Tìm `ĐKXĐ`:
\(\sqrt{\dfrac{-5}{6+x}}\)
\(\sqrt{\dfrac{-2}{6-x}}\)
\(\sqrt{\dfrac{-x+3}{-6}}\)
\(\sqrt{\dfrac{7x-1}{-9}}\)
\(\sqrt{\dfrac{x+2}{x^2+2x+1}}\)
\(\sqrt{\dfrac{x-2}{x^2-2x+4}}\)
1) \(\sqrt{x^2}=2x-5\)
2) \(\sqrt{25x^2-10x+1}=2x-6\)
3) \(\sqrt{25-10x+x^2}=2x-5\)
4) \(\sqrt{1-2x+x^2}=2x-1\)
5) \(\sqrt{4x^2+4x+1}=-x-3\)
1)\(7\sqrt{3x-7}+\left(4x-7\right)\sqrt{7-x}=32\)
2)\(4x^2-11x+6=\left(x-1\right)\sqrt{2x^2-6x+6}\)
3)\(9+3\sqrt{x\left(3-2x\right)}=7\sqrt{x}+5\sqrt{3-2x}\)
4)\(\sqrt{2x^2+4x+7}=x^4+4x^3+3x^2-2x-7\)
5)\(\frac{6-2x}{\sqrt{5-x}}+\frac{6+2x}{\sqrt{5+x}}=\frac{8}{3}\)
6)\(2\left(5x-3\right)\sqrt{x+1}+\left(x+1\right)\sqrt{3-x}=3\left(5x+1\right)\)
7)\(\sqrt{7x+7}+\sqrt{7x-6}+2\sqrt{49x^2+7x-42}=181-14x\)
1,\(\sqrt{x-5}+\sqrt{x+4}=3\)
2,\(\sqrt{x+4-4\sqrt{x}}+\sqrt{x+6-6\sqrt{x}}=1\)
3,\(\sqrt{x+3}-\sqrt{x-4}=1\)
4,\(\sqrt{15-x}+\sqrt{3-x}=6\)
5,\(\sqrt{10-x}+\sqrt{x+3}=5\)
6,\(\sqrt{2x-1}+\sqrt{x-2}=\sqrt{x+1}\)
7,\(\sqrt{x+6-4\sqrt{x+2}}+\sqrt{x+11-6\sqrt{x+2}}=1\)
8,\(\sqrt{x^2-5x+6}+\sqrt{x-2-3\sqrt{x-3}}=3\)
9,\(2x^2-x+4=2\sqrt{2x+3}\)
\(\left(5\right)\sqrt{x+3-4\sqrt{x-1}}\sqrt{x+8+6\sqrt{x-1}}=5\)
\(\left(6\right)2x^2+3x+\sqrt{2x^2+3x+9}=33\)
\(\left(7\right)\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+30}=8\)
\(\left(8\right)x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
6. giải PT
a.\(\sqrt{2x+5}=\sqrt{1-x}\)
b.\(\sqrt{x^2-x}=\sqrt{3-x}\)
c.\(\sqrt{2x^2-3}=\sqrt{4x-3}\)
giải phương trình
a) \(\sqrt{2x-2\sqrt{2x-1}}-2\sqrt{2x+3-4\sqrt{2x-1}}+3\sqrt{2x+8-\sqrt{2x-1}}=4\)
b) \(4x^2+3x+3=4x\sqrt{x+3}+2\sqrt{2x-1}\)
c) \(\sqrt{x-4}+\sqrt{6-x}=x^2-11x+27\)
d) \(\sqrt{13x^2-6x+10}+\sqrt{5x^2-13x+\frac{17}{2}}+\sqrt{17x^2-48x+36}=\frac{1}{2}\left(36x-8x^2-21\right)\)
e) \(\sqrt{\frac{6}{3-x}}+\sqrt{\frac{8}{2-x}}=6\)
Giải các PT sau:
\(a,\sqrt{x-2\sqrt{x-1}}-\sqrt{x-1}=1\)
\(b,\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}=1\)
\(c,\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
\(d,\sqrt{x+\sqrt{6x-9}}+\sqrt{x-\sqrt{6x-9}}=\sqrt{6}\)
\(\sqrt{x+2\sqrt{x-1}}=2\)
\(\sqrt{4x^2-20x+25}+2x=5\)
\(\sqrt{2x^2-3}=\sqrt{4x-3}\)
\(\sqrt{x^2-x-6}=\sqrt{x-3}\)
\(\sqrt{x^2-x}=\sqrt{3-x}\)
Phương pháp 2. Biến đổi về phương trình tích
a \(\sqrt{x^2-5x+6}+\sqrt{x+1}=\sqrt{x-2}+\sqrt{x^2-2x-3}\)
b \(2\sqrt[3]{\left(x+3\right)^2}-\sqrt[3]{\left(x-3\right)^2}=\sqrt[3]{x^2-9}\)
c \(\sqrt{2x+1}+3\sqrt{4x^2-2x+1}=3+\sqrt{8x^3+1}\)
d \(14\sqrt{x+35}+6\sqrt{x+1}=84+\sqrt{x^2+36x+35}\)