ĐKXĐ: x≠0
Ta có: \(\dfrac{x}{-15}=-\dfrac{60}{x}\)
\(\Leftrightarrow x^2=\left(-15\right)\cdot\left(-60\right)=900\)
hay \(x\in\left\{30;-30\right\}\)(nhận)
Vậy: \(x\in\left\{30;-30\right\}\)
ĐKXĐ: x≠0
Ta có: \(\dfrac{x}{-15}=-\dfrac{60}{x}\)
\(\Leftrightarrow x^2=\left(-15\right)\cdot\left(-60\right)=900\)
hay \(x\in\left\{30;-30\right\}\)(nhận)
Vậy: \(x\in\left\{30;-30\right\}\)
x/-15=-60/x
\(\dfrac{x}{-15}\) = \(\dfrac{-60}{x}\)
\(|x|+0,573=2\)
\(|x+\dfrac{1}{3}|-4=-1\)
0,01 : 2,5 =(0,75x) : 0,75
\(\dfrac{X}{3}=\dfrac{Y}{7}=\dfrac{Z}{5},X^2-Y^2+Z^2=-60\)Tìm X ,Y ,Z
Tìm x :
a)\(\dfrac{x-1}{50}+\dfrac{x-2}{49}=\dfrac{x-3}{48}+\dfrac{x-4}{47}\)
b)\(\dfrac{x+25}{6}+\dfrac{x+20}{11}+\dfrac{x+16}{15}+3=0\)
c)\(\dfrac{x-15}{6}+\dfrac{x-10}{11}=\dfrac{x-3}{18}+\dfrac{x-7}{14}\)
biết x/2=y/3 và x+y=-15,khi đó giá trị của x,y là
ghi bài làm luôn ạ
Tìm x y z
\(\dfrac{2x-y}{5}=\dfrac{2z-x}{6}=\dfrac{2x+y}{11}=\dfrac{xy}{60}\)
3*(3-x)-4*\(\left|1-x\right|\)=15
BT1: Tìm x, biết:
5) \(\text{|}x+\dfrac{1}{3}\text{|}+\text{|}x+\dfrac{1}{5}\text{|}+\text{|}x+\dfrac{1}{15}\text{|}=4x\)
Tìm x,y biết:
a)|x+4/15|-|-3.75|=-|-2.5|
b)|1/2-1/3+x|=-1/4-|y|
c)|x-y|+|y+9/25|=0
d)|x(x2-5/4)|=x