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\(a,\sqrt{1-4a+4a^2}-2a\)
\(=\sqrt{\left(1-2a\right)^2}-2a\)
\(=1-2a-2a\)
\(=1-4a\)
\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)
\(=x-2y-\sqrt{\left(x-2y\right)^2}\)
\(=x-2y-\left(x-2y\right)\)
\(=x-2y-x+2y\)
\(=0\)
\(c,x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\sqrt{\left(x^2-4\right)^2}\)
\(=x^2+x^2-4\)
\(=2x^2-4\)
Các câu còn lại tương tự nha
\(a,\sqrt{1-4a+4a^2}-2a\)
\(=\sqrt{\left(1-2a\right)^2}-2a\)
\(=\left(1-2a\right)-2a\)
\(=1-4a\)
\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)
\(=x-2y-\sqrt{\left(x-2y\right)^2}\)
\(=x-2y-\left(x-2y\right)\)
\(=x-2y-x+2y\)
\(=0\)
\(c,x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\sqrt{\left(x^2-2^2\right)^2}\)
\(=x^2+\left(x^2-4\right)\)
\(=x^2+x^2-4\)
\(=2x^2-4\)
\(d,2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}\)
\(=2x-1-\frac{\sqrt{\left(x-5\right)^2}}{x-5}\)
\(=2x-1-\frac{x-5}{x-5}\)
\(=2x-1-1\)
\(=2x-2\)
\(=2\left(x-1\right)\)
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Thôi giải lại câu 1:v (ý tưởng dồn biến là quá trâu bò! Bên AoPS em mới phát hiện ra có một cách bằng Cauchy-Schwarz quá hay!)
\(BĐT\Leftrightarrow\Sigma_{cyc}\frac{\left(a+b+c\right)^2}{2a^2+\left(a^2+b^2\right)+\left(a^2+c^2\right)}\le\frac{9}{2}\)(*)
BĐT này đúng theo Cauchy-Schwarz: \(VT_{\text{(*)}}\le\Sigma_{cyc}\left(\frac{a^2}{2a^2}+\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\right)=\frac{9}{2}\)
Ta có đpcm.
Equality holds when a = b = c = 1 (Đẳng thức xảy ra khi a = b =c = 1)
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a.
\(A=\left[\frac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x-2}\right)}\right]:\left[\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]\)
\(=\frac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\frac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}\)
\(=\frac{-4\sqrt{x}.\sqrt{x}}{-\sqrt{x}+3}=\frac{4x}{\sqrt{x}-3}\)
b.
\(A=-1\Leftrightarrow\frac{4x}{\sqrt{x}-3}=-1\)
\(\Leftrightarrow4x=-\sqrt{x}+3\)
\(\Leftrightarrow4x+\sqrt{x}-3=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(l\right)\\x=\frac{3}{4}\end{matrix}\right.\)
Vậy \(A=-1\Leftrightarrow x=\frac{3}{4}\)
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a: \(Q=\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)
\(=\dfrac{-8x}{\sqrt{x}+2}\cdot\dfrac{1}{3-\sqrt{x}}=\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
b: Để Q=-1 thì \(8x=-\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)\)
\(\Leftrightarrow8x+x-\sqrt{x}-6=0\)
\(\Leftrightarrow9x-\sqrt{x}-6=0\)
Bạn xem lại đề, nghiệm này rất xấu
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Lời giải:
a)
\(A=\left[\frac{4\sqrt{x}(\sqrt{x}-2)}{(2+\sqrt{x})(\sqrt{x}-2)}-\frac{8x}{(\sqrt{x}-2)(\sqrt{x}+2)}\right]:\left[\frac{\sqrt{x}-1}{\sqrt{x}(\sqrt{x}-2)}-\frac{2(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)}\right]\)
\(=\frac{4\sqrt{x}(\sqrt{x}-2)-8x}{(\sqrt{x}-2)(\sqrt{x}+2)}:\frac{\sqrt{x}-1-2(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)}\)
\(=\frac{-4x-8\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}(\sqrt{x}-2)}{3-\sqrt{x}}\)
\(=\frac{-4\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}(\sqrt{x}-2)}{3-\sqrt{x}}=\frac{4x}{\sqrt{x}-3}\)
b)
Để $A=-1\Leftrightarrow \frac{4x}{\sqrt{x}-3}=-1$
$\Leftrightarrow 4x+\sqrt{x}-3=0$
$\Leftrightarrow (4\sqrt{x}-3)(\sqrt{x}+1)=0$
$\Rightarrow 4\sqrt{x}-3=0$
$\Leftrightarrow x=\frac{9}{16}$ (thỏa mãn ĐKXĐ)
Vậy........
![](https://rs.olm.vn/images/avt/0.png?1311)
nhầm rồi, để làm lại
a/ \(P=\left[\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right]:\left[\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right]\)
\(=\left[\frac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right]:\left[\frac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]\)
\(=\frac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(=\frac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{-\sqrt{x}\left(2-\sqrt{x}\right)}{3-\sqrt{x}}\)
\(=\frac{4x}{\sqrt{x}-3}\)
b/ \(P=-1\Rightarrow\frac{4x}{\sqrt{x}-3}=-1\Rightarrow3-\sqrt{x}=4x\Rightarrow4x+\sqrt{x}-3=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=-1\left(l\right)\\\sqrt{x}=\frac{3}{4}\end{cases}\Rightarrow x=\frac{9}{16}}\)
Vậy x = 9/16
ĐKXĐ: x > 0 và \(x\ne4\)
a/ \(P=\left[\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right]:\left[\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right]\)
\(=\frac{4\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{\sqrt{x}\left(\sqrt{x}-1\right)-2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{8\sqrt{x}-4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-\sqrt{x}-2}\)
\(=\frac{4\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{4x}{\left(2+\sqrt{x}\right)\left(\sqrt{x}+1\right)}\)
b/ \(P=-1\Rightarrow\frac{4x}{x+3\sqrt{x}+2}=-1\Rightarrow-x-3\sqrt{x}-2=4x\)
\(\Rightarrow-5x-3\sqrt{x}-2=0\left(1\right)\), vì (1) > 0 => vô nghiệm
Vậy k có giá trị nào của x thỏa P = -1
A= 2b-√(b-2)2/(b-2)
= 2b- |b-2|/(b-2)
= 2b ( xét cả 2 th b\(\ge\)2 và b\(\le\)2)