x-1}+2=0\)

\(\Rightarrow3\left(x-2\right)-2\left(\sqrt{x-1}-1\right)=0\)

\(\Rightarrow3\left(x-2\right)-2.\dfrac{x-2}{\sqrt{x-1}+1}=0\)

\(\Rightarrow\left(x-2\right)\left[3-\dfrac{2}{\sqrt{x-1}+1}\right]=0\)

*TH1: x = 2 (t/m)

*TH2: \(3-\dfrac{2}{\sqrt{x-1}+1}=0\)

\(\Rightarrow3=\dfrac{2}{\sqrt{x-1}+1}\)

\(\Rightarrow3\sqrt{x-1}+3=2\)

\(\Rightarrow3\sqrt{x-1}=-1\) (vô lí)

Vậy S = {2}

b) \(\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\) (ĐK: \(-\dfrac{1}{4}\le x\le3\) )

\(\Rightarrow\sqrt{4x+1}-3-\sqrt{x+2}+2-\sqrt{3-x}+1=0\)

\(\Rightarrow\dfrac{4x-8}{\sqrt{4x+1}+3}-\dfrac{x-2}{\sqrt{x+2}+2}+\dfrac{x-2}{\sqrt{3-x}+1}=0\)

\(\Rightarrow\left(x-2\right)\left(\dfrac{4}{\sqrt{4x+1}+3}-\dfrac{1}{\sqrt{x+2}+2}+\dfrac{1}{\sqrt{3-x}+1}\right)=0\)

=> x = 2

\(a,3x-2\sqrt{x-1}=4\left(x\ge1\right)\\ \Leftrightarrow-2\sqrt{x-1}=4-3x\\ \Leftrightarrow4\left(x-1\right)=16-24x+9x^2\\ \Leftrightarrow9x^2-28x+20=0\\ \Leftrightarrow\left(x-2\right)\left(9x-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=\dfrac{10}{9}\left(tm\right)\end{matrix}\right.\)

\(b,\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\left(-\dfrac{1}{4}\le x\le3\right)\\ \Leftrightarrow4x+1+x+2-2\sqrt{\left(4x+1\right)\left(x+2\right)}=3-x\\ \Leftrightarrow-2\sqrt{\left(4x+1\right)\left(x+2\right)}=2-6x\\ \Leftrightarrow\sqrt{4x^2+9x+2}=3x-1\\ \Leftrightarrow4x^2+9x+2=9x^2-6x+1\\ \Leftrightarrow5x^2-15x-1=0\\ \Leftrightarrow\Delta=225+20=245\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15-\sqrt{245}}{10}=\dfrac{15-7\sqrt{5}}{10}\left(ktm\right)\\x=\dfrac{15+\sqrt{245}}{10}=\dfrac{15+7\sqrt{5}}{10}\left(tm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{15+7\sqrt{5}}{10}\)