\(\left(2\sqrt{3}+\sqrt{5}\right).\sqrt{3}-\sqrt{60}\)

\(=\sqrt{3}.\left(2+\sqrt{3}+\sqrt{5}\right)-2\sqrt{15}\)

\(=2\sqrt{3}+3+\sqrt{15}-2\sqrt{15}\)

\(=2\sqrt{3}+3-\sqrt{15}\)

Lê Trung Hiếu

sai r ạ :v như v ms đúng ạ

\(\left(2\sqrt{3}+\sqrt{5}\right)\sqrt{3}-\sqrt{60}\)

\(=2\sqrt{3}\cdot\sqrt{3}+\sqrt{5}\cdot\sqrt{3}-\sqrt{60}\)

\(=2\cdot3+\sqrt{15}-\sqrt{60}\)

\(=6+\sqrt{15}-2\sqrt{15}\)

\(=6-\sqrt{15}\)

a, 2

b, 2

c, 39

a) \(\sqrt{2x}=4\) 

 =>" x=2 

b) \(4x=8\Leftrightarrow x=2\)

c)\(\sqrt{\left(x-3\right)^2}=6\) 

\(x-3=6\) 

\(\Rightarrow x=9\) 

I I  là dấu giá trị tuyệt đối nhé

|7 + 5x| = 1 - 4x

=> \(\orbr{\begin{cases}7+5x=1-4x\left(đk:x\le\frac{1}{4}\right)\\7+5x=4x-1\left(đk:x\ge\frac{1}{4}\right)\end{cases}}\)

=> \(\orbr{\begin{cases}7-1=-4x-5x\\7+1=4x-5x\end{cases}}\)

=> \(\orbr{\begin{cases}6=-9x\\8=-x\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{2}{3}\left(tm\right)\\x=-8\left(ktm\right)\end{cases}}\)

|4x² - 2x| + 1 = 2x

=> |4x² - 2x| = 2x - 1

=> \(\orbr{\begin{cases}4x^2-2x=2x-1\left(đk:x\ge\frac{1}{2}\right)\\4x^2-2x=1-2x\left(đk:x\le\frac{1}{2}\right)\end{cases}}\)

=> \(\orbr{\begin{cases}4x^2-2x-2x+1=0\\4x^2-2x-1+2x=0\end{cases}}\)

=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\4x^2-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}2x-1=0\\x^2=\frac{1}{4}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=\pm\frac{1}{2}\end{cases}}\)(tm)

Vậy ...

a) \(\frac{3}{2\sqrt{3}}=\frac{3}{2.3^{\frac{1}{2}}}=\frac{3^{1-\frac{1}{2}}}{2}=\frac{3^{\frac{1}{2}}}{2}=\frac{\sqrt{3}}{2}\)

b) \(\frac{5}{2\sqrt{3}}=\frac{5\sqrt{3}}{2\sqrt{3}.\sqrt{3}}=\frac{5\sqrt{3}}{6}\)

Câu hỏi của Trần Thanh Phương - Toán lớp 9 | Học trực tuyến

Tự lực cánh sinh thôi...

linh ta linh tinh

a, Thay x = 9 vào biểu thức \(A=\frac{\sqrt{x}-2}{\sqrt{x}-1}\)  ta được:

\(A=\frac{\sqrt{9}-2}{\sqrt{9}-1}=\frac{\sqrt{3^2}-2}{\sqrt{3^2}-1}=\frac{3-2}{3-1}=\frac{1}{2}\)

Vậy với x = 9 thì \(A=\frac{1}{2}\)

\(b,\left(\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}-1}{x+1}\left(x\ge0;x\ne1\right)\)

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

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

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

\(=\frac{1}{\sqrt{x}+1}\)

a, Thay x=9 ta có 

\(A=\frac{\sqrt{9}-2}{\sqrt{9}-1}=\frac{3-2}{3-1}=\frac{1}{2}\)

b,\(B=\left(\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}-1}{x+1}\)

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

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

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

\(=\frac{1}{\sqrt{x}+1}\)