\(=6\sqrt{2}+3\sqrt{3}+\dfrac{4}{3}\sqrt{3}-\dfrac{2}{3}\sqrt{6}=6\sqrt{2}-\dfrac{2}{3}\sqrt{6}+\dfrac{13}{3}\sqrt{3}\)
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Các câu hỏi tương tự
1, \(x^3-x-3=2\sqrt{6x-x^2}\)
2, \(x^3+6x^2-171x-40\left(x+1\right)\sqrt{5x-1}+20=0\)
3, \(\sqrt[3]{x+3}+\sqrt[3]{x-3}=\sqrt[5]{x-5}+\sqrt[5]{x+5}\)
4. \(\left(\frac{1}{\sqrt{x}}-\frac{\sqrt{x}}{x+1}\right)^2=\frac{4\left(1+\sqrt{1+4x}\right)}{x+1+\sqrt{x^2+3x+2}}\)
Giải pt :
a) \(x^2+3x\sqrt[3]{3x+3}-12+\frac{1}{\sqrt{x}}=\frac{\sqrt{x}+8}{x}\)
b) \(\sqrt{\left(x-1\right)\left(3-x\right)}+\sqrt{x+2}=\sqrt{x-1}+\sqrt{3-x}+\frac{x}{2}\)
cho biểu thức B=\(\left(\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{8\sqrt{x}+8}{x+2\sqrt{x}}-\frac{\sqrt{x}+2}{\sqrt{x}}\right):\left(\frac{x+\sqrt{x}+3}{2+2\sqrt{x}}+\frac{1}{\sqrt{x}}\right)\)so sánh \(B^{2019}\)với 1
rut gon P=(\(\frac{3\sqrt{x}}{\sqrt{x}+2}+\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{x-\sqrt{x}}{x-4}\)):\(\left(\frac{3\sqrt{x}}{\sqrt{x}+2}\right)\)
\(\frac{1}{\sqrt{x+1}+1}+\frac{1}{\sqrt{x+4}+2}+\frac{1}{\sqrt{x+9}+3}+\frac{1}{\sqrt{x+16}+4}=\frac{1}{\sqrt{x+100}+10}\)
tìm x=?
Giải phương trình:
1) \(3\sqrt{x}+\frac{3}{2\sqrt{x}}=2\left(2x+\frac{1}{2x}\right)-3\)
2) \(\frac{x^2}{\sqrt{3x-2}}-\sqrt{3x-2}=1-x\)
1.\(\sqrt{\frac{\left(1-x\right)}{x}}=\frac{\left(2x+x^2\right)}{1+x^2}\)
2. 3(2-\(\sqrt{x+2}\))=2x+\(\sqrt{x+6}\)
3. \(\sqrt[3]{x+2}+\sqrt[3]{x+1}=\sqrt[2]{2x^2}+\sqrt[3]{2x^2+1}\)
4. \(\sqrt[3]{x+24}+\sqrt{12-x}=6\)
Toán 10 ạ, giúp em với
\(2\sqrt{\frac{x}{x^2+3}}=\frac{1+2\sqrt{1-x}}{1+\sqrt{1-x^2}}\)
Giải PT: \(\sqrt{x+1}-2\sqrt{4-x}=\frac{5\left(x-3\right)}{\sqrt{2x^2+18}}\)