Multiplication of Polynomial

\(\left\{{}\begin{matrix}r-s=4\\t-s=9\end{matrix}\right.\Rightarrow\left(t-s\right)-\left(r-s\right)=t-r=9-4=5\)

We have :

\(r^2+s^2+t^2-rs-st-rt\)

\(=\dfrac{2r^2+2s^2+2t^2-2rs-2st-2rt}{2}\)

\(=\dfrac{\left(r^2+s^2-2rs\right)+\left(r^2+t^2-2rt\right)+\left(t^2+s^2-2ts\right)}{2}\)

\(=\dfrac{\left(r-s\right)^2+\left(t-r\right)^2+\left(t-s\right)^2}{2}\)

\(=\dfrac{4^2+5^2+9^2}{2}=\dfrac{122}{2}=61\)

Ans : 61

{r−s=4t−s=9⇒(t−s)−(r−s)=t−r=9−4=5

We have :

r2+s2+t2−rs−st−rt

=2r2+2s2+2t2−2rs−2st−2rt2

=(r2+s2−2rs)+(r2+t2−2rt)+(t2+s2−2ts)2

=(r−s)2+(t−r)2+(t−s)22

=42+52+922=1222=61

Ans : 61

You must check ur problem 

Let \(A=\left(6+1\right)\left(6^2+1\right)...\left(6^{16}+1\right)\)

​ ​\(\Rightarrow A=\dfrac{1}{5}.5.\left(6+1\right)\left(6^2+1\right)...\left(6^{16}+1\right)\)

​ ​\(\Rightarrow A=\dfrac{1}{5}.\left(6-1\right).\left(6+1\right)\left(6^2+1\right)...\left(6^{16}+1\right)\)

\(\Rightarrow A=\dfrac{1}{5}.\left[\left(6-1\right)\left(6+1\right)\right]\left(6^2+1\right)...\left(6^{16}+1\right)\)

\(\Rightarrow A=\dfrac{1}{5}.\left(6^2-1\right)\left(6^2+1\right)\left(6^4+1\right)...\left(6^{16}+1\right)\)

\(...\)

\(\Rightarrow A=\dfrac{1}{5}.\left(6^{32}-1\right)=\dfrac{6^{32}-1}{5}\)

Let A=(6+1)(62+1)...(616+1)

​ ​⇒A=15.5.(6+1)(62+1)...(616+1)

​ ​⇒A=15.(6−1).(6+1)(62+1)...(616+1)

⇒A=15.[(6−1)(6+1)](62+1)...(616+1)

⇒A=15.(62−1)(62+1)(64+1)...(616+1)

...

⇒A=15.(632−1)=632−15

We have :

\(x^2-5x+1=0\)

\(\Rightarrow x^2+1=5x\) ( 1 )

\(x^2+\dfrac{1}{x^2}=\left(x^2+\dfrac{1}{x^2}+2.\dfrac{1}{x}.x\right)-2.\dfrac{1}{x}.x=\left(x+\dfrac{1}{x}\right)^2-2\)

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

( 1 )( 2 )\(\Rightarrow x^2+\dfrac{1}{x^2}=\left(\dfrac{5x}{x}\right)^2-2=25-2=23\)

Ans : 23.

We have :

x2−5x+1=0

⇒x2+1=5x

 ( 1 )

x2+1x2=(x2+1x2+2.1x.x)−2.1x.x=(x+1x)2−2

                =(x2+1x)2−2

 ( 2 )

( 1 )( 2 )⇒x2+1x2=(5xx)2−2=25−2=23

Ans : 23.

(m+n)2=52

⇒m2+n2+2mn=25

⇒m2+n2=25−2mn=25−2.6=25−12=13

⇒m3+n3=(m+n)(m2+n2−mn)=5.(13−6)=35

Ans : 35.

\(\left(m+n\right)^2=5^2\)

\(\Rightarrow m^2+n^2+2mn=25\)

\(\Rightarrow m^2+n^2=25-2mn=25-2.6=25-12=13\)

\(\Rightarrow m^3+n^3=\left(m+n\right)\left(m^2+n^2-mn\right)=5.\left(13-6\right)=35\)

Ans : 35.

The area of the square is S².

The area of the rectangle is \((S+3)(S-3)=S^2 - 9\)

So, the sum of the areas of 2 fingures is \(S^2+\left(S^2-9\right)=2S^2-9\).

The area of the square is S2.

The area of the rectangle is (S+3)(S−3)=S2−9

So, the sum of the areas of 2 fingures is S2+(S2−9)=2S2−9

.

\(=\left(1^2-2^2\right)+\left(3^2-4^2\right)+\left(5^2-6^2\right)+\left(1999^2-2000^2\right)\)

\(=\left(-3\right)+\left(-7\right)+\left(-11\right)+...+\left(-3999\right)\)

The number of  terms is \(\dfrac{\left(-3\right)-\left(-3999\right)}{4}+1=1000\) ( terms )

So the value of this expression is \(\dfrac{1000.\left[\left(-3\right)+\left(-3999\right)\right]}{2}=-2001000\)

Let m2 and n2 be a + 22 and a - 23 respectively. ( m > n > 0)

So, we have m2−n2=(a+22)−(a−23)

⇒(m−n)(m+n)=45=1.45=3.15=5.9

⋅

 If m - n =  1 and m + n = 45; then m = 23 ; n = 22; so a=m2−22=n2+23=507

⋅

 If m - n = 3 and m + n = 15; then m = 9 and n = 6, so a = 59.

⋅

 If m - n = 5 and m + n = 9; then m = 7 and n = 2; so a = 27.

Ans : a∈{27;59;507}.

Let m² and n² be a + 22 and a - 23 respectively. ( m > n > 0)

So, we have \(m^2-n^2=\left(a+22\right)-\left(a-23\right)\)

\(\Rightarrow\left(m-n\right)\left(m+n\right)=45=1.45=3.15=5.9\)

\(\cdot\) If m - n =  1 and m + n = 45; then m = 23 ; n = 22; so \(a=m^2-22=n^2+23=507\)

\(\cdot\) If m - n = 3 and m + n = 15; then m = 9 and n = 6, so a = 59.

\(\cdot\) If m - n = 5 and m + n = 9; then m = 7 and n = 2; so a = 27.

Ans : \(a\in\left\{27;59;507\right\}.\)

2002^2/2001^2+2003^2-2

2002^2/ (2002-1)^2+(2002+1)^2-2

 Have: a^2 - ( a-1)^2 = a+(a-1)

2002^2/ ( 2002^2-4003) + ( 2002^2 +4005 ) -2

2002^2/2002^2-4003+2002^2+4005-2

2002^2/2002^2+2002^2-4003+4005-2

=2002^2/2002^2+2002^2

= 2002^2/ 2002^2.2 

= 1/2

Answer: 1/2

2002^2/2001^2+2003^2-2

2002^2/ (2002-1)^2+(2002+1)^2-2

 Have: a^2 - ( a-1)^2 = a+(a-1)

2002^2/ ( 2002^2-4003) + ( 2002^2 +4005 ) -2

2002^2/2002^2-4003+2002^2+4005-2

2002^2/2002^2+2002^2-4003+4005-2

=2002^2/2002^2+2002^2

= 2002^2/ 2002^2.2 

= 1/2

Answer: 1/2

Ans :  \(x^{n+1}-1\)

a2+a=a(a+1)=0

⇔[a=0a=−1

⇔[a2001+a2000+7=0+0+7=7a2001+a2000+7=(−1)+1+7=7

Ans : 7.

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

\(\Leftrightarrow\left[{}\begin{matrix}a=0\\a=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a^{2001}+a^{2000}+7=0+0+7=7\\a^{2001}+a^{2000}+7=\left(-1\right)+1+7=7\end{matrix}\right.\)

Ans : 7.

(1−122)(1−132)...(1−1112)(1−1122)

=(1−12)(1+12)(1−13)(1+13)...(1−111)(1+111)(1−112)(1+112)

=(12.32)(23.43)....(1011.1211)(1112.1312)

=(12.23...1011.1112)(32.43...1211.1312)

=112.132=1324

\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{11^2}\right)\left(1-\dfrac{1}{12^2}\right)\)

\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)...\left(1-\dfrac{1}{11}\right)\left(1+\dfrac{1}{11}\right)\left(1-\dfrac{1}{12}\right)\left(1+\dfrac{1}{12}\right)\)

\(=\left(\dfrac{1}{2}.\dfrac{3}{2}\right)\left(\dfrac{2}{3}.\dfrac{4}{3}\right)....\left(\dfrac{10}{11}.\dfrac{12}{11}\right)\left(\dfrac{11}{12}.\dfrac{13}{12}\right)\)

\(=\left(\dfrac{1}{2}.\dfrac{2}{3}...\dfrac{10}{11}.\dfrac{11}{12}\right)\left(\dfrac{3}{2}.\dfrac{4}{3}...\dfrac{12}{11}.\dfrac{13}{12}\right)\)

\(=\dfrac{1}{12}.\dfrac{13}{2}=\dfrac{13}{24}\)

(1−122)(1−132)...(1−1112)(1−1122)

=(1−12)(1+12)(1−13)(1+13)...(1−111)(1+111)(1−112)(1+112)

=(12.32)(23.43)....(1011.1211)(1112.1312)

=(12.23...1011.1112)(32.43...1211.1312)

=112.132=1324

We have: 234567² - 234557 . 234577

= 234567² - (234567 - 10) . (234567 + 10)

=234567² - (234567² - 10²)

=10² =100

We have :

234567² - 234557.234577

= 234567² - (234567 - 10)(234567 + 10) 

= 234567² - (234567² - 10²)

= 10² = 100

The units digit of n2 is 9,so the unit digit of n can be 3 or 7 (because 32 = 9 ; 72 = 49)

The units digit of (n + 1)2 is 6,so the unit digit of n + 1 can be 4 or 6

We see 3 + 1 = 4,so the units digit of n is 3 and the units digit of n - 1 is 2.Hence the units digit of (n - 1)2 is 4

The units digit of n2 is 9,so the unit digit of n can be 3 or 7 (because 32 = 9 ; 72 = 49)

The units digit of (n + 1)2 is 6,so the unit digit of n + 1 can be 4 or 6

We see 3 + 1 = 4,so the units digit of n is 3 and the units digit of n - 1 is 2.Hence the units digit of (n - 1)2 is 4

The units digit of n² is 9,so the unit digit of n can be 3 or 7 (because 3² = 9 ; 7² = 49)

The units digit of (n + 1)² is 6,so the unit digit of n + 1 can be 4 or 6

We see 3 + 1 = 4,so the units digit of n is 3 and the units digit of n - 1 is 2.Hence the units digit of (n - 1)² is 4

S = (2 + 1)(2² + 1)(2⁴ + 1)...(2²⁰ + 1)

   = (2 - 1)(2 + 1)(2² + 1)(2⁴ + 1)...(2²⁰ + 1)

   = (2² - 1)(2² + 1)(2⁴ + 1)...(2²⁰ + 1)

   = (2⁴ - 1)(2⁴ + 1)...(2²⁰ + 1)

..........................................................

   = (2²⁰ - 1)(2²⁰ + 1)

   = 2⁴⁰ - 1

S = (2 + 1)(22 + 1)(24 + 1)...(220 + 1)

   = (2 - 1)(2 + 1)(22 + 1)(24 + 1)...(220 + 1)

   = (22 - 1)(22 + 1)(24 + 1)...(220 + 1)

   = (24 - 1)(24 + 1)...(220 + 1)

..........................................................

   = (220 - 1)(220 + 1)

   = 240 - 1

S = (2 + 1)(22 + 1)(24 + 1)...(220 + 1)

   = (2 - 1)(2 + 1)(22 + 1)(24 + 1)...(220 + 1)

   = (22 - 1)(22 + 1)(24 + 1)...(220 + 1)

   = (24 - 1)(24 + 1)...(220 + 1)

   = (220 - 1)(220 + 1)

   = 240 - 1

We have :

m4+1m4=m4+2+1m4−2=(m2+1m2)2−2

=(m2+2+1m2−2)2−2=[(m+1m)2−2]2−2

=(42−2)2−2=142−2=194

We have :

\(m^4+\dfrac{1}{m^4}=m^4+2+\dfrac{1}{m^4}-2=\left(m^2+\dfrac{1}{m^2}\right)^2-2\)

\(=\left(m^2+2+\dfrac{1}{m^2}-2\right)^2-2=\left[\left(m+\dfrac{1}{m}\right)^2-2\right]^2-2\)

\(=\left(4^2-2\right)^2-2=14^2-2=194\)