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1/
\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=0\\y-2=0\\x+y+z=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=2\\z=-3\end{matrix}\right.\)
2/ \(P=\sqrt{\left(5x-2\right)^2}+\sqrt{\left(3-5x\right)^2}\)
\(P=\left|5x-2\right|+\left|3-5x\right|\ge\left|5x-2+3-5x\right|=1\)
\(\Rightarrow P_{min}=1\) khi \(\frac{2}{5}\le x\le\frac{3}{5}\)
3/ ĐKXĐ: \(\left|x\right|\ge1\)
\(x^2-1-\sqrt{x^2-1}=0\)
\(\Leftrightarrow\sqrt{x^2-1}\left(\sqrt{x^2-1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-1}=0\\\sqrt{x^2-1}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\x^2-1=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm1\\x=\pm\sqrt{2}\end{matrix}\right.\)
Theo đề bài, ta có:
x3+y3=x2−xy+y2x3+y3=x2−xy+y2
hay (x2−xy+y2)(x+y−1)=0(x2−xy+y2)(x+y−1)=0
⇒\orbr{x2−xy+y2=0x+y=1⇒\orbr{x2−xy+y2=0x+y=1
+ Với x2−xy+y2=0⇒x=y=0⇒P=52x2−xy+y2=0⇒x=y=0⇒P=52
+ với x+y=1⇒0≤x,y≤1⇒P≤1+√12+√0+2+√11+√0=4x+y=1⇒0≤x,y≤1⇒P≤1+12+0+2+11+0=4
Dấu đẳng thức xảy ra <=> x=1;y=0 và P≥1+√02+√1+2+√01+√1=43P≥1+02+1+2+01+1=43
Dấu đẳng thức xảy ra <=> x=0;y=1
Vậy max P=4 và min P =4/3
1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4
--> Pmin=4 khi x=4
2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1
=> M=2x2-8x+\(\sqrt{x^2-4x+5}\)+6=2(t2-5)+t+6
<=> M=2t2+t-4\(\ge\)2.12+1-4=-1
Mmin=-1 khi t=1 hay x=2
\(M=\sqrt{x^2+y^2-2xy+2x-2y+10}+2y^2-8y+2024\\ =\sqrt{\left(x^2+y^2+1-2xy+2x-2y\right)+9}+\left(2y^2-8y+8\right)+2016\\ =\sqrt{\left(x-y+1\right)^2+9}+2\left(y^2-4y+4\right)+2016\\ =\sqrt{\left(x-y+1\right)^2+9}+2\left(y-2\right)^2+2016\) \(\text{Do }\left(x-y+1\right)^2\ge0\forall x;y\\ \Rightarrow\left(x-y+1\right)^2+9\ge9\forall x;y\\ \Rightarrow\sqrt{\left(x-y+1\right)^2+9}\ge3\forall x;y\\ Mà\text{ }2\left(y-2\right)^2\ge0\forall y\\ \Rightarrow\sqrt{\left(x-y+1\right)^2+9}+2\left(y-2\right)^2\ge3\forall x;y\\ M=\sqrt{\left(x-y+1\right)^2+9}+2\left(y-2\right)^2+2016\ge2019\forall x;y\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}2\left(y-2\right)^2=0\\\left(x-y+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=0\\x-y+1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=2\\x=y-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)
Vậy \(M_{Min}=2019\) khi \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
\(Q=\sqrt{25x^2-20x+4}+\sqrt{25x^2-30x+9}\\ =\sqrt{\left(5x-2\right)^2}+\sqrt{\left(5x-3\right)^2}\\ =\left|5x-2\right|+\left|5x-3\right|\\ =\left|5x-2\right|+\left|3-5x\right|\)
Áp dụng BDT: \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)
\(\Rightarrow\left|5x-2\right|+\left|3-5x\right|\ge\left|5x-2+3-5x\right|=\left|1\right|=1\)
Dấu "=" xảy ra khi:
\(\left(5x-2\right)\left(3-5x\right)\ge0\\\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}5x-2\ge0\\3-5x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}5x-2\le0\\3-5x\le0\end{matrix}\right.\end{matrix}\right. \) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}5x\ge2\\5x\le3\end{matrix}\right.\\\left\{{}\begin{matrix}5x\le2\\5x\ge3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{2}{5}\\x\le\dfrac{3}{5}\end{matrix}\right.\left(T/m\right)\\\left\{{}\begin{matrix}x\le\dfrac{2}{5}\\x\ge\dfrac{3}{5}\end{matrix}\right.\left(K^0\text{ }T/m\right)\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{2}{5}\le x\le\dfrac{3}{5}\)
Vậy \(Q_{Min}=1\) khi \(\dfrac{2}{5}\le x\le\dfrac{3}{5}\)
\(a,\dfrac{x^2+x+2}{\sqrt{x^2+x+1}}=\dfrac{x^2+x+1+1}{\sqrt{x^2+x+1}}=\sqrt{x^2+x+1}+\dfrac{1}{\sqrt{x^2+x+1}}\left(1\right)\)
Áp dụng BĐT cosi: \(\left(1\right)\ge2\sqrt{\sqrt{x^2+x+1}\cdot\dfrac{1}{\sqrt{x^2+x+1}}}=2\)
Dấu \("="\Leftrightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
câu a) rút x theo y thế vào A rồi áp dụng HĐT
b)rút xy thế vào B
c)HĐT
d)rút x theo y thé vào C
rồi dùng BĐT cô-si
e)BĐT chưa dấu giá trị tuyệt đối