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2/
a) Ta có:
\(3\sqrt{2}=\sqrt{3^2\cdot2}=\sqrt{9\cdot2}=\sqrt{18}\)
\(2\sqrt{3}=\sqrt{2^2\cdot3}=\sqrt{4\cdot3}=\sqrt{12}\)
Mà: \(12< 18\Rightarrow\sqrt{12}< \sqrt{18}\Rightarrow2\sqrt{3}< 3\sqrt{2}\)
b) Ta có:
\(4\sqrt[3]{5}=\sqrt[3]{4^3\cdot5}=\sqrt[3]{320}\)
\(5\sqrt[3]{4}=\sqrt[3]{5^3\cdot4}=\sqrt[3]{500}\)
Mà: \(320< 500\Rightarrow\sqrt[3]{320}< \sqrt[3]{500}\Rightarrow4\sqrt[3]{5}< 5\sqrt[3]{4}\)
3/
a)ĐKXĐ: \(x\ne1;x\ge0\)
b) \(A=\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(A=\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\)
\(A=\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\)
\(A=1^2-\left(\sqrt{x}\right)^2\)
\(A=1-x\)
a: \(A=\left(\dfrac{\sqrt{3}\left(x-\sqrt{3}\right)+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\right)\cdot\dfrac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\)
\(=\dfrac{x\sqrt{3}}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\cdot\dfrac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)
\(=\dfrac{1}{x-\sqrt{3}}\)
b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\)
\(=x-\sqrt{x}-x-\sqrt{x}+x+1\)
\(=x-2\sqrt{x}+1\)
c: \(C=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
Bổ sung giả thuyết x ,y \(\ge0\)
Do giả thiết x ,y \(\ge0\)
\(\sqrt{x}+\sqrt{y}\) =1
nên:
xy (x+y )\(^2\)\(\le\) \(\dfrac{1}{64}\)
<=> 64 xy (x + y )\(^2\) \(\le\)1
<=> 64 xy ( x + y)\(^2\)\(\le\)(\(\sqrt{x}+\sqrt{y}\))\(^8\)
<=> 64 xy ( x + y )\(^2\) < \((x+2\sqrt{xy}+y)^4\)
Áp dụng bất đẳng thức Cauchy cho 2 số không âm x + y và \(2\sqrt{xy}\)
ta có ;
x + y + 2\(\sqrt{xy}\) \(\ge\) \(2\sqrt{x+y}2\sqrt{xy}\)
=> ( x + y +2\(\sqrt{xy}\)) \(^4\)\(\ge\) (\(2\sqrt{x+y}2\sqrt{xy}\) )\(^4\)= 64 xy (x + y)\(^2\)
=> ĐIỀU PHẢI CHỨNG MINH
Dấu bằng xảy ra <=> x + y = \(2\sqrt{xy}\)
<=> x = y = \(\dfrac{1}{4}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{y}=b\end{matrix}\right.\), ta có:
\(A=\left[\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\times\dfrac{2}{a+b}+\dfrac{1}{a^2}+\dfrac{1}{b^2}\right]\)\(\times\dfrac{a^3+ab^2+a^2b+b^3}{ab^3+a^3b}\)
\(=\left(\dfrac{b+a}{ab}\times\dfrac{2}{a+b}+\dfrac{b^2+a^2}{a^2b^2}\right)\)\(\times\dfrac{a^2\left(a+b\right)+b^2\left(a+b\right)}{ab\left(a^2+b^2\right)}\)
\(=\dfrac{2ab+b^2+a^2}{a^2b^2}\times\dfrac{\left(a+b\right)\left(a^2+b^2\right)}{ab\left(b^2+a^2\right)}\)
\(=\dfrac{\left(a+b\right)^3}{a^3b^3}\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^3}{\sqrt{\left(xy\right)^3}}\)
1: ĐKXĐ: -2/2x-2>=0
=>2x-2<0
=>x<1
2: ĐKXĐ: 2/3x-1>=0
=>3x-1>0
=>x>1/3
3: ĐKXĐ: 2x-2/(-2)>=0
=>2x-2<=0
=>x<=1
4: ĐKXĐ: (3x-2)/5>=0
=>3x-2>=0
=>x>=2/3
5: ĐKXĐ: (x-2)/(x+3)>=0
=>x>=2 hoặc x<-3
A = (15/√x) - (11x + 2√x - 3) - (3√x - 2√x - 1) - (2√x + 3√x - 3)
Tiếp theo, kết hợp các thành phần tương tự:
A = 15/√x - 11x - 2√x + 3 + 3√x - 2√x + 1 - 2√x - 3√x + 3
Đơn giản hóa biểu thức:
A = -11x + 15/√x + 4
Để tìm giá trị lớn nhất của A, ta có thể tìm điểm đạt cực đại của hàm số A(x). Tuy nhiên, để làm điều này, cần biết thêm về giá trị của x.
Sửa đề: (3căn x-2)/căn x-1-(2căn x+3)/(căn x+3)\(A=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
\(A=\dfrac{-5\sqrt{x}-15+17}{\sqrt{x}+3}==-5+\dfrac{17}{\sqrt{x}+3}< =\dfrac{17}{3}-5=\dfrac{2}{3}\)
Dấu = xảy ra khi x=0
1) Đk: \(x\ge4\)
\(\dfrac{\sqrt{x^2-16}}{\sqrt{x-3}}+\sqrt{x-3}=\dfrac{7}{\sqrt{x-3}}\)
\(\Leftrightarrow\dfrac{\sqrt{x^2-16}}{\sqrt{x-3}}+\dfrac{x-3}{\sqrt{x-3}}=\dfrac{7}{\sqrt{x-3}}\)
\(\Leftrightarrow\dfrac{\sqrt{x^2-16}+x-10}{\sqrt{x-3}}=0\)
\(\Leftrightarrow\sqrt{x^2-16}+x-10=0\)
\(\Leftrightarrow\sqrt{x^2-16}=10-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-16=100-20x+x^2\\x\le10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}20x=116\\x\le10\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{29}{5}\left(N\right)\\x\le10\end{matrix}\right.\)
Kl: x= 29/5
2) Đk: \(x\ge-1\)
\(x^2-5x+14=4\sqrt{x+1}\)
\(\Leftrightarrow x^4+25x^2+196-10x^3-140x+28x^2=16x+16\)
\(\Leftrightarrow x^4-10x^3+53x^2-156x+180=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^3-7x^2+32x-60\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(x^2-4x+20\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x^2-4x+20=0\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow x=3\left(N\right)\)
Kl: x=3
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