a ) \(\sqrt{40+2}=8,32455532\)

\(\sqrt{40}+\sqrt{2}=7,738768883\)

Vì \(8>7\)

\(\Leftrightarrow\sqrt{40+2}>\sqrt{40}+\sqrt{2}\)

b ) \(\sqrt{26}+\sqrt{17}=9,222125139\)

Vì \(9,2...>9\)

\(\Leftrightarrow\sqrt{26}+\sqrt{17}>9\)

bình từng cái lên rồi so sánh :))

ta có : 

(2^9+2^7+1)(2^23-2^21+2^19-2^17+2^14-2... 

=2^32+(2^23+2^23-2^24)+(2^18-2^17-2^17... khi rút gọn) 

=2^32+1 

=>2^32+1 là hợp số 

chú ý biểu thức trong ngoặc đều bằng 0

=> 2^32 + 1 ko phải là số nguyên tố

t i c k nha!! 645645756785689889087070874575567856

Ta có

(2^9+2^7+1)(2^23-2^21+2^19-2^17+2^14-2)
=2^32+(2^23+2^23-2^24)+(2^18-2^17-2^17)
=2^32+1 
=>2^32+1 là hợp số 

Làm bừa thôi

Mình chƯa học âm lấy bé trừ to

20000-2888888888+2999999999=111131111

TH1: \(6-x=0\)

\(\Rightarrow x=6-0=6\)

TH2: \(6-x\ne0\)

\(\Rightarrow x=\frac{\left(6-x\right)^{2003}}{\left(6-x\right)^{2003}}=1\)

Vậy \(\orbr{\begin{cases}x=6\\x=1\end{cases}}\)

x = 6 và x = 1

t i c k nhé!!!5746756857876698796785687987698796867

\(ab=\frac{a}{b}\Rightarrow b^2=\frac{a}{a}\Rightarrow b^2=1\)

=>\(\hept{\begin{cases}b=1\\b=-1\end{cases}}\)

+) b=1

=>a-1=a

=>-1=0 (vô li)

+) b=-1

=>a-(-1)=a(-1)

=>a+1=-a

=>2a=-1

=\(a=\frac{-1}{2}\)

vậy...............

Ta có: 

1/1.2 + 1/3.4 + 1/5.6 + ... + 1/49.50

= 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... + 1/49 - 1/50

= (1 + 1/3 + 1/5 + .... + 1/49) - (1/2 + 1/4 + 1/6 + .... + 1/50)

= (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/49 + 1/50) - 2.(1/2 + 1/4 + 1/6 + ... + 1/50)

= (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/49 + 1/50) - (1 + 1/2 + 1/3 + ... + 1/25)

= 1/26 + 1/27 + 1/28 + ... + 1/50

=> đpcm

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)

\(A=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)

\(A=\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)\)

\(A< \left(\frac{1}{25}+\frac{1}{25}+...+\frac{1}{25}\right)+\left(\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)+\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)\)

                5 phân số 1/25                                 10 phân số 1/30                                 10 phân số 1/40

\(A< 5.\frac{1}{25}+10.\frac{1}{30}+10.\frac{1}{40}\)

\(A< \frac{1}{5}+\frac{1}{3}+\frac{1}{4}\)

\(A< \frac{1}{4}+\frac{1}{3}+\frac{1}{4}\)

\(A< \frac{1}{2}+\frac{1}{3}\)

\(A< \frac{5}{6}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\frac{5}{6}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\frac{5}{6}+\left(\frac{1}{5}-\frac{1}{4}\right)+\left(\frac{1}{7}-\frac{1}{6}\right)+...+\left(\frac{1}{49}-\frac{1}{48}\right)-\frac{1}{50}\)

\(\frac{1}{5}-\frac{1}{4}< 0\)

\(\frac{1}{7}-\frac{1}{6}< 0\)

\(...\)

\(\frac{1}{49}-\frac{1}{48}< 0\)

\(\frac{5}{6}\) khi cộng với các số nhỏ hơn 0 thì giá trị nó sẽ giảm, đồng thời còn bớt đi \(\frac{1}{50}\)

Do đó \(A< \frac{5}{6}\)

200000+20000+200000+2000000.2000000=4 840 000 000 000

200000+20000+200000+2000000.2000000=20000002420000