a) ( x² - 3x )( x² + 7x + 10 ) = 216

<=> x( x - 3 )( x + 2 )( x + 5 ) - 216 = 0

<=> [ x( x + 2 ) ][ ( x - 3 )( x + 5 ) ] - 216 = 0

<=> ( x² + 2x )( x² + 2x - 15 ) - 216 = 0 (1)

Đặt a = x² + 2x

(1) trở thành a( a - 15 ) - 216 = 0 <=> a² - 15a - 216 = 0 <=> ( a - 24 )( a + 9 ) = 0 <=> a = 24 hoặc a = -9

=> \(\orbr{\begin{cases}x^2+2x=24\\x^2+2x=-9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2+2x-24=0\\x^2+2x+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(x-4\right)\left(x+6\right)=0\\\left(x+1\right)^2+8>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-6\end{cases}}\)

Vậy S = { 4 ; -6 }

b) ( 2x² - 7x + 3 )( 2x² + x - 3 ) + 9 = 0

<=> ( x - 3 )( 2x - 1 )( x - 1 )( 2x + 3 ) + 9 = 0

<=> [ ( x - 3 )( 2x + 3 ) ][ ( 2x - 1 )( x - 1 ) ] + 9 = 0

<=> ( 2x² - 3x - 9 )( 2x² - 3x + 1 ) + 9 = 0

<=> ( 2x² - 3x - 4 - 5 )( 2x² - 3x - 4 + 5 ) + 9 = 0

<=> ( 2x² - 3x - 4 )² - 16 = 0

<=> x( 2x - 3 )( 2x² - 3x - 8 ) = 0

<=> x = 0 hoặc 2x - 3 = 0 hoặc 2x² - 3x - 8 = 0

<=> x = 0 hoặc x = 3/2 hoặc x = \(\frac{3\pm\sqrt{73}}{4}\)

Vậy S = { 0 ; 3/2 ; \(\frac{3\pm\sqrt{73}}{4}\)}

phải là (\(\frac{3}{\sqrt{1+a}}\)+\(\sqrt{1+a}\)): (\(\frac{3}{\sqrt{1-a^2}}\)+1) nha! LÀ dấu CỘNG , KO PHẢI TRỪ:>

F = [3/(√1 + a) + (√1 - a)] : [3/(√1 - a^2) + 1] .

=[3+√1-a^2)/√1+a]:[(3+√1-a^2/√1-a^2] 

=(3+√1-a^2/√1+a].[√1-a^2/(3+√1-a^2]

=√1-a.√1+a/√1+a=√1-a

thay a=√3/(2+√3) vào F ta được

√[1-(√3/(2+√3)]=√2/(2+√3)

=√2/(2+√3)=(√4-2√3)/4-3=√(√3-1)^2=|√3-1}

=√3-1

Ta có: \(\sqrt{x-7}\le\frac{x-7+1}{2}=\frac{x-6}{2}\)(bđt cosi)

 \(\sqrt{9-x}\le\frac{9-x+1}{2}=\frac{10-x}{2}\)

=> \(VT=\sqrt{x-7}+\sqrt{9-x}\le\frac{x-6}{2}+\frac{10-x}{2}=\frac{x-6+10-x}{2}=2\)

\(VP=x^2-16x+66=\left(x-8\right)^2+2\ge2\)

=> \(VT=VP\Leftrightarrow\hept{\begin{cases}x-7=1\\9-x=1\\x-8=0\end{cases}}\) <=> x = 8

Vậy S = {8}

\(\sqrt{x-7}+\sqrt{9-x}=x^2-16x+66\left(7\le x\le9\right)\)

Đặt \(A=\sqrt{x-7}+\sqrt{9-x}\)

\(\Rightarrow A^2=\left(\sqrt{x-7}+\sqrt{9-x}\right)^2\)

Áp dụng bất đảng thức Bunhiacopxki ta có:

\(\left(\sqrt{x-7}+\sqrt{9-x}\right)^2\le\left(x-7+9-x\right)\left(1+1\right)=4\)

=> \(A\le2\)

Ta có: \(x^2-16x+66=\left(x-8\right)^2+2\ge2\)

Dấu = xảy ra

\(\Leftrightarrow\hept{\begin{cases}\frac{\sqrt{x-7}}{1}=\frac{\sqrt{9-x}}{1}\\x-8=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-7}=\sqrt{9-x}\\x=8\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-7=9-x\\x=8\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=8\\x=8\end{cases}\left(tm\right)}\)

Vậy x = 8

Đk: \(\orbr{\begin{cases}x\ge1\\x\le-3\end{cases}}\)

Ta có: \(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)

VT = \(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}\ge0\) => >VP = 2x + 2 \(\ge\)0 => x \(\ge\)-1

mà \(\orbr{\begin{cases}x\ge1\\x\le-3\end{cases}}\) => \(x\ge1\)

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

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

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

<=> \(2x+6+x-1+2\sqrt{2\left(x-1\right)\left(x+3\right)}=4x+4\)

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

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

<=> \(\orbr{\begin{cases}\sqrt{x-1}=0\\2\sqrt{2\left(x+3\right)}=\sqrt{x-1}\end{cases}}\)

<=> \(\orbr{\begin{cases}x=1\left(tm\right)\\8x+24=x-1\end{cases}}\)

<=> \(\orbr{\begin{cases}x=1\\x=-\frac{25}{7}\left(ktm\right)\end{cases}}\)

Vậy S = {1}

\(\sqrt{3+x}+\sqrt{6-x}-\sqrt{\left(3+x\right)\left(6-x\right)}=3\)(ĐK: \(-3\le x\le6\))

Đặt \(t=\sqrt{3+x}+\sqrt{6-x}\ge0\)

\(\Leftrightarrow t^2=3+x+6-x+2\sqrt{\left(3+x\right)\left(6-x\right)}\)

\(\Leftrightarrow\sqrt{\left(3+x\right)\left(6-x\right)}=\frac{t^2-9}{2}\)

Phương trình ban đầu tương đương với: 

\(t-\frac{t^2-9}{2}=3\)

\(\Leftrightarrow\orbr{\begin{cases}t=3\left(tm\right)\\t=-1\left(l\right)\end{cases}}\)

Với \(t=3\):

\(\sqrt{3+x}+\sqrt{6-x}=3\)

\(\Leftrightarrow9+2\sqrt{\left(3+x\right)\left(6-x\right)}=9\)

\(\Leftrightarrow\left(3+x\right)\left(6-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-3\left(tm\right)\\x=6\left(tm\right)\end{cases}}\)

\(\sqrt{2x+1}-\sqrt{2-x}=x+3\)(ĐK: \(-\frac{1}{2}\le x\le2\))

Ta có: \(\sqrt{2x+1}-\sqrt{2-x}\le\sqrt{2x+1}\le\sqrt{2.2+1}=\sqrt{5}\)(vì \(-\frac{1}{2}\le x\le2\))

\(x+3\ge-\frac{1}{2}+3=2,5\)(vì \(-\frac{1}{2}\le x\le2\))

mà \(\sqrt{5}< 2,5\)

do đó phương trình vô nghiệm. 

ĐK: \(1\le x\le2\).

Với \(x=1\)thì \(\sqrt{x-1}-\sqrt{2-x}=-1\)không thỏa. 

Với \(x=2\)thì \(\sqrt{x-1}-\sqrt{2-x}=1\)thỏa. 

Với \(1< x< 2\)thì \(0< x-1< 1\Leftrightarrow0< \sqrt{x-1}< 1\)

\(0< 2-x< 1\Leftrightarrow0< \sqrt{2-x}< 1\)

Do đó \(\sqrt{x-1}-\sqrt{2-x}< 1-0=1\).

Vậy phương trình có nghiệm duy nhất \(x=2\).

ĐK : 1 ≤ x ≤ 2

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

\(\Leftrightarrow\frac{x-1-1}{\sqrt{x-1}+1}-\frac{2-x}{\sqrt{2-x}}=0\)

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

Dễ thấy với 1 ≤ x ≤ 2 thì \(\frac{1}{\sqrt{x-1}+1}+\frac{1}{\sqrt{2-x}}\)vô nghiệm

nên (1) <=> x - 2 = 0 <=> x = 2 (tm)

\(1,\sqrt{2}\left(3-2\sqrt{9}+2\sqrt{16}-\sqrt{25}\right)=\sqrt{2}\left(3-6+8-5\right)\)

\(=\sqrt{2}.0=0\)

\(2,\sqrt{2}\left(\sqrt{25}-\sqrt{9}+\sqrt{100}-\sqrt{81}\right)\)

\(=\sqrt{2}\left(5-3+10-9\right)=3\sqrt{2}\)

\(3,\sqrt{5}\left(5+\sqrt{4}-3\sqrt{9}\right)=\sqrt{5}\left(5+2-9\right)=-2\sqrt{5}\)

\(4,\sqrt{3}\left(5\sqrt{16}-4\sqrt{9}-2\sqrt{25}+\sqrt{36}\right)\)

\(=\sqrt{3}\left(20-12-10+6\right)=4\sqrt{3}\)

\(5,\sqrt{12}-\sqrt{300}-\sqrt{3}+\frac{10\sqrt{3}}{3}\)

\(\sqrt{3}\left(\sqrt{4}-\sqrt{100}-1+\frac{10}{3}\right)=\sqrt{3}\left(2-10-1+\frac{10}{3}\right)=-\frac{17\sqrt{3}}{3}\)

\(6,\sqrt{3}\left(3\sqrt{4}-4\sqrt{9}+5\sqrt{16}\right)=\sqrt{3}\left(6-12+20\right)\)

\(=14\sqrt{3}\)

\(7,\sqrt{3}\left(2+5-4\right)=3\sqrt{3}\)

\(8,\sqrt{2}\left(8+8-15\right)=\sqrt{2}\)

\(9,\sqrt{5}\left(6-6+4\right)=4\sqrt{5}\)

\(10,\sqrt{6}\left(3-6+3-5\right)=-5\sqrt{6}\)

11, \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)

\(=2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)

\(=6\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)

\(=8\sqrt{2}\)

12, \(3\sqrt{8}-4\sqrt{18}+5\sqrt{32}-\sqrt{50}\)

\(=3.2\sqrt{2}-4.3\sqrt{2}+5.4\sqrt{2}-5\sqrt{2}\)

\(=6\sqrt{2}-12\sqrt{2}+20\sqrt{2}-5\sqrt{2}\)

\(=9\sqrt{2}\)

13, \(\sqrt{125}-2\sqrt{20}-3\sqrt{80}+4\sqrt{45}\)

\(=5\sqrt{5}-2.2\sqrt{5}-3.4\sqrt{5}+4.3\sqrt{5}\)

\(=5\sqrt{5}-4\sqrt{5}-12\sqrt{5}+12\sqrt{5}\)

\(=\sqrt{5}\)

14, \(2\sqrt{28}+2\sqrt{63}-3\sqrt{175}+\sqrt{112}\)

\(=2.2\sqrt{7}+2.3\sqrt{7}-3.5\sqrt{7}+4\sqrt{7}\)

\(=4\sqrt{7}+6\sqrt{7}-15\sqrt{7}+4\sqrt{7}\)

\(=-\sqrt{7}\)

15, \(3\sqrt{2}+\sqrt{8}+\frac{1}{2}\sqrt{50}-\sqrt{32}\)

\(=3\sqrt{2}+2\sqrt{2}+\frac{1}{2}.5\sqrt{2}-4\sqrt{2}\)

\(=3\sqrt{2}+2\sqrt{2}+\frac{5}{2}\sqrt{2}-4\sqrt{2}\)

\(=\frac{7}{2}\sqrt{2}\)

16, \(3\sqrt{50}-2\sqrt{12}-\sqrt{18}+\sqrt{75}-\sqrt{8}\)

\(=3.5\sqrt{2}-2.2\sqrt{3}-3\sqrt{2}+5\sqrt{3}-2\sqrt{2}\)

\(=15\sqrt{2}-4\sqrt{3}-3\sqrt{2}+5\sqrt{3}-2\sqrt{2}\)

\(=10\sqrt{2}+\sqrt{3}\)

17, \(2\sqrt{75}-3\sqrt{12}+\sqrt{27}\)

\(=2.5\sqrt{3}-3.2\sqrt{3}+3\sqrt{3}\)

\(=10\sqrt{3}-6\sqrt{3}+3\sqrt{3}\)

\(=7\sqrt{3}\)

18, \(\sqrt{12}+\sqrt{75}-\sqrt{27}\)

\(=2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)

\(=4\sqrt{3}\)

19, \(\sqrt{27}-\sqrt{12}+\sqrt{75}+\sqrt{147}\)

\(=3\sqrt{3}-2\sqrt{3}+5\sqrt{3}+7\sqrt{3}\)

\(=13\sqrt{3}\)

20, \(2\sqrt{3}+\sqrt{48}-\sqrt{75}-\sqrt{243}\)

\(=2\sqrt{3}+4\sqrt{3}-5\sqrt{3}-9\sqrt{3}\)

\(=-8\sqrt{3}\)

21, \(6\sqrt{\frac{8}{9}}-5\sqrt{\frac{32}{25}}+14\sqrt{\frac{18}{49}}\)

\(=6.\frac{2\sqrt{2}}{3}-5.\frac{4\sqrt{2}}{5}+14.\frac{3\sqrt{2}}{7}\)

\(=4\sqrt{2}-4\sqrt{2}+6\sqrt{2}\)

\(=6\sqrt{2}\)

Điều kiện xác định: \(x\ge0\).

Lấy \(x_1>x_2\ge0\).

\(f\left(x_1\right)-f\left(x_2\right)=\sqrt{x_1}-\sqrt{x_2}=\frac{x_1-x_2}{\sqrt{x_1}+\sqrt{x_2}}>0\)

Do đó hàm số đồng biến. 

Lần lượt thế tọa độ các điểm vào hàm số ban đầu, ta thấy điểm \(C\left(9,3\right)\)thỏa mãn nên nó thuộc đồ thị của hàm số đã cho, các điểm khác không thuộc.