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\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\\ =\left(\dfrac{\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\\ =\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\\ =\dfrac{x-1}{\sqrt{x}}\)
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3: Ta có: A=B|x-4|
\(\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}-5}:\dfrac{1}{\sqrt{x}-5}=\left|x-4\right|\)
\(\Leftrightarrow\left|x-4\right|=\sqrt{x}-2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=\sqrt{x}-2\left(x\ge4;x\ne25\right)\\x-4=2-\sqrt{x}\left(0< x< 4\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{x}-2=0\\x+\sqrt{x}-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=4\left(loại\right)\end{matrix}\right.\)
1: Thay x=9 vào A, ta được:
\(A=\dfrac{3-2}{3-5}=\dfrac{-1}{-2}=\dfrac{1}{2}\)
2: Ta có: \(B=\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{x-25}\)
\(=\dfrac{3\sqrt{x}-15+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{1}{\sqrt{x}-5}\)
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Đặt \(a=\dfrac{x}{2};b=\dfrac{y}{2};c=\dfrac{z}{2}\). Khi đó \(xyz=1\).
Bất đẳng thức cần chứng minh trở thành:
\(\sum\dfrac{1}{\sqrt{8x^3+1}}\ge1\)
Ta có: \(\sum\dfrac{1}{\sqrt{8x^3+1}}=\sum\sqrt{\dfrac{b^3c^3}{8+b^3c^3}}=\sum\dfrac{b^2c^2}{\sqrt{8bc+b^4c^4}}\ge\dfrac{\left(ab+bc+ca\right)^2}{\sum\sqrt{8bc+b^4c^4}}=\dfrac{\sum b^2c^2+2\sum a}{\sum\sqrt{8bc+b^4c^4}}\ge\dfrac{\sum b^2c^2+6}{\sum\sqrt{8bc+b^4c^4}}\)
Phép chứng minh sẽ hoàn tất nếu ta chứng minh được:
\(\sum b^2c^2+6\ge\sum\sqrt{8bc+b^4c^4}\left(\cdot\right)\)
Để ý rằng, nếu ta chứng minh được \(b^2c^2+2\ge\sqrt{8bc+b^4c^4}\left(1\right)\) thì ta sẽ chứng minh được (*).
Thật vậy, bằng phép biến đổi tương đương, ta có:
\(b^2c^2+2\ge\sqrt{8bc+b^4c^4}\)
\(\Leftrightarrow b^4c^4+4b^2c^2+4\ge8bc+b^4c^4\)
\(\Leftrightarrow4\left(bc-1\right)^2\ge0\) (luôn đúng).
Vậy nhận xét (1) là đúng. Từ đây ta có điều phải chứng minh.
Dấu "=" xảy ra khi \(a=b=c=2\)
Đặt \(A=\dfrac{1}{\sqrt{a^3+1}}+\dfrac{1}{\sqrt{b^3+1}}+\dfrac{1}{\sqrt{c^3+1}}\)
\(\sqrt{a^3+1}=\sqrt{\left(a+1\right)\left(a^2-a+1\right)}\)
=>\(\sqrt{a^3+1}< =\dfrac{a+1+a^2-a+1}{2}=\dfrac{a^2+2}{2}\)
=>\(\dfrac{1}{\sqrt{a^3+1}}>=\dfrac{2}{a^2+2}\)
Chứng minh tương tự, ta được:
\(\sqrt{b^3+1}< =\dfrac{b^2+2}{2}\) và \(\sqrt{c^3+1}< =\dfrac{c^2+2}{2}\)
=>\(\dfrac{1}{\sqrt{b^3+1}}>=\dfrac{2}{b^2+2};\dfrac{1}{\sqrt{c^3+1}}>=\dfrac{2}{c^2+2}\)
=>\(\dfrac{1}{\sqrt{a^3+1}}+\dfrac{1}{\sqrt{b^3+1}}+\dfrac{1}{\sqrt{c^3+1}}>=\dfrac{2}{a^2+2}+\dfrac{2}{b^2+2}+\dfrac{2}{c^2+2}\)
=>\(A>=\dfrac{2}{a^2+2}+\dfrac{2}{b^2+2}+\dfrac{2}{c^2+2}\)
=>\(A>=\dfrac{2\left(b^2+2\right)\left(c^2+2\right)+2\left(a^2+2\right)\left(c^2+2\right)+2\left(b^2+2\right)\left(a^2+2\right)}{\left(a^2+2\right)\left(c^2+2\right)\left(b^2+2\right)}\)
=>\(A>=\dfrac{2\left(b^2c^2+2c^2+2b^2+4+a^2c^2+2a^2+2c^2+4+a^2b^2+2b^2+2a^2+4\right)}{\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)}\)
=>\(A>=\dfrac{2\left(a^2b^2+b^2c^2+c^2a^2\right)+8\left(a^2+b^2+c^2\right)+24}{\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)}\)
=>\(A>=\dfrac{2\left(a^2b^2+b^2c^2+a^2c^2\right)+4\left(a^2+b^2+c^2\right)+24+4\left(a^2+b^2+c^2\right)}{2\left(a^2b^2+b^2c^2+a^2c^2\right)+4\left(a^2+b^2+c^2\right)+8+8^2}\)(Vì abc=8)
Ta lại có: \(a^2+b^2+c^2>=3\sqrt[3]{a^2b^2c^2}=3\cdot\sqrt[3]{8^2}=12\)
=>\(4\left(a^2+b^2+c^2\right)>=48\)
=>\(A>=\dfrac{2\left(a^2b^2+b^2c^2+a^2c^2\right)+4\left(a^2+b^2+c^2\right)+24+48}{2\left(a^2b^2+b^2c^2+a^2c^2\right)+4\left(a^2+b^2+c^2\right)+8+64}\)
=>A>=1(ĐPCM)
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\(\left(xy+3\right)^2+\left(x+y\right)^2=8\)
\(\Leftrightarrow x^2y^2+x^2+y^2+1=-8xy\)
\(\dfrac{x}{x^2+1}+\dfrac{y}{y^2+1}=-\dfrac{1}{4}\Leftrightarrow\dfrac{\left(xy+1\right)\left(x+y\right)}{x^2y^2+x^2+y^2+1}=-\dfrac{1}{4}\)
\(\Rightarrow\dfrac{\left(xy+1\right)\left(x+y\right)}{-8xy}=-\dfrac{1}{4}\)
\(\Rightarrow\left(xy+1\right)\left(x+y\right)=2xy\)
\(\Rightarrow x+y=\dfrac{2xy}{xy+1}\)
Thế vào pt ban đầu:
\(\left(xy+3\right)^2+\left(\dfrac{2xy}{xy+1}\right)^2=8\)
Đặt \(xy+1=t\Rightarrow\left(t+2\right)^2+4\left(\dfrac{t-1}{t}\right)^2=8\)
\(\Rightarrow\left(t^2+2t\right)^2-4\left(t^2+2t\right)+4=0\)
\(\Leftrightarrow\left(t^2+2t-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}t=-1-\sqrt{3}\\t=-1+\sqrt{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}xy=-2-\sqrt{3}\Rightarrow x+y=1+\sqrt{3}\\xy=-2+\sqrt{3}\Rightarrow x+y=1-\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow x;y\) là nghiệm của: \(\left[{}\begin{matrix}X^2-\left(1+\sqrt{3}\right)X-2-\sqrt{3}=0\\X^2-\left(1-\sqrt{3}\right)X-2+\sqrt{3}=0\end{matrix}\right.\)
\(\Rightarrow...\)
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x-3y+4=360
nên x-3y=356
=>x=3y+356
Ta có: xy=360
nên y(3y+356)=360
\(\Leftrightarrow3y^2+356y-360=0\)
=>Bạn xem lại đề, nghiệm rất xấu
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Câu 1:
Theo hệ thức lượng trong tam giác vuông thì:
$AH^2=BH.CH=2.3=6$
$\Rightarrow AH=\sqrt{6}$ (cm)
Đáp án C.
Câu 2:
Theo hệ thức lượng trong tam giác vuông:
$AH^2=BH.CH$
$BH=\frac{AH^2}{CH}=\frac{10^2}{5}=20$ (cm)
$BC=BH+CH=20+5=25$ (cm)
Không có đáp án nào đúng.