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a) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\)
\(=\dfrac{4a^2b^3}{8\sqrt{2}a^3b^3}\)
\(=\dfrac{1}{2\sqrt{2}a}\)
\(=\dfrac{\sqrt{2}}{4a}\)
b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
chịu đấy :v
c) \(\sqrt{\dfrac{\left(x-2\right)^2}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{3-x}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{-\left(x-3\right)}+\dfrac{x^2-1}{x-3}\)
\(=-\dfrac{x-2}{x-3}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{-\left(x-2\right)+x^2-1}{x-3}\)
\(=\dfrac{-x+1+x^2}{x-3}\)
d) \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1^2\right)}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(x-1\right)^2}\)
\(=\dfrac{1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{x-1}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(\sqrt{y}-1\right)\left(x-1\right)}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{x\sqrt{y}-\sqrt{y}-x+1}\)
e) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\dfrac{\sqrt{x^2\cdot\left(x+2\right)}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\sqrt{x^2}\)
\(=4x-2\sqrt{x}+x\)
\(=5x-2\sqrt{2}\)
Bài 1:
Ta có a là số dương
\(\Leftrightarrow\left\{{}\begin{matrix}a>0\\\dfrac{1}{a}>0\end{matrix}\right.\)
Áp dụng BĐT Cauchy cho hai số dương, có:
\(a+\dfrac{1}{a}\ge2\sqrt{a.\dfrac{1}{a}}\)
\(\Leftrightarrow a+\dfrac{1}{a}\ge2\sqrt{1}\)
\(\Leftrightarrow a+\dfrac{1}{a}\ge2\left(đpcm\right)\)
Vậy ...
làm giúp mình câu 1, còn câu 2 mình biết làm rồi
Bài 2:
a: ĐKXĐ: x>0; x<>1
b: \(P=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{x-1}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2}{\sqrt{x}+1}\)
c: Khi x=1/4 thì \(P=2:\left(\dfrac{1}{2}+1\right)=2:\dfrac{3}{2}=\dfrac{4}{3}\)
a: ĐKXĐ: x>=0; \(x\notin\left\{4;9\right\}\)
b: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-1\right):\left(\dfrac{9-x+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{-3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3}{\sqrt{x}+2}\)
Thay \(x=3-2\sqrt{2}\) vào A, ta được:
\(A=\dfrac{3}{\sqrt{2}-1+2}=\dfrac{3}{\sqrt{2}+1}=3\sqrt{2}-3\)
c: Để A<1 thì A-1<0
\(\Leftrightarrow\dfrac{3-\sqrt{x}-2}{\sqrt{x}+2}< 0\)
\(\Leftrightarrow1-\sqrt{x}< 0\)
hay 0<x<1
a: \(A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{2\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}\)
\(=\dfrac{-6}{\sqrt{x}+3}\)
b: Để A<-1/2 thì A+1/2<0
\(\Leftrightarrow-\dfrac{6}{\sqrt{x}+3}+\dfrac{1}{2}< 0\)
\(\Leftrightarrow-12+\sqrt{x}+3< 0\)
=>0<x<81 và x<>9
1. \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1+\sqrt{2}-\sqrt{3}\right)\)
\(=\left(1+\sqrt{2}\right)^2-\sqrt{3}^2\)
\(=1+2\sqrt{2}+2-3\)
\(=2\sqrt{2}\)
3. \(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\cdot\left(1+\dfrac{1}{\sqrt{x}}\right)\)(1)
ĐKXĐ \(x>0,x\ne1\)
pt (1) <=> \(\left(\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\right)\cdot\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}+1+\sqrt{x}-1\right)}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}}\)
\(\Leftrightarrow\dfrac{\sqrt{x}\cdot2}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}\)
b) Để \(\sqrt{A}>A\Leftrightarrow\sqrt{\dfrac{2}{\sqrt{x}-1}}>\dfrac{2}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}>\dfrac{4}{x-2\sqrt{x}+1}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}-\dfrac{4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\cdot\left(\sqrt{x}-1\right)-4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{2}-2-4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{2}-6}{x-2\sqrt{x}+1}>0\)
Vì \(2\sqrt{2}-6< 0\Rightarrow x-2\sqrt{x}+1< 0\)
mà \(x-2\sqrt{x}+1=\left(\sqrt{x}-1\right)^2\ge0\forall x\)
Vậy không có giá trị nào của x thỏa mãn \(\sqrt{A}>A\)
(P/s Đề câu b bị sai hay sao vậy, chả có số nào mà \(\sqrt{A}>A\) cả, check lại đề giùm với nhé)
Áp dụng bất đẳng thức AM - GM, ta có:
\(S=\dfrac{1}{\left(x-1\right)^2}+\dfrac{1}{\left(2-x\right)^2}+\dfrac{1}{\left(x-1\right)\left(2-x\right)}\)
\(\ge3\sqrt[3]{\dfrac{1}{\left(x-1\right)^2}\times\dfrac{1}{\left(2-x\right)^2}\times\dfrac{1}{\left(x-1\right)\left(2-x\right)}}\)
\(=\dfrac{3}{\left(x-1\right)\left(x-2\right)}=\dfrac{3}{-x^2+3x-2}\)
Vì \(-x^2+3x-2=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
nên \(S\ge\dfrac{3}{\dfrac{1}{4}}=12\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\dfrac{1}{\left(x-1\right)^2}=\dfrac{1}{\left(2-x\right)^2}=\dfrac{1}{\left(x-1\right)\left(2-x\right)}\\x-\dfrac{3}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{3}{2}\left(\text{ nhận }\right)\)
Vậy \(Min_S=12\Leftrightarrow x=\dfrac{3}{2}\)