a.

ĐKXĐ: \(-1\le x\le1\)

Đặt \(\sqrt{1-x^2}=t\Rightarrow0\le t\le1\)

\(x^2=1-t^2\Rightarrow x^4=t^4-2t^2+1\)

Pt trở thành:

\(729\left(t^4-2t^2+1\right)+8t=36\)

\(\Leftrightarrow729t^4-1458t^2+8t+693=0\)

\(\Leftrightarrow\left(9t^2+2t-9\right)\left(81t^2-18t-77\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}9t^2+2t-9=0\\81t^2-18t-77=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{\sqrt{82}-1}{9}\\t=\dfrac{1+\sqrt{78}}{9}\end{matrix}\right.\)

\(\Rightarrow x=\pm\sqrt{1-t^2}=...\)

b.

ĐKXĐ: ...

\(-3\left(10+4x-x^2\right)-5\sqrt{10+4x-x^2}+42=0\)

Đặt \(\sqrt{10+4x-x^2}=t\ge0\)

\(\Rightarrow-3t^2-5t+42=0\)

\(\Rightarrow\left[{}\begin{matrix}t=3\\t=-\dfrac{14}{3}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{10+4x-x^2}=3\)

\(\Leftrightarrow x^2-4x-1=0\)

\(\Leftrightarrow x=...\)

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

ĐKXĐ:...

\(\Leftrightarrow\frac{36}{\sqrt{x-2}}+4\sqrt{x-2}+\frac{4}{\sqrt{y-1}}+\sqrt{y-1}=28\)

Ta có:

\(VT\ge2\sqrt{\frac{36.4\sqrt{x-2}}{\sqrt{x-2}}}+2\sqrt{\frac{4\sqrt{y-1}}{\sqrt{y-1}}}=28\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}\frac{9}{\sqrt{x-2}}=\sqrt{x-2}\\\frac{4}{\sqrt{y-1}}=\sqrt{y-1}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=11\\y=5\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x-2}=a\left(a>0\right)\\\sqrt{y-1}=b\left(b>0\right)\end{matrix}\right.\)

\(\Rightarrow\dfrac{36}{a}+\dfrac{4}{b}=28-4a-b\)

\(\Leftrightarrow\left(\dfrac{36}{a}+4a\right)+\left(\dfrac{4}{b}+b\right)=28\)

\(VT\ge2\sqrt{\dfrac{36}{a}\times4a}+2\sqrt{\dfrac{4}{b}\times b}=28\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\dfrac{36}{a}=4a\\\dfrac{4}{b}=b\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\) \(\left(a,b>0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-2}=3\\\sqrt{y-1}=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=11\\y=5\end{matrix}\right.\) (n)

Vậy . . . >3<

ĐKXĐ : \(\hept{\begin{cases}x>2\\y>1\end{cases}}\)

PT đã cho tương đương với \(\left(\frac{36}{\sqrt{x-2}}+4\sqrt{x-2}-24\right)+\left(\frac{4}{\sqrt{y-1}}+\sqrt{y+1}-4\right)=0\)

\(\Leftrightarrow\frac{\left(2\sqrt{x-2}-6\right)^2}{\sqrt{x-2}}+\frac{\left(\sqrt{y-1}-2\right)^2}{\sqrt{y-1}}=0\)

\(\Leftrightarrow\hept{\begin{cases}2\sqrt{x-2}-6=0\\\sqrt{y-1}-2=0\end{cases}}\)

Tới đây bạn tự giải được rồi :)

Câu hỏi của Thu Trần Thị - Toán lớp 9 - Học toán với OnlineMath

tham khảo nhé 

bn cần đoa

                                       Bài giải

Bạn kham khảo câu hỏi này nha bạn ! Thu Trần Thị 

Có \(4\left(\frac{9}{\sqrt{x-2}}+\sqrt{x-2}\right)\ge4.2\sqrt{\frac{9}{\sqrt{x-2}}\sqrt{x-2}}=24\)(Cô si)
\(\frac{4}{\sqrt{y-1}}+\sqrt{y-1}\ge2\sqrt{\frac{4}{\sqrt{y-1}}\sqrt{y-1}}=4\)
\(\Rightarrow\frac{4}{\sqrt{y-1}}+\sqrt{y-1}+4\left(\frac{9}{\sqrt{x-2}}+\sqrt{x-2}\right)\ge28\)
Dấu "=" xảy ra <=>\(\int^{9=x-2}_{4=y-1}\Leftrightarrow\int^{x=11}_{y=5}\)