Trả lời:

\(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right).\sqrt{7}+\sqrt{84}\)

\(=\left(\sqrt{2^2.7}-2\sqrt{3}+\sqrt{7}\right).\sqrt{7}+\sqrt{2^2.21}\)

\(=\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right).\sqrt{7}+2\sqrt{21}\)

\(=\left(3\sqrt{7}-2\sqrt{3}\right).\sqrt{7}+2\sqrt{21}\)

\(=3\sqrt{7}.\sqrt{7}-2\sqrt{3}.\sqrt{7}+2\sqrt{21}\)

\(=3.7-2.\sqrt{21}+2\sqrt{21}\)

\(=21\)

bạn ơi phải là \(5-2\sqrt{2}-\sqrt{3}\)nha bạn nếu dc thì mik lm luôn

đặt \(5-2\sqrt{2}-\sqrt{3}\)=A ta có

\(A=6-1-2\sqrt{3}-3\sqrt{2}+\sqrt{2}+\sqrt{3}\)

\(=6-1-\sqrt{12}-\sqrt{18}+\sqrt{2}+\sqrt{3}\)

\(=\left(\sqrt{6}-1\right)\left(\sqrt{6}-\sqrt{2}-\sqrt{3}+1\right)\)

\(=\left(\sqrt{2}-1\right)\left(\sqrt{3}-1\right)\left(\sqrt{6}-1\right)\)

GTBT là 10

ta có:

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

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

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

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

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

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

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