a, ĐK: \(x\le-1,x\ge3\)

\(pt\Leftrightarrow2\left(x^2-2x-3\right)+\sqrt{x^2-2x-3}-3=0\)

\(\Leftrightarrow\left(2\sqrt{x^2-2x-3}+3\right).\left(\sqrt{x^2-2x-3}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-2x-3}=-\dfrac{3}{2}\left(l\right)\\\sqrt{x^2-2x-3}=1\end{matrix}\right.\)

\(\Leftrightarrow x^2-2x-3=1\)

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

\(\Leftrightarrow x=1\pm\sqrt{5}\left(tm\right)\)

b, ĐK: \(-2\le x\le2\)

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

Khi đó phương trình tương đương:

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

\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2+x}-2\sqrt{2-x}=0\\\sqrt{2+x}-2\sqrt{2-x}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2+x=8-4x\\2+x=17-4x+12\sqrt{2-x}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\left(tm\right)\\5x-15=12\sqrt{2-x}\left(1\right)\end{matrix}\right.\)

Vì \(-2\le x\le2\Rightarrow5x-15< 0\Rightarrow\left(1\right)\) vô nghiệm

Vậy phương trình đã cho có nghiệm \(x=\dfrac{6}{5}\)

c, ĐK: \(0\le x\le9\)

Đặt \(\sqrt{9x-x^2}=t\left(0\le t\le\dfrac{9}{2}\right)\)

\(pt\Leftrightarrow9+2\sqrt{9x-x^2}=-x^2+9x+m\)

\(\Leftrightarrow-\left(-x^2+9x\right)+2\sqrt{9x-x^2}+9=m\)

\(\Leftrightarrow-t^2+2t+9=m\)

Khi \(m=9,pt\Leftrightarrow-t^2+2t=0\Leftrightarrow\left[{}\begin{matrix}t=0\\t=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}9x-x^2=0\\9x-x^2=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=9\left(tm\right)\\x=\dfrac{9\pm\sqrt{65}}{2}\left(tm\right)\end{matrix}\right.\)

Phương trình đã cho có nghiệm khi phương trình \(m=f\left(t\right)=-t^2+2t+9\) có nghiệm

\(\Leftrightarrow minf\left(t\right)\le m\le maxf\left(t\right)\)

\(\Leftrightarrow-\dfrac{9}{4}\le m\le10\)

a) ĐKXĐ: x <= 2

pt --> 4 - 2x = 25 <=> x = -21/2 (thỏa)

??

Đề kiểu gì vậy ?

b) ĐKXĐ: x >= -1

pt <=> 8sqrt(x + 1)=16 <=> sqrt(x+1)=2 --> x + 1 = 4 <=> x = 3

a, ĐKXĐ: \(x\le2\)

\(\sqrt{4-2x}=5\\ \Leftrightarrow4-2x=25\\ \Leftrightarrow2x=-21\\ \Leftrightarrow x=-10,5\left(tm\right)\)

b, ĐKXĐ: \(x\ge-1\)

\(\sqrt{25\left(x+1\right)}+\sqrt{9x+9}=16\\ \Leftrightarrow5\sqrt{x+1}+\sqrt{9\left(x+1\right)}=16\\ \Leftrightarrow5\sqrt{x+1}+3\sqrt{x+1}=16\\ \Leftrightarrow8\sqrt{x+1}=16\\ \Leftrightarrow\sqrt{x+1}=2\\ \Leftrightarrow x+1=4\\ \Leftrightarrow x=3\)

c, \(\sqrt{4x^2+12x+9}=4\Leftrightarrow4x^2+12x+9=16\\ \Leftrightarrow4x^2+12x-7=0\\ \Leftrightarrow\left(4x^2-2x\right)+\left(14x-7\right)=0\\ \Leftrightarrow2x\left(2x-1\right)+7\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

a: \(\Leftrightarrow4-2x=25\)

hay \(x=-\dfrac{21}{2}\)

c: \(\Leftrightarrow\left|2x+3\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=4\\2x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Câu 1 là \(\left(8x-4\right)\sqrt{x}-1\) hay là \(\left(8x-4\right)\sqrt{x-1}\)?

Câu 1:ĐK \(x\ge\frac{1}{2}\)

\(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)

<=> \(\left(4x^2-3x-1\right)+4\left(2x-1\right)\sqrt{x}-2\sqrt{\left(2x-1\right)\left(x+3\right)}\)

<=> \(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}\left(2\sqrt{x\left(2x-1\right)}-\sqrt{x+3}\right)=0\)

<=> \(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{8x^2-4x-x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)

<=>\(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{\left(x-1\right)\left(8x+3\right)}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)

<=> \(\left(x-1\right)\left(4x+1+2\sqrt{2x-1}.\frac{8x+3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}\right)=0\)

Với \(x\ge\frac{1}{2}\)thì \(4x+1+2\sqrt{2x-1}.\frac{8x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}>0\)

=> \(x=1\)(TM ĐKXĐ)

Vậy x=1

câu 2 ĐK \(x\ge1\)

\(\left(5x+8\right)\sqrt{2x-1}+7x\sqrt{x+3}=9x+18-\left(x+26\right)\sqrt{x-1}=0\)

<=> \(\left(5x+8\right)\left(\sqrt{2x-1}-1\right)+7x\left(\sqrt{x+3}-2\right)+\left(x+26\right)\sqrt{x-1}+10\left(x-1\right)=0\)

<=>\(\left(5x+8\right).\frac{2x-2}{\sqrt{2x-1}+1}+7x.\frac{x+3-4}{\sqrt{x+3}+2}+\left(x+26\right)\sqrt{x-1}+10\left(x-1\right)=0\)

<=> \(\sqrt{x-1}\left(\frac{2\left(5x+8\right)\sqrt{x-1}}{\sqrt{2x-1}+1}+\frac{7x\sqrt{x-1}}{\sqrt{x+3}+2}+\left(x+26\right)+10\sqrt{x-1}\right)=0\)

Với \(x\ge1\)thì cái trong ngoặc >0

=> \(x=1\)

Vậy x=1

a) Đặt $\sqrt{x+1}=a; \sqrt{9-x}=b$ thì bài toán trở thành:

Tìm max, min của $f(a,b)=a+b$ với $a,b\geq 0$ và $a^2+b^2=10$Ta có:

$f^2(a,b)=(a+b)^2=a^2+b^2+2ab=10+2ab\geq 10$ do $ab\geq 0$

$\Rightarrow f(a,b)\geq \sqrt{10}$ hay $f_{\min}=\sqrt{10}$

Mặt khác: $f^2(a,b)=(a+b)^2\leq 2(a^2+b^2)=20$ (theo BĐT AM-GM)

$\Rightarrow f(a,b)\leq \sqrt{20}=2\sqrt{5}$ hay $f_{\max}=2\sqrt{5}$

b) 

Đặt $\sqrt{x}=a; \sqrt{2-x}=b$ thì bài toán trở thành:

Tìm max, min của $f(a,b)=a+b+ab$ với $a,b\geq 0$ và $a^2+b^2=2$. Ta có:

$f(a,b)=\sqrt{(a+b)^2}+ab=\sqrt{a^2+b^2+2ab}+ab=\sqrt{2+2ab}+ab\geq \sqrt{2}$ do $ab\geq 0$

Vậy $f_{\min}=\sqrt{2}$

Lại có, theo BĐT AM-GM:

$f(a,b)=\sqrt{2+2ab}+ab\leq \sqrt{2+a^2+b^2}+\frac{a^2+b^2}{2}=\sqrt{2+2}+\frac{2}{2}=3$

Vậy $f_{\max}=3$

c) Đặt $\sqrt{8-x^2}=a$ thì bài toán trở thành tìm max, min của:

$f(x,a)=x+a+ax$ với $x,a\geq 0$ và $x^2+a^2=8$. Bài này chuyển về y hệt  như phần b. 

$f_{\min}=2\sqrt{2}$

$f_{\max}=8$

d) Tương tự:

$f_{\min}=2$ khi $x=\pm 2$

$f_{\max}=2+2\sqrt{2}$ khi $x=0$