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

\(\Leftrightarrow x^4+5x^3+9x^2-3x^3-15x^2-27x+9x^2+45x+81=9x^2\)

\(\Leftrightarrow x^4+2x^3+3x^2+18x+81=9x^2\)

\(\Leftrightarrow x^4+2x^3+3x^2+18x+81-9x^2=0\)

\(\Leftrightarrow x^4+2x^2-6x^2+18x+81=0\)

\(\Leftrightarrow\left(x^3-x^2-3x+27\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x^2-4x+9\right)\left(x+3\right)\left(x+3\right)=0\)

Vì \(x^2-4x+9\ne0\) nên: 

\(\Rightarrow x+3=0\)

     \(x=-3\)

Vậy: nghiệm phương trình là: {-3}

c: =>(x+2)(x+3)(x-5)(x-6)=180

=>(x^2-3x-10)(x^2-3x-18)=180

=>(x^2-3x)^2-28(x^2-3x)=0

=>x(x-3)(x-7)(x+4)=0

=>\(x\in\left\{0;3;7;-4\right\}\)

c: =>(x-3)(x+2)(2x+1)(3x-1)=0

=>\(x\in\left\{3;-2;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)

`1)x^4 -10x^3 +26x^2 -10x+1=0`
`x=0=>VT=1=>x=0(l)`
Chia 2 vế cho `x^2>0` ta có
`x^2-10x+26-10/x+1/x^2=0`
`=>x^2+1/x^2+26-10(x+1/x)=0`
`=>(x+1/x)^2-10(x+1/x)+24=0`
Đặt `a=x+1/x`
`pt<=>a^2-10a+24=0`
`<=>` $\left[ \begin{array}{l}a=4\\a=6\end{array} \right.$
`a=4<=>x+1/x=4<=>x^2-4x+1=0<=>` $\left[ \begin{array}{l}x=\sqrt3+2\\x=-\sqrt3+2\end{array} \right.$
`a=6<=>x+1/x=6<=>x^2-6x+1=0<=>` $\left[ \begin{array}{l}x=\sqrt8+3\\x=-\sqrt8+3\end{array} \right.$
Vậy `S={\sqrt3+2,-\sqrt3+2,\sqrt8+3,-\sqrt8+3}`

2)Do hệ số chẵn bằng=hệ số lẻ
`=>x=-1`
`pt<=>x^4+x^3+4x^3+4x^2+6x^2+6x+9x+9=0`
`<=>(x+1)(x^3+4x^2+6x+9)=0`
`<=>(x+1)(x^3+3x^2+x^2+6x+9)=0`
`<=>(x+1)[x^2(x+3)+(x+3)^2]=0`
`<=>(x+1)(x+3)(x^2+x+3)=0`
Do `x^2+x+3=(x+1/2)^2+11/4>0`
`=>` $\left[ \begin{array}{l}x=-3\\x=-1\end{array} \right.$
Vậy `S={-1,-3}`

*\(\dfrac{x-1}{x+2}\)-\(\dfrac{x}{x+2}\)=\(\dfrac{5x-2}{4-x^2}\).ĐKXĐ: x\(\ne\pm2\)

<=>\(\dfrac{\left(x-1\right)\left(2-x\right)}{4-x^2}\)-\(\dfrac{x\left(2-x\right)}{4-x^2}\)=\(\dfrac{5x-2}{4-x^2}\)

=>2x-\(x^2\)-2+x-2x+\(x^2\)=5x-2

<=>x-2=5x-2

<=>x-5x=2-2

<=>-4x=0

<=> x = 0(TM)

Vậy phương trình có tập nghiệm là S={0}

*(x+4)(5x+9)-x-4=0

<=>(x+4)(5x+9)-(x+4)=0

<=>(x+4)(5x+9-1)=0

<=>(x+4)(5x+8)=0

<=>x+4= 0 hoặc 5x+8=0

(+) x+4=0 (+)5x+8=0

<=>x=-4 <=>5x=-8

<=>x=\(\dfrac{-8}{5}\)

Vậy phương trình có tập nghiệm là S={\(-4;\dfrac{-8}{5}\)}

\(\dfrac{-1}{x+2}=\dfrac{5x-2}{4-x^2}< =>\dfrac{\left(2-x\right).\left(-1\right)}{4-x^2}=\dfrac{5x-2}{4-x^2}\)

<=> \(\dfrac{x-2}{4-x^2}=\dfrac{5x-2}{4-x^2}\) <=> x-2=5x-2<=>5x-x=-2+2 <=>4x=0

=> x vô số nghiệm

\(\left(x+4\right)\left(5x+9\right)-x-4=0\)

<=>\(5x^2+29x+36-x-4=0\)

<=>\(5x^2+28x+32=0\)

<=>\(\left(x+\dfrac{8}{5}\right)\left(x+4\right)=0\)

=> x=?

\(\Leftrightarrow\sqrt{5x^2+14x+9}-\sqrt{x^2-x-20}-5\sqrt{x+1}=0\)

\(\Rightarrow4x=-7\)

=>x=8

phương trình trên có kq: x=8

bạn trình bày ra được không

\(ĐKXĐ:x\ge-1,5\)

\(=>\left(2\sqrt{2x^3+5x^2+9x+9}\right)^2=\left(x^2+3x+6\right)^2\)

=>\(8x^3+20x^2=x^4+6x^3+21x^2\) ( Đã đc rút gọn )

=> \(x^4+6x^3+21x^2-\left(8x^3+20x^2\right)=0\)

=> \(x^4-2x^3+x^2=0\)

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

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

=> \(\left[{}\begin{matrix}\left|x\right|=\sqrt{0}\\\left|x-1\right|=\sqrt{0}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy....

a:

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

=>|x-3|=3

=>x-3=3 hoặc x-3=-3

=>x=0 hoặc x=6

b: \(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}=2\)

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

=>\(\left|\sqrt{x-1}+1\right|=2\)

=>\(\left[{}\begin{matrix}\sqrt{x-1}+1=2\\\sqrt{x-1}+1=-2\left(loại\right)\end{matrix}\right.\Leftrightarrow\sqrt{x-1}=1\)

=>x-1=1

=>x=2

c:

ĐKXĐ: x>4/5

PT \(\Leftrightarrow\sqrt{\dfrac{5x-4}{x+2}}=2\)

=>\(\dfrac{5x-4}{x+2}=4\)

=>5x-4=4x+8

=>x=12(nhận)

d: ĐKXĐ: x-4>=0 và x+1>=0

=>x>=4

PT =>\(\left(\sqrt{x-4}+\sqrt{x+1}\right)^2=5^2=25\)

=>\(x-4+x+1+2\sqrt{\left(x-4\right)\left(x+1\right)}=25\)

=>\(\sqrt{4\left(x^2-3x-4\right)}=25-2x+3=28-2x\)

=>\(\sqrt{x^2-3x-4}=14-x\)

=>x<=14 và x^2-3x-4=(14-x)^2=x^2-28x+196

=>x<=14 và -3x-4=-28x+196

=>x<=14 và 25x=200

=>x=8(nhận)

a) \(\sqrt{x^2-6x+9}=3\)

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

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

TH1: \(\left|x-3\right|=x-3\) với \(x\ge3\)

Pt trở thành:

\(x-3=3\) (ĐK: \(x\ge3\))

\(\Leftrightarrow x=3+3\)

\(\Leftrightarrow x=6\left(tm\right)\)

TH2: \(\left|x-3\right|=-\left(x-3\right)\) với \(x< 3\)

Pt trở thành:

\(-\left(x-3\right)=3\) (ĐK: \(x< 3\))

\(\Leftrightarrow x-3=-3\)

\(\Leftrightarrow x=-3+3\)

\(\Leftrightarrow x=0\left(tm\right)\)

b) \(\sqrt{x+2\sqrt{x-1}}=2\) (ĐK: \(x\ge1\))

\(\Leftrightarrow x+2\sqrt{x-1}=4\)

\(\Leftrightarrow2\sqrt{x-1}=4-x\)

\(\Leftrightarrow4\left(x-1\right)=16-8x+x^2\)

\(\Leftrightarrow4x-4=16-8x+x^2\)

\(\Leftrightarrow x^2-12x+20=0\)

\(\Leftrightarrow\left(x-10\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

c) \(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\) (ĐK: \(x\ge\dfrac{4}{5}\))

\(\Leftrightarrow\dfrac{5x-4}{x+2}=4\)

\(\Leftrightarrow5x-4=4x+8\)

\(\Leftrightarrow x=12\left(tm\right)\)