sẽ gầy

a: Theo bảng tần số, ta có: 3+y+5+7+6=25

=>y+21=25

=>y=4

=>Mốt của dấu hiệu là 24

b: Trung bình cộng là 23

=>\(15\cdot3+18\cdot4+22\cdot5+24\cdot7+x\cdot6=25\cdot23\)

=>6x+395=575

=>6x=180

=>x=30

Khó vãi lìn.Ai mà giải được,toán lớp 6cow màaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

Trung bình cộng là 10,04

=>\(7\cdot12+8\cdot9+9\cdot m+12\cdot n+13\cdot7+15\cdot5=10,04\left(12+9+m+n+7+5\right)\)

=>9m+12n+322=10,04(17+12+9+7+5)=10,04*50=502

=>9m+12n=502-322=180

=>3m+4n=60

m+n=17

=>3m+3n=51

=>3m+4n-3m-3n=60-51

=>n=9

m+n=17

=>m=17-9=8

b: Mốt là 7

a) đK: \(x\ne0;2\)

B = \(\dfrac{3x-4}{x\left(x-2\right)}.\dfrac{x\left(x-2\right)}{x^2-4-x^2}=\dfrac{3x-4}{-4}=\dfrac{4-3x}{4}\) \(\dfrac{x-4+2x}{x\left(x-2\right)}:\dfrac{\left(x-2\right)\left(x+2\right)-x^2}{x\left(x-2\right)}\)

= \(\dfrac{3x-4}{x\left(x-2\right)}.\dfrac{x\left(x-2\right)}{x^2-4-x^2}=\dfrac{4-3x}{4}\)

b) Thay x = -2 (TMDK) vào B, ta có:

\(B=\dfrac{4-3.\left(-2\right)}{4}=\dfrac{4+6}{4}=\dfrac{5}{2}\)

c) Để \(\left|B\right|-2x=5\)

<=> \(\left|\dfrac{4-3x}{4}\right|-2x=5\)

TH1: \(x\le\dfrac{4}{3}\)

<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{4-3x}{4}\)

PT <=> \(\dfrac{4-3x}{4}-2x=5\)

<=> \(\dfrac{4-3x-8x}{4}=5\)

<=> \(4-11x=20\)

<=> x = \(\dfrac{-16}{11}\) (Tm)

TH2: \(x>\dfrac{4}{3}\)

<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{3x-4}{4}\)

PT <=> \(\dfrac{3x-4}{4}-2x=5\)

<=> \(\dfrac{3x-4-8x}{4}=5\)

<=> \(-5x-4=20\)

<=> \(x=\dfrac{-24}{5}\left(l\right)\)

d) Xét (2-x)B = \(\dfrac{\left(2-x\right)\left(4-3x\right)}{4}\)  = \(\dfrac{3x^2-10x+8}{4}\)

= \(\dfrac{3\left(x-\dfrac{5}{3}\right)^2-\dfrac{1}{3}}{4}\)

Mà \(3\left(x-\dfrac{5}{3}\right)^2\ge\) 0

=> (2-x)B \(\ge\dfrac{\dfrac{-1}{3}}{4}=\dfrac{-1}{12}\)

Dấu "=" <=> x = \(\dfrac{5}{3}\left(tm\right)\)

e) Số nguyên âm lớn nhất là -1

Để B = -1

<=> \(\dfrac{4-3x}{4}=-1\)

<=> 4 - 3x = -4
<=> \(x=\dfrac{8}{3}\left(tm\right)\)

g) 

TH1: \(x\le\dfrac{4}{3}\)

<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{4-3x}{4}\)

BDT <=> \(\dfrac{4-3x}{4}< 2x-4\)

<=> \(4-3x< 8x-16\)

<=> \(x>\dfrac{20}{11}\left(l\right)\)

TH2: \(x>\dfrac{4}{3}\)

<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{3x-4}{4}\)

BDT <=> \(\dfrac{3x-4}{4}< 2x-4\)

<=> \(3x-4< 8x-16\)

<=> x > \(\dfrac{12}{5}\)

KHDK: \(x>\dfrac{12}{5}\)

Bài 1: 

Để E nguyên thì \(x+5⋮x-2\)

\(\Leftrightarrow x-2\in\left\{1;-1;7;-7\right\}\)

hay \(x\in\left\{3;1;9;-5\right\}\)

Thank you.

cai tên của mình noi lên tât cả

a) \(M=\frac{2n-7}{n-5}=\frac{2n-10}{n-5}+\frac{3}{n-5}=2+\frac{3}{n-5}\)

Để M là số nguyên thì \(\frac{3}{n-5}\) là số nguyên <=> 3 chia hết cho n-5

<=>n-5\(\in\)Ư(3)={-3;-1;1;3} <=> n\(\in\){2;4;6;8}

b)\(\left|x-3\right|=2x+4\Leftrightarrow\orbr{\begin{cases}x-3=-2x-4\\x-3=2x+4\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x=-1\\-x=7\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=-7\end{cases}}\)