a) Ta có:

1; 4; 7;...; 100 có (100 - 1) : 3 + 1 = 34 (số)

1 + 4 + 7+ ... + 100 = (100 + 1) × 34 : 2

= 101 × 17

(1 + 4 + 7 + ... + 100) : a = 17

101 × 17 : a = 17

a = 101 × 17 : 17

a = 100

b) (X - 1/2) × 5/3 = 7/4 - 1/2

(X - 1/2) × 5/3 = 5/4

X - 1/2 = 5/4 : 5/3

X - 1/2 = 3/4

X = 3/4 + 1/2

X = 5/4

a) (1 + 4 + 7 +...+ 100) : a = 17

1717 : a = 17

a = 101

b) \(\left(x-\dfrac{1}{2}\right)\times\dfrac{5}{3}=\dfrac{7}{4}-\dfrac{1}{2}\)

\(\left(x-\dfrac{1}{2}\right)\times\dfrac{5}{3}=\dfrac{10}{8}\)

\(\left(x-\dfrac{1}{2}\right)=\dfrac{10}{8}\div\dfrac{5}{3}\)

\(\left(x-\dfrac{1}{2}\right)=\dfrac{10}{8}\times\dfrac{3}{5}\)

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

\(x-\dfrac{1}{2}=\dfrac{3}{4}\)

\(x=\dfrac{3}{4}+\dfrac{1}{2}\)

\(x=\dfrac{5}{4}\)

a. (1+4+7+...+100):a=17 

=> ( 100 + 1) x 32 : 2 : a = 17

=> 1717 : a = 17 

=> a = 101

b. \(\left(X-\dfrac{1}{2}\right).\dfrac{5}{3}=\dfrac{7}{4}-\dfrac{2}{4}\)

\(\left(X-\dfrac{1}{2}\right).\dfrac{5}{3}=\dfrac{5}{4}\)

\(X-\dfrac{1}{2}=\dfrac{3}{4}\)

\(X=\dfrac{1}{4}\)

Chọn đáp án B

Ta có:

1)x=7-2y nên x+2y=7

A=3x+6y+5=3x(x+2y)+5=3x7+5=26

2)y-x=y+(-x)=-x+y=-(x-y)=-10

Vậy x-y=10

Nguyễn Thị Ngọc Ánh: Mk chưa hiểu câu 1 cho lắm, bn giải kĩ hơn đc ko?

lo chúng mày

/2x-1/=/5-x/

<=>  2x-1=5-x<=>3x=6<=>x=2

Hoặc: 2x-1=-(5-x)<=>2x-1=x-5<=>x=-4

 k mk nha 

Phá trị tuyệt đối ra, ta xét 4 trường hợp:

TH1: 2x - 1 = 5 - x

\(\Rightarrow\) 3x = 6

\(\Rightarrow\) x = 2 (1)

TH2: 2x - 1 = x - 5

\(\Rightarrow\)x = -4 (2)

TH3: 1 - 2x = 5 - x

\(\Rightarrow\) -x = 4

\(\Rightarrow\) x = -4 (3)

TH4: 1 - 2x = x - 5

\(\Rightarrow\) -3x = -6

\(\Rightarrow\) x = 2 (4)

Từ (1),(2),(3) và (4) ta suy ra có 2 giá trị của x thỏa mãn đề bài là:

 x = -4 hoặc x = 2

|x+5|=|2x+1|

\(\Rightarrow\orbr{\begin{cases}x+5=2x+1\\x+5=-\left(2x+1\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-2x=1-5\\x+5=-2x-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}-1x=-4\\x+2x=-1-5\end{cases}}\) 

\(\Leftrightarrow\orbr{\begin{cases}x=4\\3x=-6\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}\) 

Vậy x=\(\in\) {4;-2}

|x + 5| = |2x + 1| 

=> \(\orbr{\begin{cases}x+5=2x+1\\x+5=-2x-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}\)

Vậy x \(\in\left\{4;-2\right\}\)là giá trị cần tìm

\(\left|x+5\right|=\left|2x+1\right|\)

\(\Leftrightarrow\hept{\begin{cases}x+5=2x+1\\x+5=-2x-1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=4\\x=-2\end{cases}}\)

Vậy ....

a)

A=\(\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5x-5}\)

\(\Leftrightarrow\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5\left(x-1\right)}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+1\\x=0-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

MTC: 5(x-1)(x+1)

\([\dfrac{5\left(x+1\right)\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}-\dfrac{5\left(x-1\right)\left(x-1\right)}{5\left(x-1\right)\left(x+1\right)}]\div\dfrac{2x\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow[5\left(x+1\right)\left(x+1\right)-5\left(x-1\right)\left(x-1\right)]\div2x\left(x+1\right)\)

\(\Leftrightarrow[5\left(x+1\right)^2-5\left(x-1\right)^2]\div2x^2+2x\)

\(\Leftrightarrow[5\left(x^2+2x+1\right)-5\left(x^2-2x+1\right)]\div2x^2+2x\)

\(\Leftrightarrow(5x^2+10x+5-5x^2+10x-5)\div2x^2+2x\)

\(\Leftrightarrow20x\div\left(2x^2+2x\right)\)

\(\Leftrightarrow10x+10\)

a: ĐKXĐ: x<>-3

a) đkxđ: x+3\(\ne0\Rightarrow x\ne-3\)
b) ta có: M=\(\dfrac{-2}{3}\)
M= \(\dfrac{x+5}{x+3}\)=\(\dfrac{-2}{3}\)
 (x+5)3=(x+3)(-2)
 3x+15=-2x-6
 3x+2x+15=-6
 5x=-6-15
 5x=-21
 x=-21/5
Vậy x= -21/5 khi M có giá trị là -2/3
c) ko bt lm:))