X x 2+5=54
x bằng bao nhiêu
\((x-2)(x-4)(x-10)(x-5)-54x^{2}\)= 0
\((x-2)(x-4)(x-10)(x-5)-54x^{2}=0\)
\(\left(x-2\right)\left(x-4\right)\left(x-10\right)\left(x-5\right)-54x^2=0\\ \Leftrightarrow\left[\left(x-2\right)\left(x-10\right)\right]\left[\left(x-4\right)\left(x-5\right)\right]=54x^2\\ \Leftrightarrow\left(x^2-12x+20\right)\left(x^2-9x+20\right)=54x^2\)
Với x=0: ko phải nghiệm của pt
Với x≠0
\(pt\Leftrightarrow\dfrac{x^2-12x+20}{x}.\dfrac{x^2-9x+20}{x}=\dfrac{54x^2}{x^2}\\ \Leftrightarrow\left(x-12+\dfrac{20}{x}\right)\left(x-9+\dfrac{20}{x}\right)=54\left(1\right)\)
Đặt \(x+\dfrac{20}{x}=y\)
\(\left(1\right)\Leftrightarrow\left(y-12\right)\left(y-9\right)=54\\ \Leftrightarrow y^2-21y+108-54=0\\ \Leftrightarrow y^2-21y+54=0\\ \Leftrightarrow\left[{}\begin{matrix}y=3\\y=18\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{20}{x}-3=0\\x+\dfrac{20}{x}-18=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2-3x+20=0\left(vô.lí\right)\\x^2-18x+20=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=9+\sqrt{61}\\x=9-\sqrt{61}\end{matrix}\right.\)
\(\left(x-2\right)\left(x-10\right)\left(x-4\right)\left(x-5\right)-54x^2=0\)
\(\Leftrightarrow\left(x^2+20-12x\right)\left(x^2+20-9x\right)-54x^2=0\)
\(\Leftrightarrow\left(x^2+20\right)^2-21x\left(x^2+20\right)+108x^2-54x^2=0\)
\(\Leftrightarrow\left(x^2+20\right)^2-21x\left(x^2+20\right)+54x^2=0\)
\(\Leftrightarrow\left(x^2+20\right)^2-3x\left(x^2+20\right)-18x\left(x^2+20\right)+54x^2=0\)
\(\Leftrightarrow\left(x^2+20\right)\left(x^2+20-3x\right)-18x\left(x^2+20-3x\right)=0\)
\(\Leftrightarrow\left(x^2+20-3x\right)\left(x^2+20-18x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+20=0\left(vô-nghiệm\right)\\x^2-18x+20=0\end{matrix}\right.\)
\(\Rightarrow x^2-18x+81-61=0\)
\(\Rightarrow\left(x-9\right)^2=61\)
\(\Rightarrow x-9=\pm\sqrt{61}\)
\(\Rightarrow x=9\pm\sqrt{61}\)
x+1 / 2x+1 = 3/5
|x+1| + |x+2| + |x+3| +...+ |x+9|=54x
\(\dfrac{x+1}{2x+1}=\dfrac{3}{5}\)
⇔\(5\left(x+1\right)=3\left(2x+1\right)\)
⇔\(5x+5=6x+3\)
⇔\(5x+5-6x-3=0\)
⇔\(-x+2=0\)
⇔\(-x=-2\)
⇔\(x=2\)
Phân tích đa thức thành nhân tử:(x-2)(x-4)(x-5)(x-10)-54x^2
Bài 1
4) (x+3)(x-1)(x-5)(x+15)+64x^2
5) (x+2)(x-4)(x+6)(x-12)+36x^2
6) (x-2)(x-4)(x-5)(x-10)-54x^2
Phân tích thành nhân tử
4) (x+3)(x-1)(x-5)(x+15)+64x^2
5) (x+2)(x-4)(x+6)(x-12)+36x^2
6) (x-2)(x-4)(x-5)(x-10)-54x^2
Phân tích thành nhân tử
4) (x+3)(x-1)(x-5)(x+15)+64x^2
5) (x+2)(x-4)(x+6)(x-12)+36x^2
6) (x-2)(x-4)(x-5)(x-10)-54x^2
(X-2)(x-4)(x-5)(x-10)-54x2
Phân tích đa thức trên thành nhân tử
Đặt \(A=\left(x-2\right)\left(x-4\right)\left(x-5\right)\left(x-10\right)-54x^2\)
\(=\left[\left(x-2\right)\left(x-10\right)\right]\left[\left(x-4\right)\left(x-5\right)\right]-54x^2\)
\(=\left(x^2-12x+20\right)\left(x^2-9x+20\right)-54x^2\)
Đặt \(x^2-12x+20=t\)
Khi đó: \(A=t\left(t+3x\right)-54x^2\)
\(=t^2+3tx-54x^2\)
\(=t\left(t-6x\right)+9x\left(t-6x\right)\)
\(=\left(t-6x\right)\left(t+9x\right)\)
\(=\left(x^2-18x+20\right)\left(x^2-3x+20\right)\)
2 x 3 x 5 / 5 x 2 x 3 bằng bao nhiêu
Ta có:
\(\dfrac{2.3.5}{5.2.3}\)
Ta sẽ gạch những số giống nhau
= 1
Vậy kết quả bằng 1
\(\dfrac{2\times3\times5}{5\times2\times3}=1\)