\(\left(x:y\right)^2=\left(\frac{4}{3}\right)^2\)

\(=>x:y=\frac{4}{3}=>\frac{x}{y}=\frac{4}{3}\)

---> x : y = 4/3

---> 3x = 4y

lại có x^2 + y^2 = 100

---> ( 3x )^2 + ( 3y )^2 = 900

---> ( 4y )^2 + ( 3y )^2 = 900

---> 16y^2 + 9y^2 = 900

---> 25y^2 = 900

---> y^2 = 36

---> y = 6

---> x = 8

có gì ib riêng mình nhé

\(6^2+8^2=100\)

Ta có : 1³+ 2³+...+100³ = (1+2+3+...+100)²  

Đặt A = 1+2+3+...+100

⇒A= \(\dfrac{\left(100+1\right).100}{2}\) ⇔A= 5050

⇒5050² = (x-1)² ⇔ x=5051

(\(x\) + 1) + (\(x\) + 2) + (\(x\) + 3)+...+(\(x\) + 100) = 5200

Xét dãy số: 1; 2; 3; ...;100

Dãy số trên là dãy số cách đều với khoảng cách là: 2-1 = 1

Số số hạng của dãy số trên là: (100 - 1) : 1 + 1  =5200

nên ta có: 

(\(x\) + 1) + (\(x\) + 2) + (\(x+3\))+...+(\(x+100\)) = 5200

(\(x\) + \(x\) +...+\(x\)) + (1 + 2 + 3 +...+100) = 5200

\(x\) \(\times\) 100 + (100 +1) \(\times\) 100: 2 = 5200

\(x\times\) 100 + 5050 = 5200

\(x\) \(\times\) 100   = 5200 - 5050

\(x\) \(\times\) 100 = 150

\(x\) = 150 : 100

\(x\) =  1,5

(x +1) + (x+2)+ (x +3) +….+ (x + 100) = 5200

(x+x+x+...+x+x) + (1+2+3+...+99+100)= 5200  (có 100 hạng tử x )

x * 100 + (1+100) * (100:2)= 5200

x * 100 + 101 * 50= 5200

x * 100 + 5050 = 5200

x * 100 = 5200 - 5050

x * 100 = 150

x = 150:100

x= 1,5

=>100x+(1+100)+(2+99)+....+(50+51)=5200
=>100x+101x50=5200
=>100x+5050=5200
=>100x=150
=>x=1.5

\(\left|2x-3\right|=3-2x\)

\(ĐK:x\le\dfrac{3}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3-2x\\3-2x=3-2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\0=0\left(đúng\right)\end{matrix}\right.\)

Vậy \(S=\left\{x\in R;x=\dfrac{3}{2}\right\}\)

a: \(-\dfrac{2}{5}+\dfrac{4}{5}x=\dfrac{3}{5}\)

=>\(\dfrac{4}{5}x=\dfrac{3}{5}+\dfrac{2}{5}=1\)

=>\(x=1:\dfrac{4}{5}=\dfrac{5}{4}\)

b; \(-\dfrac{3}{7}-\dfrac{4}{7}:x=-2\)

=>\(\dfrac{4}{7}:x+\dfrac{3}{7}=2\)

=>\(\dfrac{4}{7}:x=2-\dfrac{3}{7}=\dfrac{11}{7}\)

=>\(x=\dfrac{4}{7}:\dfrac{11}{7}=\dfrac{4}{11}\)

\(a,\left|x-\dfrac{1}{2}\right|+\dfrac{1}{3}=\dfrac{2}{3}\\ \Rightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{1}{3}\\x-\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\\x=-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{1}{6}\end{matrix}\right.\\ b,\dfrac{x}{-2}=\dfrac{y}{5}=\dfrac{x-y}{-2-5}=\dfrac{14}{-7}=-2\\ \Rightarrow x=-2.\left(-2\right)=4;y=-2.5=-10\)

x-y=-14 nhé

a)

\(\left|x-\dfrac{1}{2}\right|+\dfrac{1}{3}=\dfrac{2}{3}\\ \left|x-\dfrac{1}{2}\right|=\dfrac{1}{3}\\ \left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{1}{3}\\x-\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{1}{3}+\dfrac{1}{2}\\x=-\dfrac{1}{3}+\dfrac{1}{2}\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{5}{6}\\x=\dfrac{1}{6}\end{matrix}\right.\)

b)

\(\dfrac{x}{-2}=\dfrac{y}{5}\)

mà `x-y=14` nên áp dụng tính chất dãy tỉ số bằng nhau ta có

\(\dfrac{x}{-2}=\dfrac{y}{5}=\dfrac{x-y}{-2-5}=\dfrac{14}{-7}=-2\\ =>\left\{{}\begin{matrix}x=-2\cdot\left(-2\right)=4\\y=-2\cdot5=-10\end{matrix}\right.\)

\(100+x=130-25\)

\(100+x=105\)

\(x=105-100\)

\(x=5\)

k cho mik nha bn

HT

TL :

x = 5

_HT_

<=> 30- x = 25

<= 30- 5 = 25

<= 5

moi nguoi hoc tot nhe