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\(0< a< 2\Rightarrow a\left(a-2\right)< 0\Rightarrow a^2< 2a\)
Tương tự: \(\left\{{}\begin{matrix}b\left(b-2\right)< 0\\c\left(c-2\right)< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b^2< 2b\\c^2< 2c\end{matrix}\right.\)
\(\Rightarrow a^2+b^2+c^2< 2\left(a+b+c\right)\)
\(\Rightarrow a^2+b^2+c^2< 2.3=6\)
\(P=2x^4+3x^2y^2+y^4+y^2\)
\(=2x^4+2x^2y^2+x^2y^2+y^4+y^2\)
\(=2x^2\left(x^2+y^2\right)+y^2\left(x^2+y^2\right)+y^2\)
\(=2x^2+y^2+y^2=2\left(x^2+y^2\right)=2\)
Ta có :
\(P\left(x\right)=2x^4+3x^2y^2+y^4+y^2\)
\(\Rightarrow P\left(x\right)=x^4+2x^2y^2+y^4+x^4+x^2y^2+y^2\)
\(\Rightarrow P\left(x\right)=\left(x^2+y^2\right)^2+x^2\left(x^2+y^2\right)+y^2\)
\(\Rightarrow P\left(x\right)=1^2+x^2.1+y^2\) Vì \(\left(x^2+y^2=1\right)\)
\(\Rightarrow P\left(x\right)=1^2+x^2+y^2=1+1=2\)
Vậy \(P\left(x\right)=2\)
Ta có:
\(M=\left(y-5\right)\left(y+8\right)-\left(y+4\right)\left(y-1\right)\\ =\left(y^2-5y+8y-40\right)-\left(y^2+4y-y-4\right)\\ =y^2+3y-40-y^2-3y+4\\ =-36\)
=> Giá trị của bt không phụ thuộc vào biến y
\(M=\left(y-5\right)\left(y+8\right)-\left(y+4\right)\left(y-1\right)\)
\(=y^2+8y-5y-40-\left(y^2-y+4y-4\right)\)
\(=y^2+3y-40-y^2-3y+4\)
=-36
=>M không phụ thuộc vào biến
\(\left(x+2\right)^2-2\left(x+2\right)\left(2x-3\right)+\left(2x-3\right)^2=25\\ < =>\left[\left(x+2\right)-\left(2x-3\right)\right]^2=25\\ < =>\left(x+2-2x+3\right)^2-25=0\\ < =>\left(-x+5\right)^2-5^2=0\\ < =>\left(-x+5-5\right)\left(-x+5+5\right)=0\\ < =>-x\left(-x+10\right)=0\\ < =>x\left(x-10\right)=0\\ < =>\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)
Vậy: ...
\(\left(x+2\right)^2-2\left(x+2\right)\left(2x-3\right)+\left(2x-3\right)^2=25\\
\Leftrightarrow\left(x+2-2x+3\right)^2=5^2\\\Leftrightarrow\left(-x+5\right)^2=5^2\\
\Leftrightarrow\left[{}\begin{matrix}-x+5=5\\-x+5=-5\end{matrix}\right.\\
\Leftrightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)
Vậy...
a:
ĐKXĐ: \(x\notin\left\{1;-3\right\}\)
\(A=\left(\dfrac{2x^2+1}{x^3-1}-\dfrac{1}{x-1}\right):\left(1-\dfrac{x^2-2}{x^2+x+1}\right)\)
\(=\left(\dfrac{2x^2+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{1}{x-1}\right):\dfrac{x^2+x+1-x^2+2}{x^2+x+1}\)
\(=\dfrac{2x^2+1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x+3}\)
\(=\dfrac{x^2-x}{\left(x-1\right)}\cdot\dfrac{1}{x+3}=\dfrac{x}{x+3}\)
b: |x-5|=2
=>\(\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)
Khi x=7 thì \(A=\dfrac{7}{7+3}=\dfrac{7}{10}\)
c: Để A nguyên thì \(x⋮x+3\)
=>\(x+3-3⋮x+3\)
=>\(-3⋮x+3\)
=>\(x+3\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{-2;-4;0;-6\right\}\)
\(\sqrt{\dfrac{2\left(4-\sqrt{7}\right)}{2}}-\sqrt{\dfrac{2\left(4+\sqrt{7}\right)}{2}}+\sqrt{2}=?\)
\(\sqrt{\dfrac{2\left(4-\sqrt{7}\right)}{2}}-\sqrt{\dfrac{2\left(4+\sqrt{7}\right)}{2}}+\sqrt{2}\)
\(=\dfrac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}-\dfrac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}+\sqrt{2}\)
\(=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}+\sqrt{2}=-\dfrac{2}{\sqrt{2}}+\dfrac{2}{\sqrt{2}}=0\)
\(\sqrt{\dfrac{8-2\sqrt{7}}{2}}-\sqrt{\dfrac{8+2\sqrt{7}}{2}}+\sqrt{2}\)
\(=\sqrt{\dfrac{\left(\sqrt{7}-1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}+1\right)^2}{2}}+\sqrt{2}\)
\(=\dfrac{\left|\sqrt{7}-1\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{7}+1\right|}{\sqrt{2}}+\sqrt{2}\)
\(=\dfrac{-2}{\sqrt{2}}+\sqrt{2}=0\)
\(M=-x^2+6x-11\\ =-\left(x^2-6x+9\right)+9-11\\ =-\left(x-3\right)^2-2\)
Ta thấy: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-3\right)^2\le0\forall x\\ \Rightarrow-\left(x-3\right)^2-2\le-2\forall x\\ \Rightarrow M\le-2\forall x\)
Dấu "=" xảy ra khi: \(x-3=0\Leftrightarrow x=3\)
Vậy \(M_{max}=-2\Leftrightarrow x=3\).
`M = -x^2 + 6x - 11`
`= -(x^2 - 6x + 11) `
`= -(x^2 - 2.3x + 3^2 + 2)`
`= -(x^2 - 2.3x + 3^2) - 2`
`= -(x-3)^2 - 2`
Do `(x-3)^2 ≥ 0` `∀x` thuộc `R`
`=> -(x-3)^2 ≤ 0` `∀x` thuộc `R`
`=> -(x-3)^2 - 2 ≤ -2` ` ∀x` thuộc `R`
Hay `M ≤ -2` `∀x` thuộc `R`
Dấu `=` có khi:
`(x-3)^2 = 0`
`<=> x - 3 = 0`
`<=> x = 3`
Vậy `M_(max) = -2 <=> x = 3`
\(x^2+4y^2-2xy+2x-14y+9=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+2\left(x-y\right)+3y^2-12y+12-3=0\)
\(\Leftrightarrow\left(x-y\right)^2+2\left(x-y\right)+1+3\left(y-2\right)^2-4=0\)
\(\Leftrightarrow\left(x-y+1\right)^2+3\left(y-2\right)^2=4\) (1)
Do \(\left(x-y+1\right)^2\ge0;\forall x;y\)
\(\Rightarrow3\left(y-2\right)^2\le4\)
\(\Rightarrow\left(y-2\right)^2\le\dfrac{4}{3}\Rightarrow\left[{}\begin{matrix}\left(y-2\right)^2=0\\\left(y-2\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y=2\\y=3\\y=1\end{matrix}\right.\)
Thế vào (1):
Với \(y=1\) \(\Rightarrow x^2=1\Rightarrow x=\pm1\)
Với \(y=2\Rightarrow\left(x-1\right)^2=4\Rightarrow x=\left\{3;-1\right\}\)
Với \(y=3\Rightarrow\left(x-2\right)^2=1\Rightarrow x=\left\{3;1\right\}\)
Vậy \(\left(x;y\right)=\left(-1;1\right);\left(1;1\right);\left(-1;2\right);\left(3;2\right);\left(1;3\right);\left(3;3\right)\)