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Đặt: \(\sqrt{x}=a\)
\(Taco:a^2-8a-9=0\Leftrightarrow a\left(a-8\right)-9=0\Leftrightarrow a\left(a-8\right)=9=1.9\)
\(\Leftrightarrow a=9\Leftrightarrow x=9^2=81\)
\(x-8\sqrt{x}-9=0\)
\(\Leftrightarrow\left(\sqrt{x}-9\right)\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=9\Leftrightarrow x=81\\\sqrt{x}=-1\left(loại\right)\end{cases}}\)
Vậy x = 81
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\(\left(x-2\right)^2-\left(-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)-\left(-x+1\right)=0\)
\(\Leftrightarrow x^2-2x-2x+4+x-1=0\)
\(\Leftrightarrow x^2-3x+3=0\)( vô nghiệm )
\(\left(x-2\right)^2-\left(-x+1\right)=0\)
\(< =>\left(x-2\right)^2-\left(1-x\right)=0\)
\(< =>x^2-4x+4-1+x=0\)
\(< =>x^2-3x-3=0\)(vô nghiệm)
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x^2-x+1/4=0
x^2-2x.1/2+(1/2)^2-(1/2)^2+1/4=0
(x-1/2)^2=0
x-1/2=0
x=1/2
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\(P=-x^2+6x+1=-\left(x^2-6x+9\right)+10=-\left(x-3\right)^2+10\le10\)Vậy \(Max_P=10\) khi \(x-3=0\Rightarrow x=3\)
b, \(P=-x^2+6x+1=-\left(x^2-6x-1\right)\)
\(=-\left(x^2-3x-3x+9-10\right)\)
\(=-\left[\left(x-3\right)^2-10\right]\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(x-3\right)^2\ge0\Rightarrow\left(x-3\right)^2-10\ge-10\)
\(\Rightarrow-\left[\left(x-3\right)^2-10\right]\ge10\)
Hay \(P\ge10\) với mọi giá trị của \(x\in R\).
Để \(P=10\) thì \(-\left[\left(x-3\right)^2-10\right]=10\)
\(\Rightarrow\left(x-3\right)^2=0\Rightarrow x=3\)
Vậy.....
Chúc bạn học tốt!!!
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\(5-9x^2=0\)
\(\Leftrightarrow9x^2=5\)
\(\Leftrightarrow x^2=\dfrac{5}{9}\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{5}{9}}\\x=-\sqrt{\dfrac{5}{9}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}}{3}\\x=-\dfrac{\sqrt{5}}{3}\end{matrix}\right.\)
\(x^2+x+\dfrac{1}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=0\)
\(\Rightarrow x+\dfrac{1}{2}=0\Rightarrow x=-\dfrac{1}{2}\)
Học tốt nha<3
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ĐKXĐ: \(x\ge0\)
Đặt \(\sqrt{x}=a\)
\(\Rightarrow a^2-2a-1=0\)
\(\Rightarrow\left(a-1\right)^2=2\)
\(\Rightarrow\orbr{\begin{cases}a-1=\sqrt{2}\\a-1=-\sqrt{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}a=\sqrt{2}+1\\a=-\sqrt{2}+1\end{cases}\Leftrightarrow}\orbr{\begin{cases}\sqrt{x}=\sqrt{2}+1\\\sqrt{x}=-\sqrt{2}+1< 0\left(v\text{ô}l\text{ý}\right)\end{cases}}}\Leftrightarrow x=\left(\sqrt{2}+1\right)^2=3+2.\sqrt{2}\)Vậy \(x=3+2.\sqrt{2}\)
P/S: Không chắc lắm