Làm

2/3 - 3/2 . ( x - 1/2 ) = 5/12

        -3/2 . ( x - 1/2 )  = 5/12 - 2/3

        - 3/2 . ( x - 1/2 ) = -1/4

                     x - 1/2   = -1/4 : -3/2

                    x - 1/2     = 1/6

                     x            = 1/6 + 1/2

                     x             = 2/3

HỌC TỐT

\(\frac{2}{3}-\frac{3}{2}.\left(x-\frac{1}{2}\right)=\frac{5}{12}\)

\(\Leftrightarrow\frac{3}{2}\left(x-\frac{1}{2}\right)=\frac{2}{3}-\frac{5}{12}\)

\(\Leftrightarrow\frac{3}{2}\left(x-\frac{1}{2}\right)=\frac{8}{12}-\frac{5}{12}\)

\(\Leftrightarrow\frac{3}{2}\left(x-\frac{1}{2}\right)=\frac{1}{4}\)

\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{4}:\frac{3}{2}\)

\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{4}.\frac{2}{3}\)

\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{6}\)

\(\Leftrightarrow x=\frac{1}{6}+\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{6}+\frac{3}{6}\)

\(\Leftrightarrow x=\frac{2}{3}\)

Vậy\(x=\frac{2}{3}\)

Linz

A = 3 + 3² + 3³ + 3⁴ + ... + 3¹⁰⁰

3A = 3( 3 + 3² + 3³ + 3⁴ + ... + 3¹⁰⁰ )

      = 3² + 3³ + 3⁴ + 3⁵ + ... + 3¹⁰¹

3A - A = ( 3² + 3³ + 3⁴ + 3⁵ + ... +¹⁰¹ ) - ( 3 + 3² + 3³ + 3⁴ + ... + 3¹⁰⁰ )

=> 2A = 3² + 3³ + 3⁴ + 3⁵ + ... +3¹⁰¹ - 3 - 3² - 3³ - 3⁴ - ... - 3¹⁰⁰

2A = 3¹⁰¹ - 3

2A + 3 = 3ⁿ 

=> 3¹⁰¹ + 3 - 3 = 3ⁿ

=> 3¹⁰¹ = 3ⁿ

=> n = 101

\(A=3+3^2+3^3+3^4+...+3^{100}\)\

\(\Rightarrow3A=3^2+3^3+3^4+3^5+...+3^{100}+3^{101}\)

\(3A-A=\left(3^2+3^3+3^4+...+3^{101}\right)-\left(3+3^2+3^3+...+3^{100}\right)\)

\(\Rightarrow2A=3^{101}-3\Leftrightarrow3^{101}-3+3=3^n\)

\(\Rightarrow3^{101}=3^n\Leftrightarrow n=101\) Vậy \(n=101\)

Ta có : \(\frac{1}{2^2}=\frac{1}{4}< \frac{1}{1.2}\)

\(\frac{1}{3^2}=\frac{1}{9}< \frac{1}{2.3}\)

\(\frac{1}{4^2}=\frac{1}{16}< \frac{1}{3.4}\)

......

\(\frac{1}{8^2}=\frac{1}{64}< \frac{1}{7.8}\)

Ta có : \(VP< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)

\(=1-\frac{1}{8}=\frac{7}{8}\)

Mà \(\frac{7}{8}< 1\)Nên \(B< 1\left(đpcm\right)\)

\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)

Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)

...

\(\frac{1}{8^2}=\frac{1}{8\cdot8}< \frac{1}{7\cdot8}\)

=> \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)

=> \(B< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)

=> \(B< \frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)

Lại có : \(\frac{7}{8}< 1\)

=> \(B< \frac{7}{8}< 1\Rightarrow B< 1\left(đpcm\right)\)

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

\(a)\)\(6-\left(15+15\right)=x-\left(15-6\right)\)

\(\Leftrightarrow6-30=x-9\)

\(\Leftrightarrow-24=x-9\)

\(\Leftrightarrow x=-24+9\)

\(\Leftrightarrow x=-15\)

Vậy\(x=-15\)

\(b)\)\(x+\frac{4}{12}=\frac{3}{-9}\)

\(\Leftrightarrow x+\frac{1}{3}=-\frac{1}{3}\)

\(\Leftrightarrow x=-\frac{1}{3}-\frac{1}{3}\)

\(\Leftrightarrow x=-\frac{2}{3}\)

Vậy\(x=\frac{-2}{3}\)

Linz

a, \(6-\left(15+15\right)=x-\left(15-6\right)\)

\(\Leftrightarrow6-30=x-9\Leftrightarrow15=x+30\Leftrightarrow x=-15\)

b, \(\frac{x+4}{12}=\frac{3}{-9}\)

\(\Leftrightarrow-9x-36=36\Leftrightarrow-9x=72\Leftrightarrow x=-8\)

A) VÌ NOM' VÀ MON LÀ 2 GÓC KỀ BÙ

NÊN NOM' +MON =180*

MÀ MON =130*

=>NOM'=180*-130*=50*

B)VÌ MON VÀ M'ON' LÀ 2 GÓC ĐỐI ĐỈNH 

NÊN MON =M'ON'

Làm

a) Ta có : góc yOx = 30° ; góc xOz = 90°

mà : góc yOx + góc xOz = góc yOz 

                 30° + 90°         = góc yOz

=> góc yOz = 120°

b) Ta có : Ot là tia phân giác góc zOx nên ta có : góc xOt = góc tOz = zOx :2

=> góc xOt = góc tOz = 90 : 2

=> góc xOt = góc tOz = 45°

Vậy góc xOt = 45° 

HỌC TỐT

x O y z tt

a. Ta có ; \(\widehat{yOz}=\widehat{xOz}-\widehat{xOy}\)

\(\Rightarrow\widehat{yOz}=90^o-30^o\)

\(\Rightarrow\widehat{yOz}=60^o\)

b.Vì Ot là phân giác  \(\widehat{yOz}\)nên \(\widehat{yOt}=\widehat{zOt}=\frac{\widehat{yOz}}{2}=\frac{60^o}{2}=30^o\)

Ta có ; \(\widehat{xOt}=\widehat{xOy}+\widehat{yOt}\)

\(\Rightarrow\widehat{xOt}=30^o+30^o\)

\(\Rightarrow\widehat{xOt}=60^o\)

c. Vì góc xOy = góc tOy = 30độ 

mà tia Oy nằm trong góc xOt nên Oy là tia pg góc xOt

Học tốt

\(a)\)\(6\frac{4}{5}-\left(1\frac{2}{3}+3\frac{4}{5}\right)\)

\(=6\frac{4}{5}-1\frac{2}{3}-3\frac{4}{5}\)

\(=\left(6\frac{4}{5}-3\frac{4}{5}\right)-1\frac{2}{3}\)

\(=\left(6+\frac{4}{5}-3-\frac{4}{5}\right)-\frac{5}{3}\)

\(=3-\frac{5}{3}\)

\(=\frac{9}{3}-\frac{5}{3}\)

\(=\frac{4}{3}\)

\(b)\)\(6\frac{5}{7}-(1\frac{3}{4}+2\frac{5}{7})\)

\(=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}\)

\(=\left(6\frac{5}{7}-2\frac{5}{7}\right)-1\frac{3}{4}\)

\(=\left(6+\frac{5}{7}-2-\frac{5}{7}\right)-\frac{7}{4}\)

\(=4-\frac{7}{4}\)

\(=\frac{16}{4}-\frac{7}{4}\)

\(=\frac{9}{4}\)

\(c)\)\(7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\)

\(=7\frac{5}{9}-2\frac{3}{4}-3\frac{5}{9}\)

\(=\left(7\frac{5}{9}-3\frac{5}{9}\right)-2\frac{3}{4}\)

\(=\left(7+\frac{5}{9}-3-\frac{5}{9}\right)-\frac{11}{4}\)

\(=4-\frac{11}{4}\)

\(=\frac{16}{4}-\frac{11}{4}\)

\(=\frac{5}{4}\)

P/s: Câu b bạn thiếu ngoặc.

Linz

 a) 6 và 4/5 - ( 1 và 2/3 + 3 và 4/5)

= 34/5- ( 5/3 + 19/5 )

= 34/5 - 5/3 - 19/5

= (34/5 - 9/5 ) - 5/3

=25/5 -5/3

= 5/1 - 5/3

= 15/3 - 5/3

=10/3

b) 6 và 5/7 - (1 và 3/4 + 2 và 5/7)

=47/7 - (7/4 + 19/7 )

= 47/7 -7/4 - 19/7

= (47/7 -19/7 ) - 7/4

= 28/7 -7/4

=4/1 - 7/4

= 16/4 - 7/4

= 9/4

c) 7 và 5/9 - ( 2 và 3/4 + 3 và 5/9 )

= 68/9 - ( 11/4 + 32/9 )

= 68/9 - 11/4 - 32/9

= (68/9 - 32/9 )- 11/4

= 36/9 -11/4

= 4/1 - 11/4

= 16/4 - 11/4

= 5/4

    Chúc bạn học tốt nha!!!!!

a, \(A=\frac{n+7}{n+2}=\frac{n+2+5}{n+2}=\frac{5}{n+2}\)

\(\Rightarrow n+2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Ta lập bảng 

b, \(B=\frac{n+5}{n-2}=\frac{n-2+7}{n-2}=\frac{7}{n-2}\)

\(\Rightarrow n-2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Ta lập bảng 

c, \(C=\frac{2n+13}{n+1}=\frac{2\left(n+1\right)+11}{n+1}=\frac{11}{n+1}\)

\(\Rightarrow n+1\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)

Ta lập bảng

d) Để D là số nguyên <=> \(\frac{3n+7}{2n+3}\)là số nguyên

<=> \(3n+7⋮2n+3\)

<=> 2(3n + 7) \(⋮\) 2n + 3

<=> 6n + 14 \(⋮\)2n + 3

<=> 3(2n + 3) + 5 \(⋮\)2n + 3

<=> 5 \(⋮\)2n + 3 (vì 3(2n + 3) \(⋮\)2n + 3)

<=> 2n + 3 \(\in\)Ư(5) = {1; -1; 5; -5}

Lập bảng:

Vậy ....