CMR : Với a,b,c khác 0 thỏa mãn : \(\frac{a^2-bc}{a}+\frac{b^2-ac}{b}+\frac{c^2-ab}{c}=0\)thì a = b = c
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x + y =1 nên x + yt = 2
t + b= = dr wn e=gv
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Xét hệ \(\hept{\begin{cases}x^2+xy^2-xy-y^3=0\left(1\right)\\x^2-3y^2+2\sqrt{x}=0\left(2\right)\end{cases}}\):
\(\left(1\right)\Leftrightarrow\left(x^2-xy\right)+\left(xy^2-y^3\right)=0\Leftrightarrow x\left(x-y\right)+y^2\left(x-y\right)=0\)\(\Leftrightarrow\left(x+y^2\right)\left(x-y\right)=0\Leftrightarrow\orbr{\begin{cases}y^2=-x\\x=y\end{cases}}\)
Trường hợp 1: \(y^2=-x\)thay vào (2), ta được: \(x^2+3x+2\sqrt{x}=0\)(*)
Đặt \(\sqrt{x}=t\left(t>0\right)\)thì (*) trở thành: \(t^4+3t^2+2t=0\Leftrightarrow t\left(t^3+3t+2\right)=0\Leftrightarrow t=0\)(Lưu ý: Phương trình \(t^3+3t+2=0\)có một nghiệm âm và hai nghiệm phức đều không thỏa mãn)
\(\Rightarrow x=y=0\)
Trường hợp 2: \(x=y\)thay vào (2), ta được: \(x^2-3x^2+2\sqrt{x}=0\Leftrightarrow2\sqrt{x}-2x^2=0\Leftrightarrow\sqrt{x}\left(1-\sqrt{x^3}\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\Rightarrow y=0\\x=1\Rightarrow y=1\end{cases}}\)
Vậy hệ có 2 nghiệm (x,y) = {(1;1) ; (0;0)}
a, Ta có : \(a=1;b=-2m;c=m+2\)
a, Để phương trình có 2 nghiệm ko âm nên : \(\hept{\begin{cases}\Delta\ge0\\S>0\\P>0\end{cases}}\)
hay \(\Delta=\left(-2m\right)^2-4\left(m+2\right)=4m^2-4m-8=\left(2m+1\right)^2-9\)
mà \(\Delta\ge0\Rightarrow\left(2m-1\right)^2-9\ge0\Rightarrow m\ge2\)
\(S>0\)mà \(S=x_1+x_2=-\frac{b}{a}\Rightarrow S=-\frac{b}{a}=2m\Rightarrow2m>0\Rightarrow m>0\)
\(P>0\)mà \(P=x_1x_2=\frac{c}{a}\Rightarrow P=\frac{c}{a}=m+2\Rightarrow m+2>0\Rightarrow m>-2\)
\(\Rightarrow\hept{\begin{cases}\Delta\ge0\\S>0\\P>0\end{cases}}\Rightarrow m\ge2\)Vậy ta có đpcm
b, Theo hệ thức Vi et : \(\hept{\begin{cases}S=-\frac{b}{a}\\P=\frac{c}{a}\end{cases}\Rightarrow\hept{\begin{cases}S=2m\\P=m+2\end{cases}}}\)
Theo bài ra ta có : \(E=\sqrt{x_1}+\sqrt{x_2}\Rightarrow E^2=\left(\sqrt{x_1}+\sqrt{x_2}\right)^2\)
\(=x_1+2\sqrt{x_1x_2}+x_2=\left(x_1+x_2\right)+2\sqrt{x_1x_2}\)
\(\Rightarrow2m+2\sqrt{m+2}=2m+\sqrt{4m+8}\)
\(\Rightarrow E=\sqrt{2m+\sqrt{4m+8}}\)
Áp dụng BĐT Bunyakovsky ta có:
\(\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\le\left(1^2+1^2+1^2\right)\left(a+b+b+c+c+a\right)\)
\(=3\cdot2\left(a+b+c\right)=3\cdot2\cdot6=36\)
\(\Rightarrow\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le6\)
Dấu "=" xảy ra khi: a = b = c = 2
Ta có: \(a^2+2b^2+3=a^2+b^2+b^2+1+2\)
\(a^2+2b^2+3\ge2\left(ab+b+1\right)\)
\(\Rightarrow\frac{1}{a^2+2b^2+3}\le\frac{1}{2\left(ab+b+1\right)}\)
Tương tự:\(\frac{1}{b^2+2c^2+3}\le\frac{1}{2\left(bc+c+1\right)}\);\(\frac{1}{c^2+2a^2+3}\le\frac{1}{2\left(ca+a+1\right)}\)
\(\Rightarrow VT\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)\)
\(\Rightarrow VT\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{abc}{bc+b+abc}+\frac{b}{abc+ab+b}\right)\)
\(\Rightarrow VT\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{ab+b+1}+\frac{b}{ab+b+1}\right)\)
\(\Rightarrow VT\le\frac{1}{2}.\frac{ab+b+1}{ab+b+1}=\frac{1}{2}\)
Dấu "=" xảy ra\(\Leftrightarrow a=b=c=1\)
đk: \(x^2-3x+1\le0\)
Ta có: \(x^4+x^2+1=\left(x^4+2x^2+1\right)-x^2=\left(x^2+1\right)^2-x^2\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\)
Đặt \(\hept{\begin{cases}\sqrt{x^2-x+1}=a>0\\\sqrt{x^2+x+1}=b>0\end{cases}}\) khi đó \(2a^2-b^2=x^2-3x+1\)
Thay vào ta được phương trình tương đương:
\(2a^2-b^2=-\frac{\sqrt{3}}{3}ab\)
\(\Leftrightarrow\left(2a^2-b^2\right)^2=\frac{1}{3}a^2b^2\)
\(\Leftrightarrow3\left(4a^4-4a^2b^2+b^4\right)=a^2b^2\)
\(\Leftrightarrow12a^4-13a^2b^2+3b^4=0\)
\(\Leftrightarrow\left(12a^4-4a^2b^2\right)-\left(9a^2b^2-3b^4\right)=0\)
\(\Leftrightarrow4a^2\left(3a^2-b^2\right)-3b^2\left(3a^2-b^2\right)=0\)
\(\Leftrightarrow\left(4a^2-3b^2\right)\left(3a^2-b^2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4a^2-3b^2=0\\3a^2-b^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}4a^2=3b^2\\3a^2=b^2\end{cases}}\Leftrightarrow\orbr{\begin{cases}4\left(x^2-x+1\right)=3\left(x^2+x+1\right)\\3\left(x^2-x+1\right)=x^2+x+1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-7x+1=0\\2x^2-4x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7\pm3\sqrt{5}}{2}\left(ktm\right)\\x=1\left(tm\right)\end{cases}}\)
Vậy x = 1
Ta có : \(\frac{a^2-bc}{a}+\frac{b^2-ac}{b}+\frac{c^2-ab}{c}=0\)
=> \(a-\frac{bc}{a}+b-\frac{ac}{b}+c-\frac{ab}{c}=0\)
=> \(a+b+c=\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\)
=> \(a+b+c=abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
=> \(\frac{a+b+c}{abc}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
=> \(\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
=> \(\frac{2}{bc}+\frac{2}{ac}+\frac{2}{ab}=\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}\)
=> \(\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}-\frac{2}{bc}-\frac{2}{ac}-\frac{2}{ac}=0\)
=> \(\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)+\left(\frac{1}{a^2}-\frac{2}{ac}+\frac{1}{c^2}\right)+\left(\frac{1}{b^2}-\frac{1}{bc}+\frac{1}{c^2}\right)=0\)
=> \(\left(\frac{1}{a}-\frac{1}{b}\right)^2+\left(\frac{1}{a}-\frac{1}{c}\right)^2+\left(\frac{1}{b}-\frac{1}{c}\right)^2=0\)
=> \(\hept{\begin{cases}\frac{1}{a}-\frac{1}{b}=0\\\frac{1}{a}-\frac{1}{c}=0\\\frac{1}{b}-\frac{1}{c}=0\end{cases}}\Rightarrow\hept{\begin{cases}\frac{1}{a}=\frac{1}{b}\\\frac{1}{a}=\frac{1}{c}\\\frac{1}{b}=\frac{1}{c}\end{cases}}\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\left(\text{đpcm}\right)\)