\(x^2=900\Leftrightarrow x^2=30^2\Rightarrow x=30\)

Chọn A

mik nghĩ A

Em ghi đề lại cho chính xác

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}

\(\dfrac{x-1}{-15}=\dfrac{-60}{x-1}\)

\(\Leftrightarrow\left(x-1\right)\left(x-1\right)=\left(-15\right)\left(-60\right)\)

\(\Leftrightarrow\left(x-1\right)^2=900\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=30^2\\\left(x-1\right)^2=\left(-30\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=30\\x-1=-30\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=31\\x=-29\end{matrix}\right.\)

Vậy .................

\(\dfrac{x-1}{-15}=\dfrac{-60}{x-1}\)

\(\Rightarrow\left(x-1\right)\left(x-1\right)=\left(-60\right)\left(-15\right)\)

\(\Rightarrow\left(x-1\right)^2=900\)

\(\Rightarrow\left(x-1\right)^2=\pm30^2\)

\(\Rightarrow\left[{}\begin{matrix}x-1=30\\x-1=-30\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=31\\x=-29\end{matrix}\right.\)

\(\dfrac{x-1}{-15}=\dfrac{-60}{x-1}\)

\(\Leftrightarrow\left(x-1\right)\left(x-1\right)=\left(-15\right)\left(-60\right)\)

\(\Leftrightarrow\left(x-1\right)^2=900\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=30^2\\\left(x-1\right)^2=\left(-30\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=30\\x-1=-30\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=31\\x=-29\end{matrix}\right.\)

Vậy ................

1)x=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 .

bn nên ghi rõ r. Bn vt thế này mk ko hiểu bn vt cái j cả

\(\frac{x-1}{15}=\frac{-60}{1-x}\)

\(x-1=\frac{900}{1-x}\)

\(\left(x-1\right)\left(1-x\right)=-900\)

\(x-x^2-1+x=-900\)

\(2x-x^2-1=-900\)

\(2x-x^2-1+900=0\)

\(2x-x^2-899=0\)

\(\left(x-31\right)\left(x+29\right)=0\)

\(\orbr{\begin{cases}x-31=0\\x+29=0\end{cases}}\)

\(\orbr{\begin{cases}x=31\\x=-29\end{cases}}\)

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

x-1/-15 = -60/x-1

=> (x-1)² = 900

=> (x-1)² = 30²

=> \(\hept{\begin{cases}x-1=-30\\x-1=30\end{cases}}\)

=>\(\hept{\begin{cases}x=-29\\x=31\end{cases}}\)

\(\frac{x-1}{-15}=\frac{-60}{x-1}\left(ĐK:x\ne1\right)\)

\(\Rightarrow\left(x-1\right)\left(x-1\right)=\left(-15\right)\left(-60\right)\\ \Rightarrow\left(x-1\right)^2=900\\ \Rightarrow\left[{}\begin{matrix}x-1=30\\x-1=-30\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=31\left(t/m\right)\\x=-29\left(t/m\right)\end{matrix}\right.\)

\(\text{Vậy }x\in\left\{31;-29\right\}\)

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 \(x\in\left\{31;-29\right\}.\)

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