Phân tích đa thức thành nhân tử
a. 7-3a (a lớn hơn hoặc =0)
b.\(14x^2-11\)
c.3x-\(6\sqrt{x}\)-6
d.\(x\sqrt{x}-3\sqrt{x}-2\)
Tìm điều kiện để các biểu thức sau có nghĩa
a, \(\sqrt{x-2}-\sqrt{4-x}\)
b, \(\dfrac{1}{\sqrt{x+1}-1}\)
c, \(\sqrt{x^2-4x+3}\)
d, \(\sqrt{-x^5}\)
e, \(\sqrt{\dfrac{x-3}{2-x}}\)
g, \(\sqrt{-\left|x-2\right|}\)
h, \(\sqrt{4x^2-4x+1}\)
Mình đang cần gấp, sắp phải nộp rồi
a)\(\sqrt{\sqrt{5}-\sqrt{3x}}\)
b) \(\sqrt{\sqrt{6x}-4x}\)
c) \(\sqrt{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}\)
d) \(\sqrt{\left(x-6\right)^6}\)
e) \(\sqrt{-12x+5}\)
f) \(2-4\sqrt{5x+8}\)
g) \(\sqrt{x^2-9}\)
d) \(x-5\sqrt{x}+6=0\)
e) \(\sqrt{x-1}+\dfrac{3}{2}\sqrt{4x-4}-\dfrac{2}{5}\sqrt{25x-25}=4\)
f) \(\sqrt{x-5}+\sqrt{4x-20}-\dfrac{1}{3}\sqrt{9x-45}=6\)
\(a,x^2-4x-6=\sqrt{2x^2-8x+12}\)
\(b,\sqrt{x^2+2-4\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}}=1\)
\(c,\sqrt{x+3+4\sqrt{x-1}}+\sqrt{8-6\sqrt{x-1}+x}=1\)
\(d,\sqrt{3-x+x^2}-\sqrt{2+x-x^2}=1\)
\(e,\sqrt{4x-9}+2\sqrt{3x^2-5x+2}=\sqrt{3x-2}+\sqrt{x-1}\)
a. \(\sqrt{\left(2x+3\right)^2}=x+1\)
b. \(\sqrt{\left(2x-1\right)^2}=x+1\)
c. \(\sqrt{x+3}=5\)
d. \(\sqrt{x+2}=\sqrt{7}\)
e. \(5\sqrt{x}=20\)
f. \(\sqrt{x+4}=7\)
g. \(\sqrt{\left(2x+1\right)^2}=3\)
a) \(\sqrt{4x^2-9}=2\sqrt{x+3}\)
b) \(\sqrt{4x+20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
c) \(\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27\sqrt{\dfrac{x-1}{81}}=4\)
d)\(5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
Giải phương trình:
a)\(\sqrt{\sqrt{5}-\sqrt{3x}}=\sqrt{8+2\sqrt{15}}\)
b)\(\sqrt{4x-20}-3\sqrt{\dfrac{x-5}{9}}=\sqrt{1-x}\)
c) \(\sqrt{4x+8}+2\sqrt{x+2}-\sqrt{9x+18}=1\)
d) \(\sqrt{x^2-6x+9}+x=11\)
e) \(\sqrt{3x^2-4x+3}=1-2x\)
f) \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
g) \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
Câu 2: Tìm x biết:
a. \(\sqrt{x-1}=2\)
b. \(\sqrt{3x+1}=\sqrt{4x-3}\)
c. \(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
d. \(\sqrt{x^2-4x+4}=\sqrt{6+2\sqrt{5}}\)