nói chuyện với mk đi

nói làm j

mình ghi nhầm nên các bạn cứ hết hai phân số là một câu nhé ví dụ như \(\dfrac{-5}{8}\):\(\dfrac{15}{4}\)

\(a.\)

\(x+5=-10\)

\(\Rightarrow x=-10-5=-15\)

a ) x +5 = -10

x = -10 -5

x = - 15

b) x - ( - 10 ) = 5

x = 5+(-10)

x = -5

c) \(\left|x\right|\) -5 = 3

\(\left|x\right|=8\)

x ϵ { -8 ; 8 }

d) 15 - ( - x ) = 20

Không có số tự nhiên x nào mà 15 ( - x ) = 20

e ) \(\left|x-4\right|=3-\left(-7\right)\\ \left|x-4\right|=10\\ \left|x\right|=14\\ x\in\left\{\pm14\right\}\)

f ) \(\left|x+5\right|=10-\left(-20\right)\\ \left|x+5\right|=30\\ \left|x\right|=25\\ x\in\left\{\pm25\right\}\)

a) Đặt \(A=\frac{7^{15}}{1+7+7^2+...+7^{14}}\)

Đặt \(B=1+7+7^2+...+7^{14}\)

\(\Rightarrow7B=7+7^2+...+7^{15}\)

\(\Rightarrow7B-B=6B=7^{15}-1\)

\(\Rightarrow B=\frac{7^{15}-1}{6}\)

\(\Rightarrow A=\frac{7^{15}-1+1}{\frac{7^{15}-1}{6}}=\left(7^{15}-1\right).\frac{6}{7^{15}-1}+\frac{6}{7^{15}-1}=6+\frac{6}{7^{15}-1}\)

Tự làm tiếp nha

bạn giải nốt đi

b, Ta có:\(\dfrac{1+3+3^2+.....+3^{10}}{1+3+3^2+.....+3^9}\) \(=\dfrac{1}{1+3+3^2+...+3^9}+\dfrac{3+3^2+...+3^{10}}{1+3+3^2+...+3^9}\)\(=\dfrac{1}{1+3+3^2+...+3^9}+\dfrac{3.\left(1+3+3^2+...+3^9\right)}{1+3+3^2+...+3^9}\)

\(=\dfrac{1}{1+3+3^2+...+3^9}+3< 4\)

\(\Rightarrow\) \(\dfrac{1+3+3^2+...+3^{10}}{1+3+3^2+...+3^9}< 4\) \(\left(1\right)\)

Ta có :\(\dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}\)

\(=\dfrac{1}{1+5+5^2+...+5^9}+\dfrac{5+5^2+...+5^{10}}{1+5+5^2+....+5^9}\)

\(=\dfrac{1}{1+5+5^2+...+5^9}+\dfrac{5.\left(1+5+5^2+...+5^9\right)}{1+5+5^2+...+5^9}\)

\(=\dfrac{1}{1+5+5^2+...+5^9}+5>5\)

\(\Rightarrow\) \(\dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}>5\) \(\left(2\right)\)

Từ \(\left(1\right)và\left(2\right)\)

\(\Rightarrow\dfrac{1+3+3^2+...+3^{10}}{1+3+3^2+...+3^9}< \dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}\)

Vậy \(\dfrac{1+3+3^2+...+3^{10}}{1+3+3^2+...+3^9}< \dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}\)

a, Đặt \(A\)\(=\dfrac{7^{15}}{1+7+7^2+...+7^{14}}\)

\(\Rightarrow\) \(\dfrac{1}{A}\) \(=\dfrac{1+7+7^2+...+7^{14}}{7^{15}}=\dfrac{1}{7^{15}}+\dfrac{7}{7^{15}}+\dfrac{7^2}{7^{15}}+...+\dfrac{7^{14}}{7^{15}}\)

\(=\dfrac{1}{7^{15}}+\dfrac{1}{7^{14}}+\dfrac{1}{7^{13}}+....+\dfrac{1}{7}\)

Đặt \(B=\dfrac{9^{15}}{1+9+9^2+...+9^{14}}\)

\(\Rightarrow\dfrac{1}{B}=\dfrac{1+9+9^2+...+9^{14}}{9^{15}}=\dfrac{1}{9^{15}}+\dfrac{9}{9^{15}}+\dfrac{9^2}{9^{15}}+...+\dfrac{9^{14}}{9^{15}}\)

\(=\dfrac{1}{9^{15}}+\dfrac{1}{9^{14}}+\dfrac{1}{9^{13}}+...+\dfrac{1}{9}\)

Mà \(\dfrac{1}{7^{15}}>\dfrac{1}{9^{15}};\dfrac{1}{7^{14}}>\dfrac{1}{9^{14}};\dfrac{1}{7^{13}}>\dfrac{1}{9^{13}};....;\dfrac{1}{7}>\dfrac{1}{9}\)

\(\Rightarrow\dfrac{1}{A}>\dfrac{1}{B}\) \(\Rightarrow A< B\)

Vậy\(\dfrac{7^{15}}{1+7+7^2+...+7^{14}}>\dfrac{9^{15}}{1+9+9^2+....+9^{14}}\)

a, \(A=\frac{10^8+2}{10^8-1}=\frac{10^8-1+3}{10^8-1}=1+\frac{3}{10^8-1}\)

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

Vì \(\frac{3}{10^8-1}< \frac{3}{10^8-3}\Rightarrow1+\frac{3}{10^8-1}< 1+\frac{3}{10^8-3}\Rightarrow A< B\)

b, \(A=\frac{10^{15}+1}{10^{16}+1}\Rightarrow10A=\frac{10\left(10^{15}+1\right)}{10^{16}+1}=\frac{10^{16}+10}{10^{16}+1}=\frac{10^{16}+1+9}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)

\(B=\frac{10^{16}+1}{10^{17}+1}\Rightarrow10B=\frac{10\left(10^{16}+1\right)}{10^{17}+1}=\frac{10^{17}+10}{10^{17}+1}=\frac{10^{17}+1+9}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)

Vì \(\frac{9}{10^{16}+1}>\frac{9}{10^{17}+1}\Rightarrow1+\frac{9}{10^{16}+1}>1+\frac{9}{10^{17}+1}\Rightarrow10A>10B0\Rightarrow A>B\)

c, giống câu b

d, giống câu b

e, \(A=\frac{10^{15}+5}{10^{15}-7}=\frac{10^{15}-7+12}{10^{15}-7}=1+\frac{12}{10^{15}-7}\)

\(B=\frac{10^{16}+7}{10^{16}-5}=\frac{10^{16}-5+12}{10^6-5}=1+\frac{12}{10^6-5}\)

Vì \(\frac{12}{10^{15}-7}>\frac{12}{10^{16}-5}\Rightarrow1+\frac{12}{10^{15}-7}>1+\frac{12}{10^{16}-7}\Rightarrow A>B\)

f, \(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)

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

Vì \(\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\Rightarrow1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}\Rightarrow A< B\)

e, Ta có: 

\(A-B=\left(\frac{-7}{10^{2013}}+\frac{-15}{10^{2014}}\right)-\left(\frac{-15}{10^{2013}}+\frac{-7}{10^{2014}}\right)\)

\(=\frac{-7}{10^{2013}}+\frac{-15}{10^{2014}}-\frac{-15}{10^{2013}}-\frac{-7}{10^{2014}}\)

\(=\frac{8}{10^{2013}}-\frac{8}{10^{2014}}>0\)

Vậy A > B

Phần a;b;c;d;e;f liên quan tới

\(\frac{a}{b}< \frac{a+c}{b+c}\forall a< b\)          \(\frac{a}{b}>\frac{a+c}{b+c}\forall a>b\)      phép trừ thì ngược lại

Giải phần g

\(A=\frac{-7}{10^{2013}}+\frac{-7}{10^{2014}}+\frac{-8}{10^{2014}}\)

\(B=\frac{-7}{10^{2013}}+\frac{-8}{10^{2013}}+\frac{-7}{10^{2014}}\)

có đcB>A

k minh nha

a, -5/6 -x = 7/12 + -1/3
⇔-10/12 - 12x/12 = 7/12 + -4/12
⇒-10 - 12x = 7 - 4
⇔-12x = 7 - 4 +10
⇔-12x = 13
⇔x = -13/12
b, x+13/-15 = 1/3
⇔-(x+13)/15 = 5/15
⇒ -x - 13 = 5
⇔-x = 5 +13
⇔-x = 18
⇔x = -18
c,-15/x-1 = -3/5
⇔-75/(x-1).5 = -3.(x-1)/5.(x-1)
⇒-75 = -3x + 3
⇔3x = 3 + 75
⇔3x = 78
⇔x = 26
d, (1/2).x + -2/5 = 1/5
⇔5x/10 + -4/10 = 1/10
⇒5x - 4 = 1
⇔5x = 1 + 4
⇔5x = 5
⇔x = 1
e, (-2/3).x + 1/5 = 1/10
⇔-20x/30 + 6/30 = 3/30
⇒-20x + 6 = 3
⇔-20x = 3 - 6
⇔-20x = -3
⇔x = 3/20
f, 4/5 - (1/2).x = 1/10
⇔8/10 - 5x/10 = 1/10
⇒8 - 5x = 1
⇔-5x = 1 - 8
⇔-5x = -7
⇔x=7/5

A = 1 + 2 + 3+ ... + 100 + 101

Dãy số trên là dãy số cách đều có 101 số hạng

Tổng của dãy số trên là:

(101 + 1) x 101 : 2 = 5151

Vậy A = 5151

B = 1 + 3+ 5+...+ 125

Xét dãy số:

1; 3; 5;..; 125

Dãy số trên là dãy số cách đều với khoảng cách là:

3 - 1 = 2

Số số hạng của dãy số trên là:

(125 - 1) : 2+ 1 = 63

Tổng B là:

(125 + 1) x 63 : 2 = 3969

C = 5 + 10 + 15 + 305

C = (3 + 305) + (10 + 15)

C = 310 + 25

C = 335

nhớ like 5 sao nha

Trước hết ta hãy so sánh :

\(\dfrac{10^{100}+1}{10^{101}+1}\)với \(\dfrac{10^{100}+1}{10^{102}+1}\)

Ta có: Cả hai phân số trên cùng tử.

\(\Rightarrow\dfrac{10^{100}+1}{10^{101}+1}>\dfrac{10^{100}+1}{10^{102}+1}\)

Tiếp đó so sánh : \(\dfrac{10^{101}+1}{10^{102}+1}\)với \(1\)

Ta được: \(\dfrac{10^{101}+1}{10^{102}+1}< 1\)

Ta lại so sánh được:\(\dfrac{10^{100}+1}{10^{102}+1}< 1\) (*)

Từ (*) suy ra \(\dfrac{10^{100}+1}{10^{101}+1}< \dfrac{10^{101}+1}{10^{102}+2}< \dfrac{10^{101}+1}{10^{102}+1}< 1\Rightarrow\dfrac{10^{100}+1}{10^{101}+1}< \dfrac{10^{101}+1}{10^{102}+1}\)

Ngoài ra còn một cách như sau:

\(\dfrac{10^{101}+1}{10^{102}+1}=\dfrac{10^{\left(100+1\right)}+1}{10^{\left(101+1\right)}+1}=\dfrac{10}{10}.\dfrac{10^{100}+1}{10^{101}+1}>\dfrac{10^{100}+1}{10^{101}+1}\) hay B > A hay A < B

Bài 1:

d)

\(\dfrac{x+5}{95}+\dfrac{x+10}{90}+\dfrac{x+15}{85}+\dfrac{x+20}{80}=-4\)

\(\Leftrightarrow\dfrac{x+5}{95}+1+\dfrac{x+10}{90}+1+\dfrac{x+15}{85}+1+\dfrac{x+20}{80}+1=-4+1+1+1+1\)

\(\Leftrightarrow\dfrac{x+100}{95}+\dfrac{x+100}{90}+\dfrac{x+100}{85}+\dfrac{x+100}{80}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{95}+\dfrac{1}{90}+\dfrac{1}{85}+\dfrac{1}{80}\right)=0\)

\(\Leftrightarrow x+100=0\) ( vì: \(\dfrac{1}{95}+\dfrac{1}{90}+\dfrac{1}{85}+\dfrac{1}{80}\ne0\))

\(\Leftrightarrow x=-100\)

a)

6x+70 =570-440

6x+70 =130

6x =130-70

6x =60

x =60:6

x =10

b)^ là mũ nhé!

(2^x+1).2^x=1000+24

(2^x+1).2^x=1024

2^x.2.2^x=1024

2^x.2^x=1024:2=512

2.2^x=512

2^x=512:2=256

2^x=2^8

x=8

bài 2 chú ý:^ là mũ nhé!

b)

= (5^15.(18+7):5^17

= (5^15.25):5^17

= (5^15.5^2):5^17

= 5^17:5^17

= 1