a) Áp dụng bđt AM-GM có:

\(\sqrt[3]{\left(9-x\right).8.8}\le\dfrac{9-x+8+8}{3}=\dfrac{25-x}{3}\)\(\Leftrightarrow\sqrt[3]{9-x}\le\dfrac{25-x}{12}\)

\(\sqrt[3]{\left(7+x\right).8.8}\le\dfrac{7+x+8+8}{3}=\dfrac{23+x}{3}\)\(\Leftrightarrow\sqrt[3]{7+x}\le\dfrac{23+x}{12}\)

Cộng vế với vế \(\Rightarrow\sqrt[3]{9-x}+\sqrt[3]{7+x}\le4\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}9-x=8\\7+x=8\end{matrix}\right.\)\(\Rightarrow x=1\)

Vậy...

b)Đk:\(x\ge2\)

Pt \(\Leftrightarrow\left(x-1\right)^2.\left(x^2-4\right)=\left(x-2\right)^2.\left(x^2-1\right)\)

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

Do \(x\ge2\Rightarrow x-1>0\)

Chia cả hai vế của pt cho x-1 ta được:

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

\(\Leftrightarrow\left(x-2\right)\left[\left(x-1\right)\left(x+2\right)-\left(x-2\right)\left(x-1\right)\right]=0\)

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

\(\Leftrightarrow\left(x-2\right)\left(4x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)

Vậy S={2}

c)Đk:\(\left\{{}\begin{matrix}9-x^2\ge0\\x^2-1\ge0\\x-3\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-3\le x\le3\\\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\\x\ge3\end{matrix}\right.\)\(\Rightarrow x=3\)

Thay x=3 vào pt thấy thỏa mãn

Vậy S={3}

4) Ta có: \(\left(x+3\right)\cdot\sqrt{10-x^2}=x^2-x-12\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\\sqrt{10-x^2}=x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\10-x^2=x^2-8x+16\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x^2-8x+16-10+x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\2x^2-8x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\2\left(x^2-4x+3\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\\left(x-1\right)\left(x-3\right)=0\end{matrix}\right.\)

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

Bài 1: ĐKXĐ: $2\leq x\leq 4$
PT $\Leftrightarrow (\sqrt{x-2}+\sqrt{4-x})^2=2$

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

$\Leftrightarrow x-2=0$ hoặc $4-x=0$

$\Leftrightarrow x=2$ hoặc $x=4$ (tm)

Bài 2:
PT $\Leftrightarrow 4x^3(x-1)-3x^2(x-1)+6x(x-1)-4(x-1)=0$

$\Leftrightarrow (x-1)(4x^3-3x^2+6x-4)=0$
$\Leftrightarrow x=1$ hoặc $4x^3-3x^2+6x-4=0$

Với $4x^3-3x^2+6x-4=0(*)$

Đặt $x=t+\frac{1}{4}$ thì pt $(*)$ trở thành:
$4t^3+\frac{21}{4}t-\frac{21}{8}=0$

Đặt $t=m-\frac{7}{16m}$ thì pt trở thành:

$4m^3-\frac{343}{1024m^3}-\frac{21}{8}=0$
$\Leftrightarrow 4096m^6-2688m^3-343=0$

Coi đây là pt bậc 2 ẩn $m^3$ và giải ta thu được \(m=\frac{\sqrt[3]{49}}{4}\) hoặc \(m=\frac{-\sqrt[3]{7}}{4}\)

Khi đó ta thu được \(x=\frac{1}{4}(1-\sqrt[3]{7}+\sqrt[3]{49})\)

Nãy mình tìm được một cách giải tương tự cho câu 2.

PT \(\Leftrightarrow\left(x-1\right)\left(4x^3-3x^2+6x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x^3-3x^2+6x-4=0\left(1\right)\end{matrix}\right.\)

Vậy pt có 1 nghiệm bằng 1.

\(\left(1\right)\Rightarrow8x^3-6x^2+12x-8=0\)

\(\Leftrightarrow7x^3+x^3-6x^2+12x-8=0\)

\(\Leftrightarrow\left(x-2\right)^3=-7x^3\)

\(\Leftrightarrow x-2=-\sqrt[3]{7}x\)

\(\Leftrightarrow x=\dfrac{2}{1+\sqrt[3]{7}}\)

Vậy pt có nghiệm \(S=\left\{1;\dfrac{2}{1+\sqrt[3]{7}}\right\}\)

Lưu ý: Nghiệm của người kia hoàn toàn tương đồng với nghiệm của mình (\(\dfrac{2}{1+\sqrt[3]{7}}=\dfrac{1}{4}\left(1-\sqrt[3]{7}+\sqrt[3]{49}\right)\))

c.

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

Đặt \(\sqrt{x^2+3}=t>0\)

\(\Rightarrow t^2-\left(3x+1\right)t+2x^2+2x=0\)

\(\Delta=\left(3x+1\right)^2-4\left(2x^2+2x\right)=\left(x-1\right)^2\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+3}=x+1\left(x\ge-1\right)\\\sqrt{x^2+3}=2x\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+3=x^2+2x+1\left(x\ge-1\right)\\x^2+3=4x^2\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow x=1\)

a.

Đề bài ko chính xác, pt này ko giải được

b.

ĐKXĐ: \(x\ge-\dfrac{7}{2}\)

\(2x+7-\left(2x+7\right)\sqrt{2x+7}+x^2+7x=0\)

Đặt \(\sqrt{2x+7}=t\ge0\)

\(\Rightarrow t^2-\left(2x+7\right)t+x^2+7x=0\)

\(\Delta=\left(2x+7\right)^2-4\left(x^2+7x\right)=49\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{2x+7-7}{2}=x\\t=\dfrac{2x+7+7}{2}=x+7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+7}=x\left(x\ge0\right)\\\sqrt{2x+7}=x+7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-7=0\left(x\ge0\right)\\x^2+12x+42=0\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow x=1+2\sqrt{2}\)

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

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

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

1.

ĐKXĐ: \(x< 5\)

\(\Leftrightarrow\sqrt{\dfrac{42}{5-x}}-3+\sqrt{\dfrac{60}{7-x}}-3=0\)

\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-9}{\sqrt{\dfrac{42}{5-x}}+3}+\dfrac{\dfrac{60}{7-x}-9}{\sqrt{\dfrac{60}{7-x}}+3}=0\)

\(\Leftrightarrow\dfrac{9x-3}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{9x-3}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}=0\)

\(\Leftrightarrow\left(9x-3\right)\left(\dfrac{1}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{1}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}\right)=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

b.

ĐKXĐ: \(x\ge2\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{x-2}-\sqrt{x+3}=0\end{matrix}\right.\)

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

\(\Rightarrow x=2\)

3.

ĐKXĐ: \(x\ge-1\)

\(x^2+x-12+12\left(\sqrt{x+1}-2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4\right)+\dfrac{12\left(x-3\right)}{\sqrt{x+1}+2}=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4+\dfrac{12}{\sqrt{x+1}+2}\right)=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)