ĐK : a,b > 0

\(=\left[\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right]\cdot\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\frac{a-b}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\cdot\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)=a-b\)

b4 : 

\(a,x-1=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

\(b,x-5=\left(\sqrt{x}-\sqrt{5}\right)\left(\sqrt{x}+\sqrt{5}\right)\)

\(c,x+2\sqrt{xy}+y=\left(\sqrt{x}+\sqrt{y}\right)^2\)

\(d,x-4\sqrt{x}\sqrt{y}+4y=\left(\sqrt{x}-2\sqrt{y}\right)^2\)

b5:

\(a,ĐK:x\ge1\)

\(\sqrt{9\left(x-1\right)}+\sqrt{4\left(x-1\right)}-\frac{4}{5}\sqrt{25\left(x-1\right)}=1\)

\(\Leftrightarrow3\sqrt{x-1}+2\sqrt{x-1}-4\sqrt{x-1}=1\)

\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x=2\left(tm\right)\)

\(b,ĐK:x\ge5\)

\(\frac{1}{3}\sqrt{9\left(x-5\right)}+\frac{1}{2}\sqrt{4\left(x-5\right)}-\frac{7}{5}\sqrt{25\left(x-5\right)}=2\)

\(\Leftrightarrow\sqrt{x-5}+\sqrt{x-5}-7\sqrt{x-5}=2\)

\(\Leftrightarrow-5\sqrt{x-5}=2\)

\(\Leftrightarrow\sqrt{x-5}=-\frac{2}{5}\left(voli\right)\)

\(c,ĐK:x>0\)

\(\sqrt{x}+\frac{9}{\sqrt{x}}=6\)

\(\Leftrightarrow x+9=6\sqrt{x}\)

\(\Leftrightarrow x-6\sqrt{x}+9=0\)

\(\Leftrightarrow\left(\sqrt{x}-3\right)^2=0\)

\(\Leftrightarrow x=9\left(tm\right)\)

\(ĐK:x\ge2\)

\(\Leftrightarrow x+2+\sqrt{x-2}-2\sqrt{x+1}=0\)

\(\Leftrightarrow\left(\sqrt{x+1}-1\right)^2+\sqrt{x-2}=0\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x+1}-1=0\\\sqrt{x-2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\end{cases}\left(voli\right)}\)

Vậy phương trình vô nghiệm

\(\Delta ABC\) vuông tại A

AM là đường trung tuyến => AM=MB=MC=\(\frac{BC}{2}\)

=> \(\Delta AMB\)cân tại M, \(\Delta AMC\) cân tại M

Xét \(\Delta AMB\) và     \(\Delta AMC\) có

     AM chung

     MB=MC

=>\(\Delta AMB=\Delta AMC\)

=>AB =AC =3 cm( 2 cạnh trương ứng)

hok tốt

AC= 3 cm ấy

4. Ta có : 2M = \(\frac{2\sqrt{x}+4}{2\sqrt{x}-3}=\frac{2\sqrt{x}-3+7}{2\sqrt{x}-3}=1+\frac{7}{2\sqrt{x}-3}\)

Để M nguyên dương thì \(\frac{7}{2\sqrt{x}-3}\)nguyên dương

=> \(2\sqrt{x}-3\in\left\{1;7\right\}\)

=> \(x\in\left\{\sqrt{2};\sqrt{5}\right\}\)

Mà  x\(\in\)Z; x\(\ge\)0

=> Không có giá trị x nguyên thỏa mãn nào để M nguyên dương

5. \(C=\frac{2\sqrt{x}-3}{\sqrt{x}-2}=\frac{2\sqrt{x}-4+1}{\sqrt{x}-2}=2+\frac{1}{\sqrt{x}-2}\)

Để C nguyên thì \(\frac{1}{\sqrt{x}-2}\)nguyên

=> \(\sqrt{x}-2\in\left\{-1;1\right\}\)

=> \(x\in\left\{1;\sqrt{3}\right\}\)

Mà x nguyên và C cần đạt GTNN nên x = 1

Vậy minC = 1 <=> x = 1

6. \(D=\frac{x-3}{\sqrt{x}+1}=\frac{x+\sqrt{x}-\sqrt{x}-1-2}{\sqrt{x}+1}\)

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

Để D nguyên thì \(\sqrt{x}\)nguyên; \(\frac{2}{\sqrt{x}+1}\)nguyên

=> \(\sqrt{x}+1\in\left\{-2;-1;1;2\right\}\)

Mà \(\sqrt{x}+1\ge1\)=> \(\sqrt{x}\in\left\{0;1\right\}\)(tm \(\sqrt{x}\)nguyên)

=> x\(\in\){ 0 ; 1 } . Mà x khác 1

=> x = 0

Vậy D nguyên khi x = 0

Rút gọn biểu thức :

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

\(=\frac{5\sqrt{5}-5\sqrt{2}+10-2\sqrt{10}}{3}\)

7/ căn bậc 2 mik ko ghi chi tiết đc thông cảm

1.46 \(a,P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)

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

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

\(P=\frac{3x-4\sqrt{x}}{x-4}\)

gọn quá