1: Ta có: \(\sqrt{-4x^2+25}=x\)
\(\Leftrightarrow-4x^2+25=x^2\)
\(\Leftrightarrow-5x^2=-25\)
\(\Leftrightarrow x^2=5\)
hay \(x=\sqrt{5}\)
2: Ta có: \(\sqrt{3x^2-4x+3}=1-2x\)
\(\Leftrightarrow3x^2-4x+3=4x^2-4x+1\)
\(\Leftrightarrow4x^2+1-3x^2-3=0\)
\(\Leftrightarrow x^2=2\)
hay \(x=-\sqrt{2}\)
Giải các phương trình sau:
a) \(\sqrt{x^2-4+4}=2-x\)
b) \(\sqrt{4x-8}-\dfrac{1}{5}\sqrt{25x-50}=3\sqrt{x-2}-1\)
c) \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)
d) \(\dfrac{1}{2}\sqrt{x-2}-4\sqrt{\dfrac{4x-8}{9}}+\sqrt{9x-18}-5=0\)
e)\(\sqrt{49-28x+4x^2}-5=0\)
f) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
g) x2 - 4x - 2\(\sqrt{2x-5}+5=0\)
h)\(\sqrt{3x-2}=\sqrt{x+1}\)
i) x + y + z + 8 = \(2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
k) \(\sqrt{x^2-3x}-\sqrt{x-3}=0\)
l)\(\sqrt{x^2-4}+\sqrt{x-2}=0\)
m) \(4\sqrt{x+1}=x^2-5x+14\)
n) \(\sqrt{x^2-6x+9}-\sqrt{4x^2+4x+1}=0\)
a:\(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}\left(b>0;a\ne4\right)\)
b:\(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\left(x\ge0;y\ge0;x\ne0\right)\)
c:\(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}\left(a>0;b\ne2\right)}\)
d:\(\dfrac{x}{\left(y-3\right)^2}.\sqrt{\dfrac{\left(y-3\right)^2}{x^2}\left(x>0;y\ne3\right)}\)
e:2x +\(\dfrac{\sqrt{1-6x+9x^2}}{3x-1}\)
Giải phương trình:
1) \(x^2-4x-2\sqrt{2x-5}+5=0\)
2)\(x+y+4=2\sqrt{x}+4\sqrt{y-1}\)
3)\(\sqrt{x-2}+\sqrt{y-3}+\sqrt{z-5}=\dfrac{1}{2}\left(x+y+z-7\right)\)
4)\(\sqrt{x-2}+\sqrt{10-x}=x^2-12x+40\)
B1: bt sau đây xác định vs gt nào của x
a)\(\sqrt{\dfrac{4}{2x+3}}\)
b)\(\sqrt{x\left(x+2\right)}\)
c)\(\sqrt{\dfrac{2x-1}{2-x}}\)
B2: tính
a)\(\sqrt{\left(\sqrt{3}-2\right)^2}\) b)\(\sqrt{4-2\sqrt{3}}\) c)\(\sqrt{3+2\sqrt{2}}\) d)\(\sqrt{9-4\sqrt{5}}\)
B3: Tìm x bik:
a) \(\sqrt{25-20x+4x^2}+2x=5\) b)\(\sqrt{x^2+\dfrac{1}{2}x+\dfrac{1}{16}}=\dfrac{1}{4}-x\) c)\(\sqrt{x-2\sqrt{x-1}}=2\)
Rút gọn:
\(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}-\frac{5}{\sqrt{3}-2\sqrt{2}}-\frac{5}{\sqrt{3}+\sqrt{8}}\)
Giải các phường trình sau:
1) \(\sqrt{2x-1}=\sqrt{5}\)
2) \(\sqrt{3}x^2-\sqrt{12}=0\)
3) \(\sqrt{x-5}=3\)
4) \(\sqrt{4x^2}-6\)
5) \(\sqrt{\left(x-3\right)^2}=9\)
6) \(\sqrt{4\left(1-x\right)^2}-6=0\)
7) \(\sqrt{9\left(x-1\right)}=21\)
8) \(\sqrt[3]{x+1}=2\)
9) \(\sqrt{4x^2+4x+1}=6\)
10) \(\sqrt{2}x-\sqrt{50}=0\)
11) \(\sqrt{\left(2x-1\right)^2}=3\)
12) \(\sqrt[3]{3-2x}=-2\)
Mọi người ơi giúp em với!!! :((((
Với giá trị nào của x thì các căn thức trên có nghĩa :
a)\(\sqrt{3x^2+1}\)
b)\(\sqrt{4x^2-4x+1}\)
c)\(\sqrt{\dfrac{3}{x+4}}\)
h)\(\sqrt{x^2-4}\)
i) \(\sqrt{\dfrac{2+x}{5-x}}\)
\(a.\sqrt{x-2}-3\sqrt{x^2-4}=0\)
b.\(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\)
Rút gọn :
\(\dfrac{\sqrt{x+\sqrt{4\left(x-1\right)}}-\sqrt{x-\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}.\left(\sqrt{x-1}-\dfrac{1}{\sqrt{x-1}}\right)\)
b)\(\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)
c)\(\left(\sqrt{5}+1\right)\left(\sqrt{7}+1\right)\left(\sqrt{35}+1\right)\left(34-4\sqrt{7}-6\sqrt{5}\right)\)
d) \(\left(\sqrt{7}+1\right)\left(2\sqrt{2}-1\right)\left(2\sqrt{14}-1\right)\left(55+12\sqrt{2}-7\sqrt{7}\right)\)
e)\(\left(3\sqrt{2}+1\right)\left(2\sqrt{3}+1\right)\left(6\sqrt{6}+1\right)\left(215-34\sqrt{3}-33\sqrt{2}\right)\)
Bài 1: Tìm điều kiện để các phân thức sau có nghĩa
a)\(\frac{x-1}{x+1}b)\frac{2x+1}{-3x+5}c)\frac{3x-1}{x^2-4}d)\frac{x-1}{x^2+4}e)\frac{x-1}{\left(x-2\right)\left(x+3\right)}g)\frac{x-1}{x+2}:\frac{x}{x+1}\)
Bài 2 :Tìm điều kiện để các căn thức sau có nghĩa:\(1)\sqrt{3x}|2)\sqrt{-x}|3)\sqrt{3x+2}|4)\sqrt{5-2x}|5)\sqrt{x^2}|6)\sqrt{-4x^2}|7)\sqrt{x-3}+\sqrt{2x+2}|8)\sqrt{\frac{-3}{x+2}}|9)\frac{3}{2x-4}\)
