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![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\dfrac{x}{1-x^2}+\dfrac{y}{1-y^2}+\dfrac{z}{1-z^2}\)
Ta có: \(2x^2.\left(1-x^2\right)\left(1-x^2\right)\le\dfrac{1}{27}\left(2x^2+1-x^2+1-x^2\right)^3=\dfrac{8}{27}\)
\(\Rightarrow x^2\left(1-x^2\right)^2\le\dfrac{4}{27}\)
\(\Rightarrow x\left(1-x^2\right)\le\dfrac{2}{3\sqrt{3}}\)
\(\Rightarrow\dfrac{x}{1-x^2}\ge\dfrac{3\sqrt{3}}{2}x^2\)
Tương tự và cộng lại:
\(P\ge\dfrac{3\sqrt{3}}{2}\left(x^2+y^2+z^2\right)=...\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow\left(5x-5\right)-\left(2\sqrt{2x^2+5x-3}-4\right)-\left(x\sqrt{2x-1}-x\right)+\left(2x\sqrt{x+3}-4x\right)=0\\ \Leftrightarrow5\left(x-1\right)-\dfrac{2\left(2x+7\right)\left(x-1\right)}{\sqrt{2x^2+5x-3}+2}-\dfrac{x\left(2x-2\right)}{\sqrt{2x-1}+1}+\dfrac{2x\left(x-1\right)}{\sqrt{x+3}+2}=0\\ \Leftrightarrow\left(x-1\right)\left(5-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+2}-\dfrac{x}{\sqrt{2x-1}+1}+\dfrac{2x}{\sqrt{x+3}+2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\5-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+2}-\dfrac{x}{\sqrt{2x-1}+1}+\dfrac{2x}{\sqrt{x+3}+2}=0\left(1\right)\end{matrix}\right.\)
Với \(x\ge\dfrac{1}{2}\Leftrightarrow\left(1\right)< 0\)
Do đó PT có nghiệm x=1
ĐKXĐ: \(x\ge\dfrac{1}{2}\)
\(2x-2\sqrt{\left(2x-1\right)\left(x+3\right)}-\left(1+x\sqrt{2x-1}-2x\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left(2x-1\right)-2\sqrt{\left(2x-1\right)\left(x+3\right)}-x\sqrt{2x-1}+2x\sqrt{x+3}=0\)
\(\Leftrightarrow\sqrt{2x-1}\left(\sqrt{2x-1}-x\right)-2\sqrt{x+3}\left(\sqrt{2x-1}-x\right)=0\)
\(\Leftrightarrow\left(\sqrt{2x-1}-x\right)\left(\sqrt{2x-1}-2\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x-1}=x\\\sqrt{2x-1}=2\sqrt{x+3}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x^2\\2x-1=4x+12\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\2x=-13\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-\dfrac{13}{2}\left(ktm\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì \(x=\sqrt{\dfrac{1}{2\sqrt{3}-2}-\dfrac{3}{2\sqrt{3}+2}}=\dfrac{\sqrt{3}-1}{2}\)
\(\Rightarrow x=\dfrac{\sqrt{3}-1}{2}\) là nghiệm của đa thức \(2x^2+2x-1\)
\(\Rightarrow B=\dfrac{2x^{2017}\left(2x^2+2x-1\right)+2x+1}{\left(2x^2+2x-1\right)+x+1}=\dfrac{2x+1}{x+1}=3-\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a. Số tiền phạt cho $20$ kg hành lý quá cước là:
$T=\frac{4}{5}.20+20=36$ (USD)
b.
$651980$ VNĐ = $\frac{651980}{23285}=28$ USD
Ta có: $28=\frac{4}{5}M+20$
$8=\frac{4}{5}M$
$M=10$ (kg)
Vậy khối lượng hành lý quá cước là $10$ kg.
![](https://rs.olm.vn/images/avt/0.png?1311)
c: Phương trình hoành độ giao điểm của \(\left(d1\right),\left(d2\right)\) là:
x-2=3x-4
\(\Leftrightarrow x-3x=-4+2\)
\(\Leftrightarrow-2x=-2\)
hay x=1
Thay x=1 vào y=x-2, ta được:
y=1-2=-1
Thay x=1 và y=-1 vào \(\left(d\right)\), ta được:
\(m^2-3m+1+m-1=-1\)
\(\Leftrightarrow m^2-2m+1=0\)
\(\Leftrightarrow m-1=0\)
hay m=1
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Theo đề, ta có:
BH+CH=25(cm)
hay BH=25-CH
Ta có: \(AH^2=HB\cdot HC\)
\(\Leftrightarrow HC\left(HC-25\right)=-144\)
\(\Leftrightarrow HC=16\left(cm\right)\)
\(\Leftrightarrow HB=9\left(cm\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}AB=\sqrt{9\cdot25}=15\left(cm\right)\\AC=\sqrt{16\cdot25}=20\left(cm\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2.
\(x^2-\left(m-2\right)x+3m-1=0\)
\(\Delta=\left(m-2\right)^2-4\left(3m-1\right)=m^2-16m+8\)
PT có hai nghiệm \(x_1,x_2\)\(\Leftrightarrow\Delta\ge0\Leftrightarrow m^2-16m+8\ge0\)
Khi đó, áp dụng hệ thức Vi - et ta có: \(\left\{{}\begin{matrix}x_1+x_2=m-2\\x_1x_2=3m-1\end{matrix}\right.\)
Để \(\left(m+1\right)x_1+x_2^2+3x_2=10\Leftrightarrow\left(m-2\right)x_1+3x_1+x_2^2+3x_2=10\)
\(\Leftrightarrow\left(x_1+x_2\right)x_1+x_2^2+3\left(x_1+x_2\right)=10\)
\(\Leftrightarrow x_1^2+x_2^2+x_1x_2+3\left(x_1+x_2\right)=10\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-x_1x_2+3\left(x_1+x_2\right)=10\)
\(\Leftrightarrow\left(m-2\right)^2-\left(3m-1\right)+3\left(m-2\right)=10\)
\(\Leftrightarrow m^2-4m-11=0\Leftrightarrow\left[{}\begin{matrix}m=2+\sqrt{15}\left(L\right)\\m=2-\sqrt{15}\left(TM\right)\end{matrix}\right.\)
Bài 8.
\(x^2-2x+m-3=0\)
\(\Delta=4-4\left(m-3\right)=-4m+16\)
a) Phương trình có 2 nghiệm phân biệt \(\Leftrightarrow\Delta>0\Leftrightarrow-4m+16>0\Leftrightarrow m< 4\)
b) Áp dụng hệ thức Vi-et ta có: \(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=m-3\end{matrix}\right.\)
Để \(x_1^2-2x_2+x_1x_2=16\Leftrightarrow x_1^2-\left(x_1+x_2\right)x_2+x_1x_2=16\)
\(\Leftrightarrow x_1^2-x_2^2=16\) \(\Leftrightarrow\left(x_1+x_2\right)\left(x_1-x_2\right)=16\Leftrightarrow x_1-x_2=8\)
Mà \(x_1+x_2=2\) \(\Rightarrow x_1=5;x_2=-3\Rightarrow x_1x_2=m-3=-15\Rightarrow m=-12\) (TM).