Phương trình bậc hai một ẩn - Hỏi đáp

\(\left[\left(\dfrac{1}{a^2+1}\right).\dfrac{1}{a^2+2a+1}+\dfrac{2}{\left(a+1\right)^3}.\left(\dfrac{1}{a}+1\right)\right]:\dfrac{a-1}{a^3}\)

\(=\left(\dfrac{1}{\left(a^2+1\right)\left(a^2+2a+1\right)}+\dfrac{2}{\left(a+1\right)^3}\cdot\dfrac{1+a}{a}\right)\cdot\dfrac{a^3}{a-1}\)

\(=\left(\dfrac{1}{\left(a^2+1\right)\left(a+1\right)^2}+\dfrac{2}{a\left(a+1\right)^2}\right)\cdot\dfrac{a^3}{a-1}\)

\(=\dfrac{a+2\left(a^2+1\right)}{a\left(a^2+1\right)\left(a+1\right)^2}\cdot\dfrac{a^3}{a-1}\)

\(=\dfrac{a+2a^2+2}{\left(a^2+1\right)\left(a+1\right)^2}\cdot\dfrac{a^2}{a-1}\)

\(=\dfrac{a^2\left(a+2a^2+2\right)}{\left(a^2+1\right)\left(a+1\right)^2\cdot\left(a-1\right)}\)

\(=\dfrac{a^3+2a^4+2a^2}{\left(a^3-a^2+a-1\right)\left(a+1\right)^2}\)

nhầm là x⁴₁+x⁴₂+x⁴₃+x⁴_{4=68. giúp mình với}

ta có : \(\dfrac{x}{2}=\dfrac{y}{-5}\) \(\Leftrightarrow\) \(-5x=2y\) \(\Leftrightarrow\) \(-5x-2y=0\) (1)

mà ta có \(x+y=-12\) (2)

từ (1) và (2) ta có hệ phương trình : \(\left\{{}\begin{matrix}-5x-2y=0\\x+y=-12\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}-5x-2y=0\\2x+2y=-24\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}-3x=-24\\x+y=-12\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=8\\8+y=-12\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=8\\y=-20\end{matrix}\right.\)

vậy ta có \(x=8\) ; \(y=-20\)

\(\Rightarrow\) \(x-y\) \(\Leftrightarrow\) \(8-\left(-20\right)\) = \(8+20=28\)

vậy hiệu \(x-y=28\)

Áp dụng t/c của dãy tỉ số = nhau ta có:

\(\dfrac{x}{2}=\dfrac{y}{-5}=\dfrac{x+y}{2+\left(-5\right)}=\dfrac{-12}{-3}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4\cdot2=8\\y=4\cdot\left(-5\right)=-20\end{matrix}\right.\)

\(\Rightarrow x-y=8-\left(-20\right)=8+20=28\)

Vậy...........

\(x^4+\left(x-1\right)\left(x^2-2x+2\right)=0\)

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

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

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

Ta có: \(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)nên

\(\Rightarrow x^2+2x-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1-\sqrt{3}\\x=-1+\sqrt{3}\end{matrix}\right.\)

Còn ít nhất 2 cách giải nữa. Các bạn vẫn còn cơ hội. Đây chỉ là một phương trình đưa ra về phương trình bậc hai thôi.

\(\Delta\) = 4m² - 4.(m²).(-3) = 4m² + 12m² = 16m² \(\ge\) 0 \(\forall\)m

\(\Rightarrow\) phương trình có 2 nghiện \(\sqrt{\Delta}\) = \(\sqrt{16m^2}\) = \(\sqrt{\left(4m\right)^2}\) = 4m

x₁ = \(\dfrac{2m+4m}{2m^2}\) = \(\dfrac{6m}{2m^2}\) = \(\dfrac{3}{m}\)

x₂ = \(\dfrac{2m-4m}{2m^2}\) = \(\dfrac{-2m}{2m^2}\) = \(\dfrac{-1}{m}\)

\(\left|x+5\right|=2x-18\)

\(\Rightarrow\left|x+5\right|-2x=-18\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5-2x=-18\left(đk:x+5\ge0\right)\\-\left(x+5\right)-2x=-18\left(đk:x+5< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-x+5=-18\\-x-5-2x=-18\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=-18-5\\-3x-5=-18\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=-23\\-3x=-18+5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=-23\\-3x=-13\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=23\left(đk:x\ge-5\right)\\x=\dfrac{13}{3}\left(đk:x< -5\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=23\\x\in\varnothing\end{matrix}\right.\)

Vậy \(x=23\)

\(\left|x+5\right|=2x-18\)

+ ) \(\left|x+5\right|=x+5.\) Khi \(x+5\ge0\Leftrightarrow x\ge-5\)

\(\Leftrightarrow x+5=2x-18\)

\(\Leftrightarrow-x=-23\Leftrightarrow x=23\left(TM\right)\)

+ ) \(\left|x+5\right|=-x-5.\)Khi \(-x-5< 0\Leftrightarrow-x< 5\Leftrightarrow x>5\)

\(\Leftrightarrow-3x=-13\Leftrightarrow x=\dfrac{13}{3}\left(KTM\right)\)

Vậy ...

đầu tiên đưa pt về dạng ax²+bx+c=0

tiếp theo tính \(\Delta\) hoặc \(\Delta'\)

nếu \(\Delta\) hoặc \(\Delta'\)<0 pt vô nghiệm

nếu \(\Delta\) hoặc \(\Delta'\)\(\ge0\) thì ta tính nghiệm theo công thức nghiệm

?

pt có 2 no trái dấu khi ac<0

=> m-1>0=>m>1

để phương trình có nghiệm thì: \(\Delta'\ge0\Leftrightarrow3+m\ge0\Leftrightarrow m\ge-3\)

để ptrình có nghiệm trái dấu thì \(m-1>0\Rightarrow m>1\)

ta có:

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)

\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{1}{x+y+z}=0\)

\(\Leftrightarrow\dfrac{x+y}{xy}+\dfrac{x+y+z-z}{z\left(x+y+z\right)}=0\)

\(\Leftrightarrow\left(x+y\right)\left(\dfrac{1}{xy}+\dfrac{1}{z\left(x+y+z\right)}\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(\dfrac{xz+yz+z^2+xy}{xyz\left(x+y+z\right)}\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(\dfrac{\left(y+z\right)\left(x+z\right)}{xyz\left(x+y+z\right)}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\\dfrac{\left(y+z\right)\left(x+z\right)}{xyz\left(x+y+z\right)}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\x+z=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^8=\left(-y\right)^8\\y^9=\left(-z\right)^9\\z^{10}=\left(-x\right)^{10}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x^8-y^8=0\\y^9+z^9=0\\x^{10}-z^{10}=0\end{matrix}\right.\)\(\Rightarrow\left(x^8-y^8\right)\left(y^9+z^9\right)\left(z^{10}-x^{10}\right)=0\)

\(\Rightarrow M=\dfrac{3}{4}\)