\(\sqrt{3x+1}+1=3x\)
\(\Leftrightarrow\sqrt{3x+1}=3x-1\left(x\ge-\dfrac{1}{3}\right)\)
\(\Leftrightarrow3x+1=\left(3x-1\right)^2\)
\(\Leftrightarrow3x+1=9x^2-6x+1\)
\(\Leftrightarrow9x^2-9x=0\)
\(\Leftrightarrow9x\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\left(TM\right)\\x=1\left(TM\right)\end{matrix}\right.\)
KL....
cho bieu thuc A=\(\left(\dfrac{6x+4}{3\sqrt{3x^3}-8}-\dfrac{\sqrt{3x}}{3x+2\sqrt{3x}+4}\right)\left(\dfrac{1+3\sqrt{3x^3}}{1+\sqrt{3x}}-\sqrt{3x}\right)\)
a.rut gon bieu thuc A
b. tim cac gia tri nguyen x de A nguyen
\(P=\left(\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{\sqrt{x}}\right).\left(\dfrac{1}{1-\sqrt{x}}-1\right)\)
a, rut gon
b, tim x de P=\(\sqrt{x}\)
Giải phương trình vô tỉ:
a) \(\sqrt{\left(1+x\right)\left(2-x\right)}=1+2x-2x^2\)
b) \(\sqrt{3x+8+6\sqrt{3x-1}}+\sqrt{3x+8-6\sqrt{3x-1}}=3x+4\)
c) \(2x\sqrt{x^2-x-1}+4x\sqrt{3x+1}=2x^2+2x+6\)
Tìm x:
1, \(\sqrt{1-2x}+\sqrt{1+2x}=\sqrt{4+x}\)
2,\(\sqrt{3x+1}-\sqrt{2-x}=\sqrt{3x-2}\)
3, \(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}=4\)
4, \(3x^2-6x-4=4\left(x-1\right)\sqrt{3x+1}\)
Giải phương trình vô tỉ:
1/ \(\sqrt{x^2+12}+5=3x+\sqrt{x^2+15}\)
2/ \(\sqrt{3x^2-5x+1}-\sqrt{x^2-2}=\sqrt{3\left(x^2-x+1\right)}-\sqrt{x^2-3x+4}\)
3/ \(\sqrt[5]{x-1}+\sqrt[3]{x+8}=-x^3+1\)
4/ \(\sqrt{5-x^6}+\sqrt[3]{3x^4-2}=1\)
\(\sqrt{3x^2-5x+1}-\sqrt{x^2-2}=\sqrt{3\left(x^2-x-1\right)}-\sqrt{x^2-3x+4}\)
1)giải pt: 1+\(\dfrac{2}{3}\sqrt{x-x^2}=\sqrt{x}+\sqrt{1-x}\)
2)giải pt: \(\dfrac{x^2}{\sqrt{3x-2}}-\sqrt{3x-2}=1-x\)
Tìm điều kiện xác định
\(A=\sqrt{x^2-5x+6}\)
\(B=\dfrac{x}{\sqrt{7x^2-8}}\)
\(C=\sqrt{-9x^2+6x-1}-\dfrac{1}{\sqrt{x^2+x+2}}\)
\(D=\sqrt{3-x^2}-\sqrt{\dfrac{2021}{3x+2}}\)
\(E=\sqrt{\dfrac{3x^2}{2x+1}-1}\)
\(F=\sqrt{25x^2-10x+1}+\dfrac{1}{1-5x}\)
Cho P=\(\left(\dfrac{6x+4}{3\sqrt{3x^3}}-\dfrac{\sqrt{3x}}{3x+2\sqrt{3x}+4}\right).\left(\dfrac{1+3\sqrt{3x^3}}{1+\sqrt{3x}}-\sqrt{3x}\right)\)
a) Rút gọn P.
b) Tìm các giá trị nguyên của x để P nhận giá trị nguyên.
