Đặt:\(\sqrt{x+9}=v;\sqrt{x+6}=u\)

Ta có: \(v+5u=5+vu\Leftrightarrow-v\left(u-1\right)+5\left(u-1\right)\Leftrightarrow\left(5-v\right)\left(u-1\right)\)

\(\left\{{}\begin{matrix}5-v=0\Leftrightarrow5=\sqrt{x+9}\Leftrightarrow x=16\left(N\right)\\u-1=0\Leftrightarrow\sqrt{x+6}=1\Leftrightarrow x=-5\left(N\right)\end{matrix}\right.ĐKXĐ:x>=-6\)

\(S=\left\{16,-5\right\}\)

Câu trên mình quên -5>-6

Đặt: \(\sqrt{x+9}=v;\sqrt{x+6}=u\)

Ta có: \(v+5u=5+vu\)

\(\Leftrightarrow v+5u-5-uv=0\)

\(\Leftrightarrow-v\left(u-1\right)+5\left(u-1\right)\)

\(\Leftrightarrow\left(5-v\right)\left(u-1\right)\)

\(\left\{{}\begin{matrix}5-v=0\Leftrightarrow5=\sqrt{x+9}\Leftrightarrow x=16\left(N\right)\\u-1=0\Leftrightarrow\sqrt{x+6}=1\Leftrightarrow x=-5\left(L\right)\end{matrix}\right.\)          ĐKXĐ:\(x>=-6\)

\(S=\left\{16\right\}\)

\(\sqrt{x+9}+5\sqrt{x+\text{6}}=5+\sqrt{\left(x+9\right)\left(x+\text{6}\right)}\Leftrightarrow\sqrt{x+9}+5=5+\sqrt{x+9}\Leftrightarrow\sqrt{x+9}-\sqrt{x+9}=0\Leftrightarrow x+9-x-9=0\Leftrightarrow0=0\)

Vậy x vô số nghiệm

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) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)

\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)

b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)

\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)

\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)

\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)

\(\Leftrightarrow x\left(6-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

Vậy: S={0;6}

c) Ta có: \(3x-15=2x\left(x-5\right)\)

\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)

d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)

\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)

\(\Leftrightarrow30-6x=6x-8\)

\(\Leftrightarrow30-6x-6x+8=0\)

\(\Leftrightarrow-12x+38=0\)

\(\Leftrightarrow-12x=-38\)

\(\Leftrightarrow x=\dfrac{19}{6}\)

Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)

e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)

\(\Leftrightarrow6x+4-3x-1=12x+10\)

\(\Leftrightarrow3x+3-12x-10=0\)

\(\Leftrightarrow-9x-7=0\)

\(\Leftrightarrow-9x=7\)

\(\Leftrightarrow x=-\dfrac{7}{9}\)

Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)

\(8,1-\left(x-6\right)=4\left(2-2x\right)\)

\(\Leftrightarrow1-x+6=8-8x\)

\(\Leftrightarrow-x+8x=8-1-6\)

\(\Leftrightarrow7x=1\)

\(\Leftrightarrow x=\dfrac{1}{7}\)

\(9,\left(3x-2\right)\left(x+5\right)=0\)

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

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

\(10,\left(x+3\right)\left(x^2+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)

`8)1-(x-5)=4(2-2x)`

`<=>1-x+5=8-6x`

`<=>5x=2<=>x=2/5`

`9)(3x-2)(x+5)=0`

`<=>[(x=2/3),(x=-5):}`

`10)(x+3)(x^2+2)=0`

  Mà `x^2+2 > 0 AA x`

 `=>x+3=0`

`<=>x=-3`

`11)(5x-1)(x^2-9)=0`

`<=>(5x-1)(x-3)(x+3)=0`

`<=>[(x=1/5),(x=3),(x=-3):}`

`12)x(x-3)+3(x-3)=0`

`<=>(x-3)(x+3)=0`

`<=>[(x=3),(x=-3):}`

`13)x(x-5)-4x+20=0`

`<=>x(x-5)-4(x-5)=0`

`<=>(x-5)(x-4)=0`

`<=>[(x=5),(x=4):}`

`14)x^2+4x-5=0`

`<=>x^2+5x-x-5=0`

`<=>(x+5)(x-1)=0`

`<=>[(x=-5),(x=1):}`

\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

Rảnh rỗi thật sự .-.

d) Ta có: \(D=\left(\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2}{\sqrt{x}+3}\right):\left(1+\dfrac{6}{x-9}\right)\)

\(=\dfrac{5\sqrt{x}-6-2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{x-9+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{5\sqrt{x}-6-2\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{x-3}\)

\(=\dfrac{3\sqrt{x}}{x-3}\)

f) Ta có: \(\left(\dfrac{3}{\sqrt{1+x}}+\sqrt{1-x}\right):\left(\dfrac{3}{\sqrt{1-x^2}}+1\right)\)

\(=\dfrac{3+\sqrt{1-x^2}}{\sqrt{1+x}}:\dfrac{3+\sqrt{1-x^2}}{\sqrt{1-x^2}}\)

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

a: =>10x-4=15-9x

=>19x=19

hay x=1

b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)

=>30x+9=36+32x+24

=>30x-32x=60-9

=>-2x=51

hay x=-51/2

c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)

=>3x=6/5

hay x=2/5

d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)

\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)

=>21x-120x+1080=80x+60

=>-179x=-1020

hay x=1020/179

e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)

=>35x-5+60x=96-6x

=>95x+6x=96+5

=>x=1

f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)

=>6x+24-30x+120=10x-15x+30

=>-24x+96=-5x+30

=>-19x=-66

hay x=66/19

giúp mik mik đang cần gấp

nhưng phả có lời giải đừng cho mỗi đáp án

a:Ta có: \(\left(x-9\right)^7=\left(x-9\right)^4\)

\(\Leftrightarrow\left(x-9\right)^4\cdot\left[\left(x-9\right)^3-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-9=0\\x-9=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=10\end{matrix}\right.\)

b: ta có: \(\left(3x-15\right)^{15}=\left(3x-15\right)^{10}\)

\(\Leftrightarrow\left(3x-15\right)^{10}\cdot\left[\left(3x-15\right)^5-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-15=0\\3x-15=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{16}{3}\end{matrix}\right.\)

bài này hôm qua giải rồi mà!!

\(1,2\left(x-3\right)+1=2\left(x+1\right)-9\\ \Rightarrow2x-6+1=2x+2-9\\ \Rightarrow2x-5=2x-7\\ \Rightarrow-2=0\left(vô.lí\right)\)

\(2,\dfrac{5-x}{2}=\dfrac{3x-4}{6}\\ \Rightarrow30-6x=6x-8\\ \Rightarrow12x=38\\ \Rightarrow x=\dfrac{19}{6}\)

\(3,\left(x-1\right)^2+\left(x+2\right)\left(x-2\right)=\left(2x+1\right)\left(x-3\right)\\ \Rightarrow x^2-2x+1+x^2-4=2x^2-6x+x-3\\ \Rightarrow2x^2-2x-3=2x^2-5x-3\\ \Rightarrow3x=0\\ \Rightarrow x=0\)

\(4,\left(x+5\right)\left(x-1\right)-\left(x+1\right)\left(x+2\right)=1\\ \Rightarrow x^2+5x-x-5-x^2-2x-x-2=1\\ \\ \Rightarrow x-7=1\\ \Rightarrow x=8\)

\(5,\dfrac{6x-1}{15}-\dfrac{x}{5}=\dfrac{2x}{3}\\ \Rightarrow\dfrac{6x-1}{15}-\dfrac{3x}{15}=\dfrac{10x}{15}\\ \Rightarrow6x-1-3x=10x\\ \Rightarrow3x-1=10x\\ \Rightarrow7x=-1\\ \Rightarrow x=\dfrac{-1}{7}\)

\(6,\dfrac{5\left(x-2\right)}{2}-\dfrac{x+5}{3}=1-\dfrac{4\left(x-3\right)}{5}\\ \Rightarrow\dfrac{75\left(x-2\right)}{30}-\dfrac{10\left(x+5\right)}{30}=\dfrac{30}{30}-\dfrac{24\left(x-3\right)}{30}\\ \Rightarrow75\left(x-2\right)-10\left(x+5\right)=30-24\left(x-3\right)\\ \Rightarrow75x-150-10x-50=30-24x+72\\ \Rightarrow65x-200=102-24x\\ \Rightarrow89x=302\\ \Rightarrow x=\dfrac{320}{89}\)