Đặt x^2-x=a(a>=0)
=>căn a=a-2
=>a=(a-2)^2
=>a^2-4a+4-a=0
=>a^2-5a+4=0
=>(a-1)(a-4)=0
=>a=1 hoặc a=4
=>x^2-x=1 hoặc x^2-x=4
=>\(x\in\left\{\dfrac{1\pm\sqrt{5}}{2};\dfrac{1\pm\sqrt{17}}{2}\right\}\)
GIẢI CÁC PT SAU:
\(\sqrt{x^2+5x+1}=\sqrt{x+1}\)
\(\sqrt{x^2+2x+4}=\sqrt{2-x}\)
\(\sqrt{2x+4}-\sqrt{2-x}=0\)
GIẢI PT SAU:
\(\sqrt{3x-3}-\sqrt{5-x}=\sqrt{2x-4}\)
\(x^2-6x+9=4\sqrt{x^2-6x+6}\)
\(x^2-x+8-4\sqrt{x^2-x+4}=0\)
\(\sqrt{x^2-1}+\sqrt{x^2-3x+2}\le2\sqrt{x^2-x}\)
Giải phương trình:
a) \(5x^2-10x=4\left(x-1\right)\sqrt{x^2-2x+2}\)
b) \(\sqrt{2x^2+22x+29}-x-2=2\sqrt{2x+3}\)
c) \(x^3-7x^2+9x+12=\left(x-3\right)\left(x-2+5\sqrt{x-3}\right)\left(\sqrt{x-3}-1\right)\)
Mọi người giúp gấp với ạ.
Giải pt:
\(\sqrt{x^2+10x+21}=3\sqrt{x+3}+2\sqrt{x+7}-6\)
\(4\left(x+1\right)^2=\left(2x+10\right)\left(1-\sqrt{3+2x}\right)^2\)
\(\frac{1}{1-\sqrt{1-x}}-\frac{1}{1+\sqrt{1-x}}=\frac{\sqrt{3}}{x}\)
\(\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{x^2+4x+3}\)
\(\sqrt{x-2}+\sqrt{4-x}=x^2-6x+11\)
a.\(\sqrt{5x^2+14x+9}=5\sqrt{x+1}+\sqrt{x^2-8x-20}\)
b.\(\left(x+1\right)\left(x-3\right)+3\sqrt{x^2-x-3}=4\)
c. \(\sqrt{4x^2+x-1}=x+1\)
cứu với mọi ng ơi ☘ 
Giải các pt sau bằng cách đặt ẩn phụ:
a/\(-4\sqrt{\left(4-x\right)\left(2+x\right)}=x^2-2x-12\)
b/\(\left(x-3\right)^2+3x-22=\sqrt{x^2-3x+7}\)
c/\(\frac{\sqrt{x+4}+\sqrt{x-4}}{2}=x+\sqrt{x^2-16}-6\)
d/\(\sqrt{x-1}+\sqrt{x+3}+2\sqrt{\left(x-1\right)\left(x+3\right)}=4-2x\)
e/\(\sqrt{x+7}+\sqrt{7x-6}+\sqrt{49x^2+7x-42}=181-14x\)
f/\(5\sqrt{x}+\frac{5}{2\sqrt{x}}=2x+\frac{1}{2x}+4\)
\(\left\{{}\begin{matrix}\sqrt{x+y}=2+\sqrt{x-y}\\\sqrt{x^2+y^2+1}-\sqrt{x^2-y^2}=3\end{matrix}\right.\)
\(\sqrt{x+\sqrt{x^2-1}}=\dfrac{27\sqrt{2}}{8}\left(x-1\right)^2\sqrt{x-1}\)
Giải pt: \(x+\sqrt[3]{x^3-x^2}+\sqrt[3]{x^3-x}=\sqrt[3]{x^2+x+\frac{1}{3}}+\sqrt[3]{x^2+\frac{1}{3}}+\sqrt[3]{x+\frac{1}{3}}\)
