ĐK: \(12x\ge0\Leftrightarrow x\ge0\)
\(-3-\frac{7}{2}\sqrt{12x}+\frac{5}{2}\sqrt{12x}=-3\sqrt{12x}\\ \Leftrightarrow-\frac{7}{2}\sqrt{12x}+\frac{5}{2}\sqrt{12x}+3\sqrt{12x}=3\\ \Leftrightarrow2\sqrt{12x}=3\\ \Leftrightarrow4\sqrt{3x}=3\\\Leftrightarrow \sqrt{3x}=\frac{3}{4}\\ \Leftrightarrow3x=\frac{9}{16}\\ \Leftrightarrow x=\frac{3}{16}\left(tm\right)\)
Vậy \(x=\frac{3}{16}\)
Bài 1:Tìm x biết
1.\(\frac{2}{3}\sqrt{12x}+\sqrt{12x}-9=\frac{1}{3}\sqrt{3x}\)
2.\(\sqrt{2x}-3\sqrt{8x}+4\sqrt{18x}=14\)
Tính giá trị của A=\(\left(4x^5+12x^4+7x^3-1\right)^{2016}+2016\)
biết x =\(\sqrt{\frac{2}{16-5\sqrt{7}}}-\frac{1}{18}\left(37+2\sqrt{7}\right)+\frac{\sqrt{2}}{2}\)
\(\sqrt{12-\frac{12}{x^2} }+\sqrt{x^2+\frac{12}{x} }=x^2+\frac{25}{2} \)
\(\frac{\sqrt[3]{10-x}+\sqrt[3]{8-x}}{\sqrt[3]{10-x}-\sqrt[3]{8-x}}=9-x \)
\(4x^2+\sqrt{2x+1}+5=12x\)
1) Tìm x biết
a) \(\sqrt{4x-20}+\sqrt{x-5}-\frac{1}{3}=\sqrt{9x-45}=4\)
b) \(\sqrt{16x-32x}-\sqrt{12x}=\sqrt{3x}+\sqrt{9-18x}\)
c) \(\sqrt{4x-20}-3\sqrt{\frac{x-5}{9}}=\sqrt{1-x}\)
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A=\(\frac{\sqrt{2x^2+5x+2}}{\sqrt{2x^3+9x^2+12x+4}}\)
Giải phương trình
1.\(\sqrt{2x-3}-\sqrt{5-2x}=3x^2-12x+14\)
2.\(x^2+2x+15=6\sqrt{4x+5}\)
3.\(x^2-5x-8=2\sqrt{x-2}\)
4.\(\sqrt{x+1+\sqrt{x+\frac{3}{4}}}=x+1\)
Giải phương trình sau:
a) \(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
b) \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
c) \(2x-x^2+\sqrt{6x^2-12x+7}=0\)
d) \(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6\)
1) Tìm đk để BT sau có nghĩa
a) \(\sqrt{6x+1}\)
b) \(\frac{\sqrt{-3}}{2+x}\)
b) \(\frac{3}{\sqrt{12x-1}}\)
