ĐKXĐ: ....

\(\Leftrightarrow x^2-8x+16+3x+4-8\sqrt{3x+4}+16=0\)

\(\Leftrightarrow\left(x-4\right)^2+\left(\sqrt{3x+4}-4\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-4=0\\\sqrt{3x+4}-4=0\end{matrix}\right.\) \(\Rightarrow x=4\)

Lời giải:

a) ĐKXĐ: $x\ge \frac{-4}{3}$

Ta có:

PT \(\Leftrightarrow x^2-5x+36-8\sqrt{3x+4}=0\)

\(\Leftrightarrow (x^2-8x+16)+(3x+4-8\sqrt{3x+4}+16)=0\)

\(\Leftrightarrow (x-4)^2+(\sqrt{3x+4}-4)^2=0\)

Dễ thấy \((x-4)^2\geq 0; (\sqrt{3x+4}-4)^2\geq 0, \forall x\geq \frac{-4}{3}\)

Do đó để tổng của chúng bằng $0$ thì \((x-4)^2=(\sqrt{3x+4}-4)^2=0\Leftrightarrow x=4\) (thỏa mãn)

Vậy..........

b) ĐK: $x\geq \frac{2}{3}$

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

\(\Rightarrow 3x-2=(2-\sqrt{3})^2=7-4\sqrt{3}\)

\(\Rightarrow x=\frac{7-4\sqrt{3}+2}{3}=\frac{9-4\sqrt{3}}{3}\) (thỏa mãn)

Vậy.......

Lời giải:

a) ĐKXĐ: $x\ge \frac{-4}{3}$

Ta có:

PT \(\Leftrightarrow x^2-5x+36-8\sqrt{3x+4}=0\)

\(\Leftrightarrow (x^2-8x+16)+(3x+4-8\sqrt{3x+4}+16)=0\)

\(\Leftrightarrow (x-4)^2+(\sqrt{3x+4}-4)^2=0\)

Dễ thấy \((x-4)^2\geq 0; (\sqrt{3x+4}-4)^2\geq 0, \forall x\geq \frac{-4}{3}\)

Do đó để tổng của chúng bằng $0$ thì \((x-4)^2=(\sqrt{3x+4}-4)^2=0\Leftrightarrow x=4\) (thỏa mãn)

Vậy..........

b) ĐK: $x\geq \frac{2}{3}$

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

\(\Rightarrow 3x-2=(2-\sqrt{3})^2=7-4\sqrt{3}\)

\(\Rightarrow x=\frac{7-4\sqrt{3}+2}{3}=\frac{9-4\sqrt{3}}{3}\) (thỏa mãn)

Vậy.......

dk \(x\ge-\frac{4}{3}\)

\(x^2-5x+4=8\sqrt{3x+4}-32\)

\(\Leftrightarrow\left(x-1\right)\left(x-4\right)=8\left(\sqrt{3x+4}-4\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x-4\right)-8\frac{\left(\sqrt{3x+4}-4\right)\left(\sqrt{3x+4}+4\right)}{\sqrt{3x+4}+4}=0\)

\(\left(x-1\right)\left(x-4\right)-8.\frac{3\left(x-4\right)}{\sqrt{3x+4}+4}=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-1-\frac{24}{\sqrt{3x+4}+4}=0\right)\)

đến đây để rồi tự làm nhé ^^

bài toán của mk k có -23 ở vế sau ạ

a) \(\sqrt{1+x}-\sqrt{8-x}+\sqrt{\left(1+x\right)\left(8-x\right)}=3\)

đặt t \(=\sqrt{1+x}-\sqrt{8-x}\)

\(\Leftrightarrow t^2=1+x-2\sqrt{\left(1+x\right)\left(8-x\right)}+8-x\)

\(\Leftrightarrow\sqrt{\left(1+x\right)\left(8-x\right)}=\dfrac{9-t^2}{2}\)

pt \(\Rightarrow t+\dfrac{9-t^2}{2}=3\)

\(\Leftrightarrow t^2-2t-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-1\\t=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{1+x}-\sqrt{8-x}=-1\\\sqrt{1+x}-\sqrt{8+x}=3\end{matrix}\right.\)

suy ra tìm đc x

câu b đặt t =\(3x^2+5x+8\)

ta có pt \(\Leftrightarrow\sqrt{t}-\sqrt{t-7}=1\)

\(\Rightarrow t=16\)

\(\Leftrightarrow3x^2+5x+8=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{8}{3}\end{matrix}\right.\)

\(x^2-5x+36=8\sqrt{3x+4}\)

\(\Leftrightarrow x^2-5x+36-8\sqrt{3x+4}=0\)

\(\Leftrightarrow\left(-8\sqrt{3x+4}+32\right)+\left(x^2-5x+4\right)=0\)

\(\Leftrightarrow-8\left(\sqrt{3x+4}-4\right)+\left(x-1\right)\left(x-4\right)=0\)

\(\Leftrightarrow-8.\frac{3x+4-16}{\sqrt{3x+4}+4}+\left(x-1\right)\left(x-4\right)=0\)

\(\Leftrightarrow-8.\frac{3x-12}{\sqrt{3x+4}+4}+\left(x-1\right)\left(x-4\right)=0\)

\(\left(x-4\right)\left(\frac{-24}{\sqrt{3x+4}+4}+x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\\frac{-24}{\sqrt{3x+4}+4}+x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\-\frac{24}{\sqrt{3x+4}+4}+3+x-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\-3.\frac{16-3x-4}{\left(\sqrt{3x+4}+4\right)^2}+\left(x-4\right)=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\\left(x-4\right)\left[\frac{9}{\left(\sqrt{3x+4}+4\right)^2}+1\right]=0\end{cases}}\)

Mà \(\frac{9}{\left(\sqrt{3x+4}+4\right)^2}+1>0\forall x\) nên \(x-4=0\Rightarrow x=4\)

Vật PT có nghiệm duy nhất là \(x=4\)

cảm ơn bạn

6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)

Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)

Phương trình sẽ trở thành là: a^2+a-42=0

=>(a+7)(a-6)=0

=>a=-7(loại) hoặc a=6(nhận)

=>2x^2+3x+9=36

=>2x^2+3x-27=0

=>2x^2+9x-6x-27=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)