a,\(\frac{11}{12}-\left(\frac{5}{42}-x\right)=\frac{15}{28}-\frac{11}{12}\)

\(\Leftrightarrow\frac{11}{12}-\frac{5}{42}+x=\frac{15}{28}-\frac{11}{12}\)

\(\Leftrightarrow x=\frac{15}{28}-\frac{11}{12}-\frac{11}{12}+\frac{5}{42}\)

\(\Leftrightarrow x=\left(\frac{15}{28}+\frac{5}{42}\right)-\left(\frac{11}{12}+\frac{11}{12}\right)\)

\(\Leftrightarrow x=\frac{55}{84}-\frac{11}{6}\)

\(\Leftrightarrow x=\frac{-33}{28}\)

b, \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

phan AVIET DAY so them mot it di

Ta có:

A = 1 + 3 + 5 + 7 +... + 101

A = \(\frac{102.51}{2}=2601\)

M = 16 - 18 + 20 - 22 + 24 - 26 + .. + 64 - 66 + 68

M = ( 16 - 18 ) + ( 20 - 22 ) + ( 24 - 26 ) + ... + ( 64 - 66 ) + 68

M = (- 2 + - 2 + -2  + ... + - 2 ) + 68

M = 25/2 . ( - 2 ) + 68

M = -25 + 68

M = 43

H = ( 1 + 2 + 3 +...+ 99 ) x ( 13 x 15 - 12 x 15 - 15 )

H = ( 1 + 2 + 3 +...+ 99 ) x { (13 - 12 - 1) x 15 }

H = ( 1 + 2 + 3 +...+ 99 ) x 0

H = 0

G = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + 10 - 11 - 12 + 13 + 14 - ... + 301 + 302

G = ( 1 + 2 ) + ( -3 - 4 ) + ( 5 + 6 ) + ( -7 - 8 ) + ( 9 + 10 ) + ( - 11 - 12 ) + ( 13 + 14 ) -...+ ( 301 + 302 )

G = ( 3 - 7 ) + ( 11 - 15 ) + ( 19 - 23 ) + 27 - ... + 603

G = -4 + - 4 + -4 + 27 - ... + 603

G = 75 x ( -4 ) + 603

G = -300 + 603

G = 303

2.

a) 1 + 2 + 3 + 4 +...+ 99 + 100 + 2 x X = 5052

= > \(\frac{100.101}{2}\)+ 2 x X = 5052

= > 5050 + 2 x X = 5052

= > 2X = 2

= > X = 1

bui thi huyen chi nào vậy

Ai đấy tôi ko biết

Trên đời có mỗi tui Là Bùi Thị Huyền Chi thoy

Ko có ng thứ 2 đâu

\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-3x-4x+12\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left[x\left(x-3\right)-4\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)

Vậy tập nghiệm của ohuowng trình là \(S=\left\{\dfrac{1}{3};3;4\right\}\)

nhiều quá

nếu bạn muốn có người giải thì cho ít ít thôi

nhiều quá

bn muốn gủi

thì gửi tùng

bài 1 thôi

MÌNH VẼ CHO BẠN 1 CÁCH:

MỒI LÀN GỬI 1 CÂU SẼ CÓ NGƯỜI TRẢ LỜI NGAY

NẾU NHIỀU CÂU THÌ 1 LÀN GỬI 2 CẦU

K MÌNH NHA

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

vì \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)

=> x+1 =0 => x =-1

a: \(\dfrac{x-6}{7}+\dfrac{x-7}{8}+\dfrac{x-8}{9}=\dfrac{x-9}{10}+\dfrac{x-10}{11}+\dfrac{x-11}{12}\)

\(\Leftrightarrow\left(\dfrac{x-6}{7}+1\right)+\left(\dfrac{x-7}{8}+1\right)+\left(\dfrac{x-8}{9}+1\right)=\left(\dfrac{x-9}{10}+1\right)+\left(\dfrac{x-10}{11}+1\right)+\left(\dfrac{x-11}{12}+1\right)\)

=>x+1=0

hay x=-1

c: |x-2|=13

=>x-2=13 hoặc x-2=-13

=>x=15 hoặc x=-11

d: \(\Leftrightarrow3\left|x-2\right|+4\left|x-2\right|=2-\dfrac{1}{3}=\dfrac{5}{3}\)

=>7|x-2|=5/3

=>|x-2|=5/21

=>x-2=5/21 hoặc x-2=-5/21

=>x=47/21 hoặc x=37/21

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

=> x + 1 = 0 ( vì 1/10 + 1/11 + 1/12 - 1/13 - 1/14  khac 0 )  

=> x = -1 

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

mà 1/10 > 1/13; 1/11>1/14

\(\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)

=> x + 1 = 0

x = -1

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right)\left(\frac{1}{13}+\frac{1}{14}\right)\)

\(\Rightarrow x+1=0\)( do \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\ne\frac{1}{13}+\frac{1}{14}\))

\(\Rightarrow x=-1\)