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

\(ĐK:\left\{{}\begin{matrix}1-x\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x\ge2\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

Vậy PT vô nghiệm

a, ĐKXĐ : x ≠ 4

b,

\(\Leftrightarrow3x+2=2\left(x-4\right)\)

\(\Leftrightarrow3x+2=2x-8\)

\(\Leftrightarrow x=-10\) (N)

Vậy : ...

`a)` Ptr xác định `<=>x-4 \ne 0<=>x \ne 4`

`b)[3x+2]/[x-4]=2`         `ĐK: x \ne 4`

`<=>3x+2=2(x-4)`

`<=>3x+2=2x-8`

`<=>3x-2x=-8-2`

`<=>x=-10` (t/m)

Vậy `S={-10}`

a: Sửa đề: PT x^2-2x-m-1=0

Khi m=2 thì Phương trình sẽ là:

x^2-2x-2-1=0

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

=>(x-3)(x+1)=0

=>\(\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

b:

\(\text{Δ}=\left(-2\right)^2-4\left(-m-1\right)\)

\(=4+4m+4=4m+8\)

Để phương trình có hai nghiệm dương thì

\(\left\{{}\begin{matrix}4m+8>0\\2>0\\-m-1>0\end{matrix}\right.\Leftrightarrow-2< m< -1\)

\(\sqrt{x_1}+\sqrt{x_2}=2\)

=>\(x_1+x_2+2\sqrt{x_1x_2}=4\)

=>\(2+2\sqrt{-m-1}=4\)

=>\(2\sqrt{-m-1}=2\)

=>-m-1=1

=>-m=2

=>m=-2(loại)

a: ĐKXĐ: \(x\in R\)

b: ĐKXĐ: \(x\in R\)

\(\Leftrightarrow x+9=49\)

hay x=40

ĐK x khác 0

X=-2

\(\sqrt{x+9}=7\)

Đk: \(x+9\ge0\Leftrightarrow x\ge-9\)

Pt\(\Rightarrow x+9=49\Rightarrow x=40\left(tmđk\right)\)

giải pt thì chịu còn điều kiện thì biết

x^2-x+1>0

x^2-x-2>0

ĐK: `x \ne 3; x \ne -3`

`A=3/(x-3)-(6x)/(9-x^2)+x/(x+3)`

`=3/(x-3)+(6x)/(x^2-9)+x/(x+3)`

`=3/(x-3)+(6x)/((x-3)(x+3))+x/(x+3)`

`=(3(x+3)+6x+x(x-3))/((x-3)(x+3))`

`=(3x+9+6x+x^2-3x)/((x+3)(x-3))`

`=(x^2+6x+9)/((x-3)(x+3))`

`=((x+3)^2)/((x-3)(x+3))`

`=(x+3)/(x-3)`

`x=5 => A=(5+3)/(5-3)=4`

ĐKXĐ:\(\left\{{}\begin{matrix}x-3\ne0\\9-x^2\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x^2\ne9\\x\ne-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)

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

Thay x=5 vào \(\dfrac{x+3}{x-3}=\dfrac{5+3}{5-3}=\dfrac{8}{2}=4\)