1.3 Giải phương trình: 

a) \(\sqrt{2x+3}=1+\sqrt{2}\)(ĐK: \(x\ge-\frac{3}{2}\)) 

\(\Leftrightarrow2x+3=\left(1+\sqrt{2}\right)^2=3+2\sqrt{2}\)

\(\Leftrightarrow2x=2\sqrt{2}\)

\(\Leftrightarrow x=\sqrt{2}\)(tm) 

b) \(\sqrt{x+1}=\sqrt{5}+3\)(ĐK: \(x\ge-1\)) 

\(\Leftrightarrow x+1=\left(\sqrt{5}+3\right)^2=14+6\sqrt{5}\)

\(\Leftrightarrow x=13+6\sqrt{5}\)(tm) 

c) \(\sqrt{3x-2}=2-\sqrt{3}\)(ĐK: \(x\ge\frac{2}{3}\))

\(\Leftrightarrow3x-2=\left(2-\sqrt{3}\right)^2=7-4\sqrt{3}\)

\(\Leftrightarrow x=\frac{9-4\sqrt{3}}{3}\)(tm) 

1.4: Phân tích thành nhân tử: 

a) \(ab+b\sqrt{a}+\sqrt{a}+1=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)=\left(b\sqrt{a}+1\right)\left(\sqrt{a}+1\right)\)

b) \(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}=x\sqrt{x}-y\sqrt{y}+x\sqrt{y}-y\sqrt{x}\)

\(=\left(x-y\right)\left(\sqrt{x}+\sqrt{y}\right)\)

x=-3 

nhớ tít cho mình nha

x= 1  pp: bình phương 2 vế

b nha

Đặt bth đã cho là A, ta có:

A²=3−√5+3+√5+2√3−√5.√3+√53−5+3+5+23−5.3+5

A²=6+2√(3−√5)(3+√5)6+2(3−5)(3+5)

A^{2=6+2√9−56+29−5}

A²=6+4=10

 ( Tôi giúp ng ae rồi đấy, ok thì kb nhoa) =33

A=√10

=\(\frac{\sqrt{2}\left(\sqrt{3+\sqrt{5}}\right)}{\sqrt{2}}+\frac{\sqrt{2}\left(\sqrt{3-\sqrt{5}}\right)}{\sqrt{2}}\)

=\(\frac{\sqrt{6+2\sqrt{5}}}{\sqrt{2}}+\frac{\sqrt{6-2\sqrt{5}}}{\sqrt{2}}\)

=\(\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}\)

=\(\frac{\sqrt{5}+1+\sqrt{5}-1}{\sqrt{2}}\)

=\(\frac{2\sqrt{5}}{\sqrt{2}}\)

=\(\sqrt{2}\sqrt{5}\)=\(\sqrt{10}\)

Trả lời:

\(\frac{a-\sqrt{a}}{1-a}\)

\(=\frac{\sqrt{a}.\sqrt{a}+\sqrt{a}}{1^2-\left(\sqrt{a}\right)^2}\)

\(=\frac{\sqrt{a}.\left(\sqrt{a}+1\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\)

\(=\frac{\sqrt{a}}{1-\sqrt{a}}\)

Trả lời:

Bài 1:

a, \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}\)

\(=3\sqrt{2}-4\sqrt{3^2.2}+2\sqrt{4^2.2}\)

\(=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}\)

\(=-\sqrt{2}\)

b, \(\sqrt{\left(2+\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)

\(=\left|2+\sqrt{5}\right|+\sqrt{5-6\sqrt{5}+9}\)

\(=2+\sqrt{5}+\sqrt{\left(\sqrt{5}\right)^2-2.\sqrt{5}.3+3^2}\)

\(=2+\sqrt{5}+\sqrt{\left(\sqrt{5}-3\right)^2}\)

\(=2+\sqrt{5}+\left|\sqrt{5}-3\right|\)

\(=2+\sqrt{5}+\left(3-\sqrt{5}\right)\)

\(=2+\sqrt{5}+3-\sqrt{5}=5\)

c, \(\frac{2}{\sqrt{3}-1}+\frac{2}{\sqrt{3}+1}\)

\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}+\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)

\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}\right)^2-1^2}+\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}\right)^2-1^2}\)

\(=\frac{2\left(\sqrt{3}+1\right)}{3-1}+\frac{2\left(\sqrt{3}-1\right)}{3-1}\)

\(=\frac{2\left(\sqrt{3}+1\right)}{2}+\frac{2\left(\sqrt{3}-1\right)}{2}\)

\(=\sqrt{3}+1+\sqrt{3}-1=2\sqrt{3}\)

Bài 2:

a, \(\sqrt{2x+1}-3=1\left(ĐK:x\ge-\frac{1}{2}\right)\)

\(\Leftrightarrow\sqrt{2x+1}=4\)

\(\Leftrightarrow2x+1=16\)

\(\Leftrightarrow2x=15\)

\(\Leftrightarrow x=\frac{15}{2}\left(tm\right)\)

Vậy x = 15/2