Tuyển Cộng tác viên Hoc24 nhiệm kì 30 tại đây: https://forms.gle/T8wg79khMGHeXQuDA
ĐKXĐ: x>=-2/3
Ta có: \(3x-5\sqrt{3x+2}=-6\)
=>\(3x+6-5\sqrt{3x+2}=0\)
=>\(3x+2-5\cdot\sqrt{3x+2}+4=0\)
=>\(\left(\sqrt{3x+2}-4\right)\left(\sqrt{3x+2}-1\right)=0\)
TH1: \(\sqrt{3x+2}-4=0\)
=>\(\sqrt{3x+2}=4\)
=>3x+2=16
=>3x=14
=>\(x=\frac{14}{3}\) (nhận)
TH2: \(\sqrt{3x+2}-1=0\)
=>\(\sqrt{3x+2}=1\)
=>3x+2=1
=>3x=-1
=>\(x=-\frac13\) (nhận)
\(\sqrt{2X^2+3X-2}-3\sqrt{X+6}=4-\sqrt{2X^2+11X-6}+3\sqrt{X+2}\)
\(\sqrt{3X^2-7X+3}-\sqrt{X^2-2}=\sqrt{3X^2-5X-1}-\sqrt{X^2-3X+4}\)
\(8x^2+\sqrt{3x^2+6x+5}=74-\sqrt{36x-5}\)
\(\left(5\right)\sqrt{x+3-4\sqrt{x-1}}\sqrt{x+8+6\sqrt{x-1}}=5\)
\(\left(6\right)2x^2+3x+\sqrt{2x^2+3x+9}=33\)
\(\left(7\right)\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+30}=8\)
\(\left(8\right)x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
1) ( \(3\sqrt{50}-5\sqrt{18}+3\sqrt{8}\) ).\(\sqrt{2}\)
2. Tìm điều kiện:
1)\(\sqrt{-10x}\) 6) \(\sqrt{\dfrac{3x+21}{-5}}\)
2) \(\sqrt{6-3x}\) 7) \(\sqrt{\left(x^2+1\right).2x}\)
3) \(\dfrac{-4}{\sqrt{4x+6}}\) 8) \(\sqrt{\left(-x^2-2\right)}.3x\)
4) \(\dfrac{5}{\sqrt{2x}}\) 9) \(\sqrt{\dfrac{1}{\left(x-1\right)^2}}\)
5) \(\sqrt{\dfrac{-1}{2x-6}}\)
hộ mk với tí nx pk nộp r ;-; help
A=\(\left(3x^3+3x^2+2\right)^{1998}\) với x=\(\frac{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}\)
1\(\sqrt{5+2\sqrt{8}}-\sqrt{5-2\sqrt{8}}\) 2)\(\dfrac{\sqrt{x^2+2\sqrt{3x}+3}}{x^2-3}\) 3) \(\dfrac{\sqrt{x^2-5x+6}}{\sqrt{x-2}}\) 4)\(\dfrac{\sqrt{\left(x-4\right)^2}}{x^2-5x+4}\) 5) \(\dfrac{3x+1}{\sqrt{9x^2+6x+1}}\)
a)\(\sqrt{-3x+5}=7\)
b)\(\sqrt{\dfrac{1}{2}x+6}=\sqrt{2}\)
c)\(\sqrt{2x+1}=1+2\sqrt{3}\)
d)\(\sqrt{\left(3x-1\right)^2}=8\)
giúp mình với ạ mình cần gấp
Giải phương trình, x>0
\(\frac{\left(x^3+3x^2\sqrt{x^3-3x+6}\right)\left(3x-x^3-2\right)}{2+\sqrt{x^3-3x+6}}=4\left[2\sqrt{\left(x^3-3x+6\right)^3}-\left(x^3-3x+6\right)^2\right]\)
Giải phương trình, x>0
\(\frac{\left(x^3+3x^2\sqrt{x^3-3x+6}\right)\left(3x-x^3-2\right)}{2+\sqrt{x^3-3x+6}}=4\left[2\sqrt{\left(x^3-3x+6\right)^3}-\left(x^3-3x+6\right)^2\right]\)
\(\sqrt{x^2+3x}+2\sqrt{x+2}=2x+\sqrt{x+\dfrac{6}{x}+5}\)
