ĐK: `{(3x+1 >=0),(4+x)>=0):} <=> x>=-1/3`
PT `<=>\sqrt(3x+1)=\sqrt(x+4)+1`
`<=>3x+1=x+4+1+2\sqrt(x+4)`
`<=>2x-4=2\sqrt(x+4)`
`<=>x-2=\sqrt(x+4)`
`<=>x^2-4x+4 =x+4 (x>=2)`
`<=> x^2-5x=0`
`<=>[(x=0(L)),(x=5 (TM)):}`
Vậy `x=5`.
ĐK: `{(3x+1 >=0),(4+x)>=0):} <=> x>=-1/3`
PT `<=>\sqrt(3x+1)=\sqrt(x+4)+1`
`<=>3x+1=x+4+1+2\sqrt(x+4)`
`<=>2x-4=2\sqrt(x+4)`
`<=>x-2=\sqrt(x+4)`
`<=>x^2-4x+4 =x+4 (x>=2)`
`<=> x^2-5x=0`
`<=>[(x=0(L)),(x=5 (TM)):}`
Vậy `x=5`.
a)\(\sqrt{3x-1}=2\) c) \(\sqrt{x^2-4x+4}=3x-1\)
b)\(\sqrt{x+}=2-x\) d)\(\sqrt{x^2+4}=\sqrt{3x+8}\)
\(\sqrt{10x+1}+\sqrt{3x+5}=\sqrt{9x+4}+\sqrt{2x-2}\)
\(\sqrt{2x^2+x-1}+\sqrt{3x^2+x-1}=\sqrt{x^2+4x-3}-\sqrt{x^2-3x+4}\)
\(\frac{x^2}{\left(1+\sqrt{x+1}\right)^2}>x-4\)
tìm ĐKXĐ
1, \(\sqrt{6x+1}\)
2,\(\dfrac{\sqrt{3}-4}{\sqrt{3x-5}}\)
3, \(\sqrt{\dfrac{2\sqrt{15}-\sqrt{59}}{x-7}}\)
4,\(\sqrt{\dfrac{-3x}{1-\sqrt{2}}}\)
5, \(\sqrt{\sqrt{5}-\sqrt{3}x}\)
\(\left(5\right)\sqrt{x+3-4\sqrt{x-1}}\sqrt{x+8+6\sqrt{x-1}}=5\)
\(\left(6\right)2x^2+3x+\sqrt{2x^2+3x+9}=33\)
\(\left(7\right)\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+30}=8\)
\(\left(8\right)x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
Giúp mình với cảm ơn ạ
Giải các pt vô tỉ sau
1)\(\sqrt{21-x}+1=x\)
2)\(\sqrt{8-x}+2=x\)
3)\(1+\sqrt{3x+1}=3x\)
4)\(2+\sqrt{3x-5}=\sqrt{x+1}\)
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) \(\dfrac{3x+1}{\sqrt{9x^2+6x+1}}\)
giải phương trình sau:
a) \(4x^2+\left(8x-4\right).\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)
b) \(8x^3-36x^2+\left(1-3x\right)\sqrt{3x-2}-3\sqrt{3x-2}+63x-32=0\)
c) \(2\sqrt[3]{3x-2}-3\sqrt{6-5x}+16=0\)
d) \(\sqrt[3]{x+6}-2\sqrt{x-1}=4-x^2\)
\(7+12\sqrt{x+1}=x+4\sqrt{x^2+3x+2}\)
\(\sqrt{x^2+x+2}=\dfrac{3x^2+3x+2}{3x+1}\)
Với giá trị nào của x thì mỗi căn thức sau đây có nghĩa:
a) \(\sqrt{\dfrac{x}{3}}\)
b) \(\sqrt{-5x}\)
c) \(\sqrt{4-x}\)
d) \(\sqrt{3x+7}\)
e) \(\sqrt{-3x+4}\)
f) \(\sqrt{\dfrac{1}{-1+x}}\)
g) \(\sqrt{1+x^2}\)
h) \(\sqrt{\dfrac{5}{x-2}}\)
1) \(x+1+\sqrt{x^2-4x+1}=3\sqrt{x}\)
2) \(4x^3+x-\left(x+1\right)\sqrt{2x+1}=0\)
3) \(x-\sqrt{x}=1-\sqrt{2\left(x^2-x+1\right)}\)
4) \(\sqrt{x+1}+\sqrt{4-x}+\sqrt{\left(x+1\right)\left(4-x\right)}=5\)
5) \(\sqrt{3x-2}+\sqrt{x-1}=4x-9+2\sqrt{3x^2-5x+2}\)
6) \(3\sqrt{x+2}-6\sqrt{2-x}+4\sqrt{4-x^2}=10-3x\)