=> x< 2 và x>-9

Đây là lớp 6 nhé ko phải lớp 8 đâu. Tại mình ghi nhầm

\(\Leftrightarrow x\in\left\{0;43\right\}\)

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)

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

x=6 hoặc -6

x=3 hoặc x=-3

9-\(X^2\)=0

9-\(X^2\)=0:2

9-\(X^2\)=0

X=0+9

X=9

9=3.3

=\(3^2\)

VẬY X=3

MIK LM KO LOGIC CHO LẮM ĐÂU

\(\Rightarrow18:x=10-1=9\Rightarrow x=2\)

\(\Leftrightarrow10-18:x=1\)

\(\Leftrightarrow18:x=9\)

hay x=2

\(10-18:x=2020^0\)

\(10-18:x=1\)

\(18:x=10-1\)

\(x=18:9\)

\(x=2\)

\(\left(x^2+1\right)\left(x-5\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x^2+1=0\left(vô.lí.vì.x^2\ge0,1>0\right)\\x-5=0\end{matrix}\right.\\ \Rightarrow x=5\)

\(\left(x^2+1\right)\left(x-5\right)=0\)

TH1 : x^2 + 1 = 0 ( vô lí vì x^2 + 1 > 0 ) 

TH2 : x - 5 = 0 <=> x = 5 

Vậy x = 5 

`5/(-x^2+5x-6)+(x+3)/(2-x)=0`

Đk:`x ne 2,x ne 3`

`pt<=>-5/(x^2-5x+6)-(x+3)/(x-2)=0`

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

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

`<=>x^2-9=-5`

`<=>x^2-4=0`

`<=>(x-2)(x+2)=0`

`x ne 2=>x-2 ne 0`

`<=>x+2=0`

`<=>x=-2`

Vậy `S={-2}`

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

\(\Leftrightarrow\dfrac{2\sqrt{x}\sqrt{x}}{2\sqrt{x}}+\dfrac{1}{2\sqrt{x}}=\dfrac{3.2\sqrt{x}}{2\sqrt{x}}\)

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

\(\Leftrightarrow2x-6\sqrt{x}+1=0\)

\(\Leftrightarrow...\)

\(\Rightarrow2x+1=6\sqrt{x}\)

\(\Rightarrow2x-6\sqrt{x}+1=0\)

\(\Rightarrow\sqrt{x}=\dfrac{3\pm\sqrt{7}}{2}\)

\(\Rightarrow x=\left(\dfrac{3\pm\sqrt{7}}{2}\right)^2=\dfrac{8\pm3\sqrt{7}}{2}\)

\(\Leftrightarrow x-\left[3-x+3+x-2\right]=0\)

=>x=4