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\(\Leftrightarrow\sqrt{x+3}-2-2\sqrt{x}+2=\sqrt{2x+2}-2+2-\sqrt{3x+1}\)
=>\(\dfrac{x+3-4}{\sqrt{x+3}+2}-2\left(\sqrt{x}-1\right)=\dfrac{2x+2-4}{\sqrt{2x+2}+2}+\dfrac{4-3x-1}{2+\sqrt{3x+1}}\)
=>\(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x+3}+2}-2\left(\sqrt{x}-1\right)=\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{2x+2}+2}-\dfrac{3\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2+\sqrt{3x+1}}\)
=>\(\left(\sqrt{x}-1\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x+3}+2}-2-\dfrac{2\sqrt{x}+2}{\sqrt{2x+2}+2}+\dfrac{3\sqrt{x}+3}{2+\sqrt{3x+1}}\right)=0\)
=>căn x-1=0
=>x=1
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a) Có \(x+1< x+2\)
\(\Rightarrow\sqrt{x+1}< \sqrt{x+2}\)
\(\Leftrightarrow\frac{\sqrt{x+1}}{\sqrt{x+2}}< 1\)
b) Vì \(\sqrt{x+1}< \sqrt{x+2}\)
\(\Rightarrow\sqrt{x+1}.\sqrt{x+1}.\sqrt{x+2}< \sqrt{x+2}.\sqrt{x+1}.\sqrt{x+1}\)
\(\Leftrightarrow\sqrt{x+1}^2.\sqrt{x+2}< \sqrt{x+2}^2.\sqrt{x+1}\)
\(\Rightarrow\frac{\sqrt{x+1}^2}{\sqrt{x+2}^2}< \frac{\sqrt{x+1}}{\sqrt{x+2}}\)
hay \(\frac{\sqrt{x+1}}{\sqrt{x+2}}>\frac{\sqrt{x+1}^2}{\sqrt{x+2}^2}\)
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a, \(x.\sqrt{\frac{2}{5}}\) = \(\sqrt{x^2}.\sqrt{\frac{2}{5}}\) = \(\sqrt{\frac{x^2.2}{5}}\)
b, \(\left(x-5\right).\sqrt{\frac{x}{25-x^2}}\)= \(\sqrt{\left(x-5\right)^2}\). \(\sqrt{\frac{x}{\left(5-x\right)\left(5+x\right)}}\) = \(\sqrt{\frac{\left(x-5\right)^2.x}{\left(x-5\right)\left(x+5\right)}}\)= \(\sqrt{\frac{x.\left(x-5\right)}{x+5}}\)
c,\(x.\sqrt{\frac{7}{x^2}}\) = \(\sqrt{x^2}\). \(\sqrt{\frac{7}{x^2}}\) = \(\sqrt{\frac{x^2.7}{x^2}}\) = \(\sqrt{7}\)
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1:|3x-1|-x=2
|3x-1| =2+x
=> 3x-1=2+x hay 3x-1=-2-x
3x-x=2+1 hay 3x+x=-2+1
2x=3 hay 4x=-1
x=3/2 hay x=-1/4
Vậy x=3/2; x=-1/4
2
a 4\(\sqrt{x}=8\)
=>\(\sqrt{x}=2\\ =>x=4\)
b
\(2\sqrt{x}>3\\ \sqrt{x}>\dfrac{3}{2}\\ x>\dfrac{9}{4}\)
c,\(4\sqrt{x}< 13\\ \sqrt{x}< \dfrac{13}{4}\\ x< \dfrac{1703}{16}\)
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Bạn cần viết lại đề bằng công thức toán (gõ công thức trong hộp có biểu tượng $\sum$) để được hỗ trợ tốt hơn. Nhìn đề thế này rối mắt quá.
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a, \(x^2-6=x^2-\sqrt{6^2}=\left(x-\sqrt{6}\right)\left(x+\sqrt{6}\right)\)
b, \(x^2+2\sqrt{3}x+3=x^2+2\sqrt{3}x+\sqrt{3}=\left(x+\sqrt{3}\right)^2=\left(x+\sqrt{3}\right)\left(x+\sqrt{3}\right)\)
c, \(x^2-2\sqrt{5}x+5=x^2-2\sqrt{5}x+\sqrt{5}=\left(x-\sqrt{5}\right)^2=\left(x-\sqrt{5}\right)\left(x-\sqrt{5}\right)\)
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a: (x+1)^3-x(x-2)^2+x-1=0
=>x^3+3x^2+3x+1-x(x^2-4x+4)+x-1=0
=>x^3+3x^2+4x-x^3+4x^2-4x=0
=>7x^2=0
=>x=0
b: =>x^3-3x^2+3x-1-x^3-27+3x^2-12=2
=>3x=2+1+27+12=39+3=42
=>x=14