a) x = 2

b) x = 7.

c) x = 12

d) x = 45

e) x = 18.

f) x  = 10

a) x = 2.               

b) x = 7.

c) x= 12.

d)  x= 45.

e) x = 18.

f) x  = 10.

đáp số = 0 đó các bạn!

quá dễ!

100+100+200-600+400-90-64+50+4-80+20-6+3+3-50+10+200-90-90-20+123-100-20-3+50-49-1+60-30-30+741-700-40-1=108

chắc 100% luôn mik dùng máy tính mà mik nhanh nhất phải k mik đó các abnj ko được cặp bi đâu phải tính nếu đúng thì tích mik nha

mệt quá!

a) = (-15)+(-10)

= -25

b) = (-222)-100-80

= (-322)-80

= -402

c) = 229+219-401+12

= 448-401+12

=47+12

= 59

d) = 500-(-120)+18

= 620+18

= 638

A,-25

B,-402

C,59

D,638

A, -25

B, -402

C, 59

D, 638

a, Ta có: \(\frac{2001}{2002}=\frac{2002-1}{2002}=\frac{2002}{2002}-\frac{1}{2002}=1-\frac{1}{2002}\)

\(\frac{2000}{2001}=\frac{2001-1}{2001}=\frac{2001}{2001}-\frac{1}{2001}=1-\frac{1}{2001}\)

Vì \(\frac{1}{2002}< \frac{1}{2001}\Rightarrow1-\frac{1}{2002}>1-\frac{1}{2001}\Rightarrow\frac{2001}{2002}>\frac{2000}{2001}\)

b, Ta có: \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{81}\right)^7=\left(\frac{1}{3^4}\right)^7=\left(\frac{1}{3}\right)^{28}=\frac{1}{3^{28}}\)

\(\left(\frac{1}{243}\right)^6=\left(\frac{1}{3^5}\right)^6=\left(\frac{1}{3^5}\right)^6=\frac{1}{3^{30}}\)

Vì \(\frac{1}{3^{28}}>\frac{1}{3^{30}}\Rightarrow\left(\frac{1}{81}\right)^7>\left(\frac{1}{243}\right)^6\Rightarrow\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^6\)

c, Ta có: \(\left(\frac{3}{8}\right)^5=\frac{3^5}{\left(2^3\right)^5}=\frac{243}{2^{15}}>\frac{243}{3^{15}}>\frac{125}{3^{15}}=\frac{5^3}{\left(3^5\right)^3}=\frac{5^3}{243^3}=\left(\frac{5}{243}\right)^3\)

Vậy \(\left(\frac{3}{8}\right)^5>\left(\frac{5}{243}\right)^3\)

d, Ta có: \(\frac{2011}{2012}>\frac{2011}{2012+2013}\)

\(\frac{2012}{2013}>\frac{2012}{2012+2013}\)

\(\Rightarrow\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2012+2013}+\frac{2012}{2012+2013}=\frac{2011+2012}{2012+2013}\)

e, \(C=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=\frac{20^{10}-1}{20^{10}-1}+\frac{2}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)

\(D=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=\frac{20^{10}-3}{20^{10}-3}+\frac{2}{2^{10}-3}=1+\frac{2}{2^{10}-3}\)

Vì \(\frac{2}{10^{10}-1}< \frac{2}{10^{10}-3}\Rightarrow1+\frac{2}{10^{10}-1}< 1+\frac{2}{10^{10}-3}\Rightarrow C< D\)

g, \(G=\frac{10^{100}+2}{10^{100}-1}=\frac{10^{100}-1+3}{10^{100}-1}=\frac{10^{100}-1}{10^{100}-1}+\frac{3}{10^{100}-1}=1+\frac{3}{10^{100}-1}\)

\(H=\frac{10^8}{10^8-3}=\frac{10^8-3+3}{10^8-3}=\frac{10^8-3}{10^8-3}+\frac{3}{10^8-3}=1+\frac{3}{10^8-3}\)

Vì \(\frac{3}{10^{100}-1}< \frac{3}{10^8-3}\Rightarrow1+\frac{3}{10^{100}-1}< 1+\frac{3}{10^8-3}\Rightarrow G< H\)

h, Vì E < 1 nên:

\(E=\frac{98^{99}+1}{98^{89}+1}< \frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}=\frac{98\left(98^{98}+1\right)}{98\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}=F\)

Vậy E = F

có các đáp án A; C; E.

Nguyễn Khắc Vinh nói xạo quá

mới 9 tuổi

^{dễ nên tự làm nhé} bạn _{Nguyễn Quỳnh Nga}

biết thì giúp người ta đi lại còn lắm mồm

a) Ta có: 10 < x < 99

=> x \(\in\){11; 12; 13; 14; .....; 98}

=> A = {11; 12; 13; 14; ...; 98}

b) Ta có : x \(⋮\)4 => x \(\in\)B(4) = {0; 4; 8; 12; 16; ....;96}

Do 0 < x < 100 => x \(\in\){4; 8; 12; 16; ....; 96}

=> B = {4; 8; 12; 16; ...; 96}

c) Ta có: x + 5 = 10 => x = 10 - 5 = 5

=> C = {5}

d) Ta có: 2x + 5 = 17

=> 2x = 17 - 5

=> 2x = 12

=> x = 12 : 2 = 6

=> D = {6}