X-1/-15=-60/X-1
15: nếu \(\dfrac{x}{-15}\)=\(\dfrac{-60}{x}\) thì kết quả x bằng:
A) x=30 B) x=30 hoặc x=-1 C) x=3= hoặc x=-30 D) x=\(\dfrac{60}{15}\)
\(x^2=900\Leftrightarrow x^2=30^2\Rightarrow x=30\)
Chọn A
x-1/ -15 = -60 / x - 1
\(\dfrac{x-1}{-15}=\dfrac{-60}{x-1}\\ \Rightarrow\left(x-1\right)^2=-15\cdot\left(-60\right)\\ \Rightarrow\left(x-1\right)^2=900\\ \Rightarrow\left[{}\begin{matrix}x-1=30\\x-1=-30\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=31\\x=-29\end{matrix}\right.\)
tìm x biết :
1x - ( x-1/3 ) = 1/6
x-1/-15 = -60/x-1
1\(x\) - (\(x\) - \(\dfrac{1}{3}\)) = \(\dfrac{1}{6}\)
\(x\) - \(x\) + \(\dfrac{1}{3}\) = \(\dfrac{1}{6}\)
\(\dfrac{1}{3}\) = \(\dfrac{1}{6}\) (vô lí)
Vậy không có giá trị nào của \(x\) thỏa mãn đề bài
\(\dfrac{x-1}{-15}\) = - \(\dfrac{60}{x-1}\)
(\(x\) - 1).(\(x\) - 1) = (-60).(-15)
(\(x\) - 1)2 = 900
\(\left[{}\begin{matrix}x-1=-30\\x-1=30\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-29\\x=31\end{matrix}\right.\)
Vậy \(x\in\) {-29; 31}
X-1/-15=-60/X-1
\(\dfrac{x-1}{-15}=\dfrac{-60}{x-1}\)
\(\Leftrightarrow\left(x-1\right)\left(x-1\right)=\left(-15\right)\left(-60\right)\)
\(\Leftrightarrow\left(x-1\right)^2=900\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=30^2\\\left(x-1\right)^2=\left(-30\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=30\\x-1=-30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=31\\x=-29\end{matrix}\right.\)
Vậy .................
1 ) x + 15 = 60 Tìm x
2 ) x + 60 = 180
1)x=60-15=45
2) x=180-60=120 Mà ảnh đại diện của bn đẹp z
1) x + 15 = 60
x = 60 - 15
x = 45
2) x + 60 = 180
x = 180 - 60
x = 120
1 ) x + 15 = 60
x = 60 - 15
x = 45 .
2 ) x + 60 = 180
x = 180 - 60
x = 120 .
tim x
x-1/15=-60/1-x
bn nên ghi rõ r. Bn vt thế này mk ko hiểu bn vt cái j cả
\(\frac{x-1}{15}=\frac{-60}{1-x}\)
\(x-1=\frac{900}{1-x}\)
\(\left(x-1\right)\left(1-x\right)=-900\)
\(x-x^2-1+x=-900\)
\(2x-x^2-1=-900\)
\(2x-x^2-1+900=0\)
\(2x-x^2-899=0\)
\(\left(x-31\right)\left(x+29\right)=0\)
\(\orbr{\begin{cases}x-31=0\\x+29=0\end{cases}}\)
\(\orbr{\begin{cases}x=31\\x=-29\end{cases}}\)
Tìm Y:
Y x 16/64 + Y x 25/100 + Y x 1/4 + Y x 15/60 - 13/15 = 17/15
y \(\times\) \(\dfrac{16}{64}\) + y \(\times\) \(\dfrac{25}{100}\) + y \(\times\) \(\dfrac{1}{4}\) + y \(\times\) \(\dfrac{15}{60}\) - \(\dfrac{13}{15}\) = \(\dfrac{17}{15}\)
y \(\times\) ( \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)) - \(\dfrac{13}{15}\) = \(\dfrac{17}{15}\)
y = \(\dfrac{17}{15}\) + \(\dfrac{13}{15}\)
y = \(\dfrac{30}{15}\)
y = 2
Cái này bằng rủ em chơi oẳn tù tì ,em ra lá còn anh ra hôn má em
me nhìn công thức của cô Hoài moà đâu đầu nun á
Tìm x, y biết:
x-1/ -15= -60/ x-1
x-1/-15 = -60/x-1
=> (x-1)2 = 900
=> (x-1)2 = 302
=> \(\hept{\begin{cases}x-1=-30\\x-1=30\end{cases}}\)
=>\(\hept{\begin{cases}x=-29\\x=31\end{cases}}\)
Tìm x
x-1/-15=-60/x-1
\(\frac{x-1}{-15}=\frac{-60}{x-1}\left(ĐK:x\ne1\right)\)
\(\Rightarrow\left(x-1\right)\left(x-1\right)=\left(-15\right)\left(-60\right)\\ \Rightarrow\left(x-1\right)^2=900\\ \Rightarrow\left[{}\begin{matrix}x-1=30\\x-1=-30\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=31\left(t/m\right)\\x=-29\left(t/m\right)\end{matrix}\right.\)
\(\text{Vậy }x\in\left\{31;-29\right\}\)
x-1/-15=-60/x-1
(x-1)²=-15×(-60)
(x-1)²=900
=>[x-1=30
[x-1=-30
=>[x=31
[x=-29
\(\frac{x-1}{-15}=\frac{-60}{x-1}\)
\(\Rightarrow\left(x-1\right).\left(x-1\right)=\left(-15\right).\left(-60\right)\)
\(\Rightarrow\left(x-1\right)^2=900\)
\(\Rightarrow\left(x-1\right)^2=\left(\pm30\right)^2\)
\(\Rightarrow x-1=\pm30.\)
\(\Rightarrow\left[{}\begin{matrix}x-1=30\\x-1=-30\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=30+1\\x=\left(-30\right)+1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=31\\x=-29\end{matrix}\right.\)
Vậy \(x\in\left\{31;-29\right\}.\)
Chúc bạn học tốt!