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a: \(P=\dfrac{2x-18-2x-6\sqrt{x}+5\sqrt{x}+20}{x-9}:\dfrac{\sqrt{x}+3-5}{\sqrt{x}+3}\)
\(=\dfrac{-\sqrt{x}+2}{x-9}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}-2}=\dfrac{-1}{\sqrt{x}-3}\)
b: Để P<-1/2 thì P+1/2<0
=>\(\dfrac{-1}{\sqrt{x}-3}+\dfrac{1}{2}< 0\)
=>\(\dfrac{-2+\sqrt{x}-3}{2\left(\sqrt{x}-3\right)}< 0\)
=>\(\dfrac{\sqrt{x}-5}{2\left(\sqrt{x}-3\right)}< 0\)
=>3<căn x<5
=>9<x<25
c: \(Q=\dfrac{-1}{\sqrt{x}-3}\cdot\sqrt{x}\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)=-x+5\sqrt{x}\)
\(=-\left(x-5\sqrt{x}+\dfrac{25}{4}-\dfrac{25}{4}\right)=-\left(\sqrt{x}-\dfrac{5}{2}\right)^2+\dfrac{25}{4}< =\dfrac{25}{4}\)
Dấu = xảy ra khi x=25/4
Điều kiện: x khác 0
\(=\sqrt{\frac{x^4-6x^2+9+12x^2}{x^2}}+\sqrt{x^2+4x+4-8x}\)
\(=\sqrt{\frac{x^4+6x^2+9}{x^2}}+\sqrt{x^2-4x+4}=\sqrt{\frac{\left(x^2+3\right)^2}{x^2}}+\sqrt{\left(x-2\right)^2}\)
\(=\left|\frac{x^2+3}{x}\right|+\left|x-2\right|=\frac{x^2+3}{\left|x\right|}+\left|x-2\right|\)
\(\sqrt{\frac{\left(x^2-3\right)^2+12x^2}{x^2}}+\sqrt{\left(x+2\right)^2-8x}\)
=\(\frac{\sqrt{x^4-6x+9+12x^2}}{\sqrt{x^2}}+\sqrt{x^2+4x+4-8x}\)
=\(\frac{\sqrt{x^4+6x+9}}{x}+\sqrt{x^2-4x+4}\)
=\(\frac{\sqrt{\left(x^2+3\right)^2}}{x}+\sqrt{\left(x-2\right)^2}\)
=\(\frac{\sqrt{\left(x^2+3\right)^2}}{x}+\left|x-2\right|\)
=\(\frac{x^2+3}{x}+\left|x-2\right|\)
TH1: x\(\ge\)2 =>|x-2|=x-2
=>\(\frac{x^2+3}{x}+\left|x-2\right|\)
=\(\frac{x^2+3}{x}+x-2\)
=\(\frac{x^2+3}{x}+\frac{x^2-2x}{x}=\frac{2x^2-2x+3}{x}\)
TH2:x\(\le\)2 =>|x-2|=2-x
=>\(\frac{x^2+3}{x}+\left|x-2\right|\)
=\(\frac{x^2+3}{x}+2-x\)
=\(\frac{x^2+3}{x}+\frac{2x-x^2}{x}=\frac{2x+3}{x}\)
\(a.A=\sqrt{\dfrac{\left(x^2-3\right)^2+12x^2}{x^2}}+\sqrt{\left(x+2\right)^2-8x}=\sqrt{\dfrac{x^4+6x^2+9}{x^2}}+\sqrt{x^2-4x+4}=\left|\dfrac{x^2+3}{x}\right|+\left|x-2\right|=\left|x+\dfrac{3}{x}\right|+\left|x-2\right|\left(x\ne0\right)\)
\(b.\) Để : \(A\in Z\Leftrightarrow\left(x+\dfrac{3}{x}\right)\in Z\Leftrightarrow x\in\left\{\pm1;\pm3\right\}\)
a, Với \(x>0;x\ne4;x\ne9\)
\(A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(=\left(\frac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\left(\frac{8\sqrt{x}-4x+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{3-\sqrt{x}}{\sqrt{x}\left(2-\sqrt{x}\right)}=\frac{4\sqrt{x}}{2-\sqrt{x}}.\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{3-\sqrt{x}}=\frac{4x}{3-\sqrt{x}}\)
b, Ta có : A = -2 hay
\(\frac{4x}{3-\sqrt{x}}=-2\Rightarrow4x=-6+2\sqrt{x}\)
\(\Leftrightarrow4x+6-2\sqrt{x}=0\Leftrightarrow2\left(2x+3-\sqrt{x}\right)=0\)
\(\Leftrightarrow2x+3-\sqrt{x}=0\Leftrightarrow\sqrt{x}=2x+3\)
bình phương 2 vế ta có :
\(x=\left(2x+3\right)^2=4x^2+12x+9\)
\(\Leftrightarrow-4x^2-11x-9=0\)giải delta ta thu được : \(x=-\frac{11\pm\sqrt{23}i}{8}\)
\(a,A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)
\(=\frac{4\sqrt{x}.\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{\sqrt{x}-1-2.\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{8\sqrt{x}-4x+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)
\(=\frac{\left(4x+8\sqrt{x}\right)\left(\sqrt{x}\right)\left(\sqrt{x}-2\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)
\(=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}\right)\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)
\(=\frac{4x}{\sqrt{x}-3}\)
\(M=\dfrac{4x-8\sqrt{x}+8x}{x-4}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{12x-8\sqrt{x}}{x-4}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}\left(3\sqrt{x}-2\right)}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}}{3-\sqrt{x}}=\dfrac{4x\left(3\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(3-\sqrt{x}\right)}\)
ĐKXĐ: x>0, x≠0;x≠4
\(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)=\left(\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)=\dfrac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}=\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)\sqrt{x}}{\left(\sqrt{x}+2\right)\left(3-\sqrt{x}\right)}=\dfrac{4x}{\sqrt{x}-3}\)
Bài 1:
\(\sqrt{24+8\sqrt{15}-\sqrt{9-4\sqrt{5}}}\)
\(=\sqrt{24+8\sqrt{15}-\left(\sqrt{5}-2\right)}\)
\(=\sqrt{26+8\sqrt{15}-\sqrt{5}}\)
Bài 2:
\(A=\sqrt{\frac{\left(x^2-3\right)^2+12x^2}{x^2}}+\sqrt{\left(x+2\right)^2-8x}\)
\(A=\sqrt{\frac{x^4+6x^2+9}{x^2}}\)
\(A=\frac{\sqrt{x^4+6x^2+9}}{\sqrt{x^2}}\)
\(A=\frac{\sqrt{\left(x^2+3\right)^2}}{x}\)
\(A=\frac{x^2+3}{x}\)
\(A=\frac{x^2+3}{x}+x-2\)
\(A=\frac{2x^2+3}{x}-2\)
wrecking ball sai rồi \(\frac{\sqrt{\left(x^2+3\right)^2}}{x}=\frac{trituyetdoix^2+3}{x}\) bằng
B1:
\(C=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
\(=\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)\)
\(=\sqrt{3^2-\left(\sqrt{5}\right)^2}\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)\)
\(=\sqrt{2}\left(\sqrt{3-\sqrt{5}}.\sqrt{2}+\sqrt{3+\sqrt{5}}.\sqrt{2}\right)\)
\(=\sqrt{2}\left(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\right)\)
\(=\sqrt{2}\left(\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}\right)\)
\(=\sqrt{2}\left(\sqrt{5}-1+\sqrt{5}+1\right)=2\sqrt{10}\)