a) \(x^3\)+\(x^2\)=36

\(\Leftrightarrow\)\(x^3\)+\(x^2\)\(-36=0\)

\(\Leftrightarrow\)\(x^3\)\(-3x^2\)\(+4x^2\)\(-12x\)\(+12x-36=0\)

\(\Leftrightarrow\)\(x^2\left(x-3\right)+4x\left(x-3\right)+12\left(x-3\right)=0\)

\(\Leftrightarrow\)\(\left(x-3\right)\left(x^2+4x+12\right)=0\)

Suy ra: \(x-3=0\) hoặc \(x^2+4x+12=0\)

Vậy \(x=3\)

giờ mình đi học mai sẽ làm nốt phần còn lại

C) \(\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)

\(\Leftrightarrow\) \(\left(\left(x^2-5x\right)^2+2.\left(x^2-5x\right).5+25\right)-1=0\)

\(\Leftrightarrow\) \(\left(x^2-5x+5\right)^2-1=0\)

\(\Leftrightarrow\) \(\left(x^2-5x+5-1\right).\left(x^2-5x+5+1\right)=0\)

\(\Leftrightarrow\) \(\left(x^2-5x+4\right).\left(x^2-5x+6\right)=0\)

Suy ra \(x^2-5x+4=0\) hoặc \(x^2-5x+6=0\)

Nếu \(x^2-5x+4=0\) \(\Leftrightarrow\) \(\left(x-1\right).\left(x-4\right)=0\) \(\Leftrightarrow\) \(\left[\begin{array}{} x=1\\ x=4 \end{array} \right.\)

Nếu \(x^2-5x+6=0\) \(\Leftrightarrow\) \(\left(x-2\right).\left(x-3\right)=0\) \(\Leftrightarrow\) \(\left[\begin{array}{} x=2\\ x=3 \end{array} \right.\)

Vậy x \(\in\) {1;2;3;4}

49x = 441 + 1029 = 1470 =. x = 1470 : 49 = 30

bấm máy cũng được!!!

49x(x-21)=441

      x-21 = 441 : 49

      x-21 = 9

      x     = 9 + 21

      x     = 30

\(=\left(x^3+x^2\right)-\left(7x^2+7x\right)+\left(19x+19\right)=\left(x+1\right)\left(x^2-7x+19\right)=0\)

Ta thấy:  \(x^2-7x+19=x^2-2\times\frac{7}{2}x+\frac{7}{2}^2+\frac{27}{4}=\left(x-\frac{7}{2}\right)^2+\frac{27}{4}\ge\frac{27}{4}\)lớn hơn 0

\(\Rightarrow x+1=0\Rightarrow x=-1\)

\(x^3-6x^2+12x+19=0\)

\(\Leftrightarrow\left(x^3+x^2\right)-\left(7x^2+7x\right)+\left(19x+19\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-7x+19\right)=0\)

Mà \(x^2-7x+19>0\)với \(\forall x\)

\(\Rightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy \(x=-1\)

\(\dfrac{1}{2}x^3-49x=0\)

\(\dfrac{1}{2}x^2.x-49x=0\)

\(x.\left(\dfrac{1}{2}x^2-49\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\\dfrac{1}{2}x^2-49=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\dfrac{1}{2}x^2=49\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=98\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=7\sqrt{2}\end{matrix}\right.\)

Vậy \(x\in\left\{0,7\sqrt{2}\right\}\)

49. x³-x=0

<=> 49. x(x²-1)=0

<=> 49. x(x²-1²)=0

<=> 49.x(x-1)(x+1)=0

=> x=0 hoặc x-1=0 hoặc x+1=0

=> x=0 hoặc x=1 hoặc x= -1

49. x³-x=0

<=> 49. x(x²-1)=0

<=> 49. x(x²-1²)=0

<=> 49.x(x-1)(x+1)=0

=> x=0 hoặc x-1=0 hoặc x+1=0

=> x=0 hoặc x=1 hoặc x= -1

à tìm x, y nha mng:>