1) \(\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{4x+15}{9-x^2}\)

ĐKXĐ : \(x\ne\pm3\)

\(\Leftrightarrow\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{-4x-15}{x^2-9}\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\frac{x^2-4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{x^2+3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\frac{x^2-4x+3-x^2-3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow-7x+3=-4x-15\)

\(\Leftrightarrow-7x+4x=-15-3\)

\(\Leftrightarrow-3x=-18\)

\(\Leftrightarrow x=6\)( tmđk )

Vậy x = 6 là nghiệm của phương trình

2) 2x + 3 < 6 - ( 3 - 4x )

<=> 2x + 3 < 6 - 3 + 4x

<=> 2x - 4x < 6 - 3 - 3

<=> -2x < 0

<=> x > 0

Vậy nghiệm của bất phương trình là x > 0

a) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

có bị viết nhầm thì thông cảm nha!

la`thu'hai nga`y 19 nhe

Phân tích  : x²-3x +2=(x-1)(x-2) , x²-4x +3 = (x-1 )(x-3) ,  điều kiện  : x # 1, x # 2 ,x # 3

pt tương đương với  : \(\frac{x+4}{\left(x-1\right)\left(x-2\right)}=\frac{2x+5+x+1}{\left(x-1\right)\left(x-3\right)}\)

                               <=> \(\frac{x+4}{\left(x-1\right)\left(x-2\right)}=\frac{3\left(x+2\right)}{\left(x-1\right)\left(x-3\right)}\)       

                              <=> \(\frac{\left(x+4\right)\left(x-3\right)-3\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)

                             <=> \(\frac{x\left(1-2x\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)

                             <=> x=0 hoặc x=1/2

2:

a: =>2x^2-4x-2=x^2-x-2

=>x^2-3x=0

=>x=0(loại) hoặc x=3

b: =>(x+1)(x+4)<0

=>-4<x<-1

d: =>x^2-2x-7=-x^2+6x-4

=>2x^2-8x-3=0

=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)

1)3(x^3+1) + 2x(x+1)=8

suy ra : 3(x+1)(x^2 +x+1) + 2x(x+1) =8

suy ra : (x+1) ( 3( x^2+x+1)  +2x) -8 =0

suy ra : (x+1) ( 3x^2 +3x+3+2x) -8 =0

mk ko bt nữa

\(x^2+2x-28+8-\sqrt{2x^2+4x+8}=0\)

\(x^2+2x-28+\frac{64-2x^2-4x-8}{8+\sqrt{2x^2+4x+8}}=0\)

\(x^2+2x-28+\frac{-2\left(x^2+2x-28\right)}{8+\sqrt{2x^2+4x+8}}=0\)

\(\left(x^2+2x-28\right)\left(1-\frac{2}{8+\sqrt{2x^2+4x+8}}\right)=0\)

mà \(1-\frac{2}{8+\sqrt{2x+4x+8}}\ne0\Rightarrow x^2+2x-28=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1-\sqrt{29}\\x=-1+\sqrt{29}\end{cases}}\)

phần b nx bạn ơi

a) đk: \(x\ge-2\)

Ta có: \(\sqrt{x+2}-\sqrt{4x+8}+\frac{3}{4}\sqrt{9x+18}=3\)

\(\Leftrightarrow\sqrt{x+2}-2\sqrt{x+2}+\frac{9}{4}\sqrt{x+2}=3\)

\(\Leftrightarrow\frac{5}{4}\sqrt{x+2}=3\)

\(\Leftrightarrow\sqrt{x+2}=\frac{12}{5}\)

\(\Leftrightarrow x+2=\frac{144}{25}\)

\(\Rightarrow x=\frac{94}{25}\) (tm)

b) đk: \(x\ge\frac{3}{2}\)

Ta có: \(\sqrt{x^2-4x+4}=2x-3\)

\(\Leftrightarrow\left|x-2\right|=2x-3\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=2x-3\\x-2=3-2x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\left(ktm\right)\\x=\frac{5}{3}\left(tm\right)\end{cases}}\)

a) \(\sqrt{x+2}-\sqrt{4x+8}+\frac{3}{4}\sqrt{9x+18}=3\)

ĐKXĐ : x ≥ -2

⇔ \(\sqrt{x+2}-\sqrt{2^2\left(x+2\right)}+\frac{3}{4}\sqrt{3^2\left(x+2\right)}=3\)

⇔ \(\sqrt{x+2}-2\sqrt{x+2}+\frac{3}{4}\cdot3\sqrt{x+2}=3\)

⇔ \(-\sqrt{x+2}+\frac{9}{4}\sqrt{x+2}=3\)

⇔ \(\frac{5}{4}\sqrt{x+2}=3\)

⇔ \(\sqrt{x+2}=\frac{12}{5}\)

⇔ \(x+2=\frac{144}{25}\)

⇔ \(x=\frac{94}{25}\left(tmđk\right)\)

b) \(\sqrt{x^2-4x+4}=2x-3\)

⇔ \(\sqrt{\left(x-2\right)^2}=2x-3\)

⇔ \(\left|x-2\right|=2x-3\)(1)

Với x < 2

(1) ⇔ -( x - 2 ) = 2x - 3

     ⇔ 2 - x = 2x - 3

     ⇔ -x - 2x = -3 - 2

     ⇔ -3x = -5

     ⇔ x = 5/3 ( tm )

Với x ≥ 2

(1) ⇔ x - 2 = 2x - 3

     ⇔ x - 2x = -3 + 2

     ⇔ -x = -1

     ⇔ x = 1 ( ktm )

Vậy x = 5/3

\(\Leftrightarrow\frac{2^{3x^2-3x+1}}{3^{x^2-x+1}}.\frac{3^{2x^2-3x+2}}{5^{2x^2-3x+2}}.\frac{5^{3x^2-4x+3}}{7^{3x^2-4x+3}}.\frac{7^{4x^2-5x+4}}{2^{4x^2-5x+4}}=210^{\left(x-1\right)^2}\)

\(\Leftrightarrow\frac{\left(3.5.7\right)^{x^2-x+1}}{2^{x^2-2x+1}}=2^{\left(x-1\right)^2}.\left(3.5.7\right)^{\left(x-1\right)^2}\)

\(\Leftrightarrow105^x=2^{2\left(x-1\right)^2}\)

Lấy Logarit cơ số 2 hai vế, ta được :

\(2\left(x-1\right)^2=\left(\log_2105\right)x\)

\(\Leftrightarrow2x^2-\left(4+\log_2105\right)x+2=0\)

\(\Leftrightarrow x=\frac{\left(2+\log_2105\right)\pm\sqrt{\log^2_2105+8\log_2105}}{4}\)

Vậy phương trình đã cho có 2 nghiệm