Đặt \(A=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)\left(11-\sqrt{113}\right)....\left(11-\sqrt{104}\right)\)

\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-\sqrt{121}\right)....\left(11-\sqrt{104}\right)\)

\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-11\right)....\left(11-\sqrt{104}\right)\)

\(=0\)

Do đó biểu thức trên đầu bài bằng 0

bạn ơi, trong dãy này không có số \(\sqrt{121}\)đâu

\(A=2.\left|\left(-3\right)\right|^3+2.\left(-2\right)^2-4\left|\left(-2\right)^3\right|\)

\(=54+8-32=30\)

\(B=\left|\sqrt{2}-2\right|+\left|\sqrt{2}-3\right|=2-\sqrt{2}+3-\sqrt{2}\)

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

\(C=\left|3-\sqrt{3}\right|-\left|1+\sqrt{3}\right|=3-\sqrt{3}-1-\sqrt{3}\)

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

\(D=\left|5+\sqrt{6}\right|-\left|\sqrt{6}-5\right|=5+\sqrt{6}-5+\sqrt{6}\)

\(=2\sqrt{6}\)

\(E=\sqrt{15^2}-\sqrt{5^2}=15-5=10\)

`A=2sqrt{(-3)^6}+2sqrt{(-2)^4}-4sqrt{(-2)^6}=2|(-3)^3|+2|(-2)^2|-4|(-2)^3|=54+8-32=30` $\\$ `B=sqrt{(sqrt2-2)^2}+sqrt{(sqrt2-3)^2}=2-sqrt2+3-sqrt2=5-2sqrt2` $\\$ `C=sqrt{(3-sqrt3)^2}-sqrt{(1+sqrt3)^2}=3-sqrt3-sqrt3-1=2-2sqrt3` $\\$ `D=sqrt{(5+sqrt6)^2}-sqrt{(sqrt6-sqrt5)^2}=5+sqrt6-5+sqrt6=2sqrt6` $\\$ `E=sqrt{17^2-8^2}-sqrt{3^2+4^2}=sqrt{289-64}-sqrt{9+16}=sqrt(225)-sqrt{25}=15-5=10`

\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\\ =\left|2-\sqrt{5}\right|+\left|2\sqrt{2}-\sqrt{5}\right|\\ =\sqrt{5}-2+2\sqrt{2}-\sqrt{5}\\ =-2+\sqrt{2}\)

\(\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)}\\ =\left|\sqrt{7}-2\sqrt{2}\right|+\left|3-2\sqrt{2}\right|\\ =2\sqrt{2}-\sqrt{7}+3-2\sqrt{2}\\ =3-\sqrt{7}\)

\(\sqrt{\left(x-3\right)^2}\\ =\left|x-3\right|\\ =x-3\left(vì.x>3\right)\)

\(\sqrt{\left(1-x\right)^2}\\ =\left|1-x\right|\\ =x-1\left(vì.x>1\right)\)

\(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}\\ =\left|3a^2\right|\\ =3a^2\)

\(\sqrt{100a^2}\\ =\sqrt{\left(10a\right)^2}\\ =\left|10a\right|\\ =-10a\left(vì.a< 0\right)\)

Lời giải:

a. $=|2-\sqrt{5}|+|2\sqrt{2}-\sqrt{5}|$

$=(\sqrt{5}-2)+(2\sqrt{2}-\sqrt{5})=-2+2\sqrt{2}$
b. $=|\sqrt{7}-2\sqrt{2}|+|3-2\sqrt{2}|=2\sqrt{2}-\sqrt{7}+(3-2\sqrt{2})$

$=3-\sqrt{7}$

c.

$=|x-3|=x-3$
d.

$=|1-x|=x-1$

$=\sqrt{(3a^2)^2}=|3a^2|=3a^2$
e.

$=\sqrt{(10a)^2}=|10a|=-10a$

Câu 1,2,3 Ez quá rồi :3

Câu 4:

Tổng quát:

\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v

Câu 5 ko khác câu 4 lắm :v

Câu 5: 

Tổng quát:

\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v

Sao làm hổng ai bảo đú.n/g vậy :(((

\(S=\frac{-1+\sqrt{2}}{2-1}+\frac{-\sqrt{2}+\sqrt{3}}{3-2}+...+\frac{-\sqrt{99}+\sqrt{100}}{100-99}\)

\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-....-\sqrt{99}+\sqrt{100}\)

\(=-1+\sqrt{100}\)

\(\hept{\begin{cases}a=\left(x^2-x+1\right)^2\\b=x^2\end{cases}}\)

\(a^2-\left(b+1\right)a+b=0\Leftrightarrow\left(a-1\right)\left(a-b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=1\\a=b\end{cases}\Leftrightarrow}\orbr{\begin{cases}\left(x^2-x+1\right)^2=1\\\left(x^2-x+1\right)^2=x^2\end{cases}}\)(easy)

Nhận xét: \(\left[\sqrt{n^2}\right]=n\); \(\left[\sqrt{a}\right]=n-1\) với (n - 1)² < a < n²

=>  \(\left[\sqrt{1}\right]+\left[\sqrt{2}\right]+\left[\sqrt{3}\right]=1+1+1=1.3\)

\(\left[\sqrt{4}\right]+...+\left[\sqrt{8}\right]=2.5\)

\(\left[\sqrt{9}\right]+...+\sqrt{15}=3.7\)

\(\left[\sqrt{16}\right]+...+\left[\sqrt{24}\right]=4.9\)

Tương tự, nhóm các số có phần nguyên là 5; 6; 7; 8 ;9 ; 10

=> B = 1.3 + 2.5 + 3.7 + 4.9 + 5.11 + 6.13 + 7 .15 + 8.17 + 9.19 + 10.21

B = 825