=Ta có : $16x^2-8x+5=0$

$=>(16x^2-8x+1)+4=0$

$=>(4x-1)^2+4=0$

$=>(4x-1)^2=-4$ (vô lý vì $(4x-1)^2 \geq 0$)

Vậy pt vô nghiệm.

Ta có : $3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)$

$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)$

$=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)$

$=(2^8-1)(2^8+1)(2^{16}+1)(2^{32}+1)$

$=(2^{16}-1)(2^{16}+1)(2^{32}+1)$

$=(2^{32}-1)(2^{32}+1)$

$=2^{64}-1$

a) $A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}$

$=>A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$

$=>A=(1+\dfrac{1}{3}+...+\dfrac{1}{99})-(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100})$

$=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100})-(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}.2)$

$=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100})-(1+\dfrac{1}{2}+...+\dfrac{1}{50})$

$=>A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}$

b) Ta có : $A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$

$=>A=(1-\dfrac{1}{2}+\dfrac{1}{3})-(\dfrac{1}{4}-\dfrac{1}{5})-...-(\dfrac{1}{98}-\dfrac{1}{99})-\dfrac{1}{100}$

$=>A<1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}$

Bài 1 :

Để $x\in Z$ thì $\dfrac{a+11}{a}\in Z$.

$=>a+11\vdots a$

$=>11\vdots a$

\(=>a\in\left\{-11;-1;1;11\right\}\)

A)=>x + 1/2011+ x + 1/2012 - x + 1/2013 - x + 1/2014 =0

<=>(x + 1) . ( 1/2011+ 1/2012-1/2013 - 1/2014) = 0

=>x + 1 = 0 (vì 1/2011+ 1/2012 - 1/2013 - 1/2014 khác 0)

=>x     =  -1

vậy x   =  -1

B)x-100/24+x-98/26+x-96/28=3

<=>x - 100/24 - 1 + x - 98/26 - 1 + x - 96/28 =0

<=>x - 124/24 + x - 124/26 + x - 124/28 = 0

<=>(x-124).( 1/24 + 1/26 + 1/28 ) = 0

mà 1/24 + 1/26 +1/28 khác 0

=>x - 124 = 0

<=>x       = 124 

$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$=(a^3+ab^2+ac^2-a^2b-abc-a^2c)+(ba^2+b^3+bc^2-ab^2-b^2c-abc)+(ca^2+cb^2+c^3-abc-bc^2-c^2a)$

$=a^3+b^3+c^3-3abc$

$1-4+7-10+...-X=-75$

$=>(1-4)+(7-10)+...+(X-3-X)=-75$

$=>-3+(-3)+...+(-3)=-75$

$=>-3.k=-75$ (k là số cặp, k thuộc N*)

$=>k=25$

$X=4+(25-1).6=148$

Ý bạn là $a\in Z$?

Để $x\in Z$ thì $\dfrac{a+11}{2}\in Z$

$=>a+11\vdots 2$

=> a chia 2 dư 1.

Vậy để $x\in Z$ thì a chia 2 dư 1 và $a\in Z$

xã hội

quần đảo

người lao động

quê hương
 

a) $3^8:3^6=3^{8-6}=3^2$

$19^7:19^3=19^{7-3}=19^4$

$2^{10}:8^3=2^{10}:(2^3)^3=2^{10}:2^9=2^{10-9}=2^1$

$12^7:6^7=(12:6)^7=2^7$

$27^5:81^3=(3^3)^5:(3^4)^3=3^{15}:3^{12}=3^{15-12}=3^3$

b) $10^6:10=10^{6-1}=10^5$

$5^8:25^2=5^8:(5^2)^2=5^8:5^4=5^{8-4}=5^4$

$4^9:64^2=4^9:(4^3)^2=4^9:4^6=4^{9-6}=4^3$

$2^25:32^4=2^{25}:(2^5)^4=2^{25}:2^{20}=2^{25-20}=2^5$

$18^3:9^3=(18:9)^3=2^3$