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\(12x^2+13y^2=25xy\)
\(\Leftrightarrow12x^2-25xy+13y^2=0\)
\(\Leftrightarrow12x^2-12xy-13xy+13y^2=0\)
\(\Leftrightarrow12x\left(x-y\right)-13y\left(x-y\right)=0\)
\(\Leftrightarrow\left(12x-13y\right)\left(x-y\right)=0\)
\(\Rightarrow\orbr{\begin{cases}12x-13y=0\\x-y=0\end{cases}}\)
Mà để A xác định \(\Leftrightarrow x-y\ne0\) Do đó \(12x-13y=0\Leftrightarrow12x=13y\Rightarrow x=\frac{13}{12}y\)
\(\Rightarrow A=\frac{\frac{13}{12}y+y}{\frac{13}{12}y-y}=\frac{y\left(\frac{13}{12}+1\right)}{y\left(\frac{13}{12}-1\right)}=\left(\frac{13}{12}+1\right):\left(\frac{13}{12}-1\right)=\frac{25}{12}:\frac{1}{12}=25\)
1. Đặt : x = a + \(\dfrac{1}{3}\) ; y = b + \(\dfrac{1}{3}\) ; z = \(c+\dfrac{1}{3}\)
Ta có : x + y + z = 1
⇒ a + b + c = 0
Ta có : x2 + y2 + z2 = ( a + \(\dfrac{1}{3}\))2 + ( b + \(\dfrac{1}{3}\))2 + ( c + \(\dfrac{1}{3}\))2
= a2 + \(\dfrac{2}{3}a+\dfrac{1}{9}+b^2+\dfrac{2}{3}b+\dfrac{1}{9}+c^2+\dfrac{2}{3}c+\dfrac{1}{9}\)
= \(\dfrac{1}{3}+\dfrac{2}{3}\left(a+b+c\right)+a^2+b^2+c^2\)
= \(\dfrac{1}{3}+a^2+b^2+c^2\) ≥ \(\dfrac{1}{3}\)
Dâu "=" xảy ra khi và chỉ khi : a = b = c = 0 ⇔ x = y = z = \(\dfrac{1}{3}\)
Mk xin lỗi nha, câu c sai đề
c) (x+6)4 + (x+8)4 = 272
1: =>x^3-5x^2+x^2-5x+3x-15=0
=>(x-5)(x^2+x+3)=0
=>x-5=0
=>x=5
2: =>x^3+6x^2+12x+35=0
=>x^3+5x^2+x^2+5x+7x+35=0
=>(x+5)(x^2+x+7)=0
=>x+5=0
=>x=-5
3: \(\Leftrightarrow\left(\dfrac{x+43}{57}+1\right)+\left(\dfrac{x+46}{54}+1\right)=\left(\dfrac{x+49}{51}+1\right)+\left(\dfrac{x+52}{48}+1\right)\)
=>x+100=0
=>x=-100
2/
\(x^3-2x+1=0\)
\(\Rightarrow x^3-x-x+1=0\)
\(\Rightarrow x\left(x^2-1\right)-\left(x-1\right)=0\)
\(\Rightarrow x\left(x-1\right)\left(x+1\right)-\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x^2+x-1\right)\)
\(\Rightarrow x=1\)
Vậy S = {1}