=>x-9>0 hoặc x-3<0

=>x<3 hoặc x>9

cụ thể hơn được ko bạn mình chưa hiều lắm

a) 2/3 + 11/3 = 13/3

b) 4/5 - 2/3 = 12/15 - 10/15 = 2/15

c) 21/9 x 3/4 = 7/3 x 3/4 = 21/12 = 7/4

d) 6/5 : 6/9 = 6/5 x 9/6 = 54/30 = 9/5

chúc bạn học tốt!

\(\dfrac{-3}{5}-x=\dfrac{21}{10}\)

\(x=\dfrac{-3}{5}-\dfrac{21}{10}\)

\(x=\)-\(\dfrac{27}{10}\)

\(x:\dfrac{2}{9}=\dfrac{9}{2}\)

\(x.\dfrac{9}{2}=\dfrac{9}{2}\)

\(x=\dfrac{9}{2}:\dfrac{9}{2}\)

\(x=1\)

\(\dfrac{x}{9}=\dfrac{5}{3}\)

\(x.3=5.9\)

\(x.3=45\)

\(x=45:3=15\)

\(x:\left(\dfrac{2}{5}\right)^3=\left(\dfrac{5}{2}\right)^3\)

\(x:\dfrac{8}{125}=\dfrac{125}{8}\)

\(x.\dfrac{125}{8}=\dfrac{125}{8}\)

\(x=\dfrac{125}{8}:\dfrac{125}{8}=1\)

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

\(\Leftrightarrow\dfrac{x}{3+x}-\dfrac{1}{x-3}=\dfrac{x^2+9}{\left(x-3\right)\left(x+3\right)}\left(1\right)\)

Đk : (x - 3)(x + 3) \(\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)

(1) \(\Leftrightarrow\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+9}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow x^2-3x-x-3=x^2+9\)

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

\(\Leftrightarrow-4x=12\Leftrightarrow x=-3\left(KTMĐK\right)\)

Vậy phương trình trên vô nghiệm \(S=\varnothing\)

u(x)=        -3x⁴ -2x² +6x-9

+v(x)=         3x⁴+2x² +6x-9

u(x)+v(x)=12x -18

u(x)-v(x)=-6x⁴-4x 

câu u(x)-v(x)   bạn tự trình bày như trên thêm dấu+,- vào giữa 2 đa thức nhoa!!

chúc bạn yhi tốt nha!!!!

a) (x - 7)⁵ + (x - 7)³ = 0

=> (x - 7)³ [(x - 7)² + 1] = 0

=> \(\orbr{\begin{cases}\left(x-7\right)^3=0\\\left(x-7\right)^2+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\\left(x-7\right)^2=-1\end{cases}}\)

Mà \(\left(x-7\right)^2\ge0\)

\(-1< 0\)

=> x = 7 thỏa mãn

Vậy x = 7

b) \(\left(\frac{9}{8}\right)^6:\left(\frac{3}{4}\right)^6=\left(\frac{3}{2}\right)^x\)

=> \(\left(\frac{9}{8}:\frac{3}{4}\right)^6=\left(\frac{3}{2}\right)^x\)

=> \(\left(\frac{9}{8}\cdot\frac{4}{3}\right)^6=\left(\frac{3}{2}\right)^x\)

=> \(\left(\frac{3}{2}\right)^6=\left(\frac{3}{2}\right)^x\)

=> x = 6

\(\Leftrightarrow x+\dfrac{2}{3}=\dfrac{6}{7}\\ \Leftrightarrow x=\dfrac{6}{7}-\dfrac{2}{3}=\dfrac{4}{21}\)

cop y chang ở trên

a, \(3x+2\left(x-5\right)=6-\left(5x-1\right)\)

\(\Leftrightarrow3x+2x-10=6-5x+1\)

\(\Leftrightarrow-15\ne0\)Vậy phương trình vô nghiệm 

b, \(x^3-3x^2-x+3=0\)

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

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x+1\right)=0\Leftrightarrow x=3;\pm1\)

Vậy tập nghiệm của phương trình là S = { 1 ; -1 ; 3 }

c, \(\frac{1}{x-3}+\frac{x}{x+3}=\frac{2}{x^2-9}ĐK:x\ne\pm3\)

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

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

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

Vậy ... 

1) Ta có: \(\dfrac{a-6\sqrt{a}+9}{5\sqrt{a}-15}\)

\(=\dfrac{\left(\sqrt{a}-3\right)^2}{5\left(\sqrt{a}-3\right)}\)

\(=\dfrac{\sqrt{a}-3}{5}\)

2) Ta có: \(5x-\sqrt{x^2-10x+25}\)

\(=5x-\left|x-5\right|\)

\(=5x-5+x\)

=6x-5

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

\(=\dfrac{\left|x-1\right|}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{\pm1}{x+1}\)

4) Ta có: \(3\sqrt{5}-\sqrt{46-6\sqrt{5}}\)

\(=3\sqrt{5}-3\sqrt{5}+1\)

=1