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a) \(2x^2+2x+1=0\)
\(\Rightarrow2x^2+2x=-1\)
\(\Rightarrow2x\left(x+1\right)=-1\)
⇒ Pt vô nghiệm
a: \(2x^2+2x+1=0\)
\(\text{Δ}=2^2-4\cdot2\cdot1=4-8=-4< 0\)
Vì Δ<0 nên phương trình vô nghiệm
2:
a: =>x-1=0 hoặc 3x+1=0
=>x=1 hoặc x=-1/3
b: =>x-5=0 hoặc 7-x=0
=>x=5 hoặc x=7
c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)
d: =>x=0 hoặc x^2-1=0
=>\(x\in\left\{0;1;-1\right\}\)
Bài 2:
a: Ta có: \(2x^2+y^2-2xy+x+2=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{7}{4}=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\left(vôlý\right)\)
b: Ta có: \(-x^2-26y^2+10xy-20y-150=0\)
\(\Leftrightarrow x^2-10xy+25y^2+y^2+20y+100+50=0\)
\(\Leftrightarrow\left(x-5y\right)^2+\left(y+10\right)^2+50=0\left(vôlý\right)\)
Bài 1:
\(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\Leftrightarrow2\left(ab+bc+ca\right)=0-1=-1\)hay \(ab+bc+ca=-\dfrac{1}{2}\Leftrightarrow\left(ab+bc+ca\right)^2=\dfrac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2=\dfrac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\dfrac{1}{4}\)Ta có: \(P=a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)=1-2.\dfrac{1}{4}=\dfrac{1}{2}\)Vậy \(P=\dfrac{1}{2}\)
\(5x^2+10y^2-6xy-4x-10y+14\)
\(=\left(4x^2-4x+1\right)+\left(x^2-6xy+9y^2\right)+\left(y^2-10y+25\right)-12\)
\(=\left(2x-1\right)^2+\left(x-3y\right)^2+\left(y-5\right)^2-12\ge-12\) đề có nhầm không bạn?
a, \(2x\left(x-3\right)-15+5x=0\\ \Rightarrow2x\left(x-3\right)-\left(15-5x\right)=0\\ \Rightarrow2x\left(x-3\right)-5\left(3-x\right)=0\\ \Rightarrow\left(2x+5\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=3\end{matrix}\right.\)
b, \(x^3-7x=0\\ \Rightarrow x\left(x^2-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm7\end{matrix}\right.\)
c, \(\left(2x-3\right)^2-\left(x+5\right)^2=0\\ \Rightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\\ \Rightarrow\left(x-8\right)\left(3x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Xem lại đề câu d
\(x^2-6x+9=-y^2-10y-20.\)
\(\left(x-3\right)^2=-y^2-10y-20\)
\(\left(x-3\right)^2=-y^2-10y-20\)
\(\left(x-3\right)^2=-\left(y^2+2.5y+25\right)+5\)
\(\left(x-3\right)^2=-\left(y+5\right)^2+5\)
\(\hept{\begin{cases}x=3\\y+5=\sqrt{5}\Leftrightarrow y=\sqrt{5}-5\end{cases}}\)
b)
\(\left(4x^2-4x+1\right)=-y^2-x^2-2xy\)
\(\left(2x-1\right)^2=-\left(y+x\right)^2\)
\(x=\frac{1}{2}\Leftrightarrow y=-\frac{1}{2}\)