Bài 2:

a: ĐKXĐ: x>=0

\(\sqrt{3x}-5\sqrt{12x}+7\cdot\sqrt{27x}=12\)

=>\(\sqrt{3x}-5\cdot2\sqrt{3x}+7\cdot3\sqrt{3x}=12\)

=>\(12\sqrt{3x}=12\)

=>\(\sqrt{3x}=1\)

=>3x=1

=>x=1/3(nhận)

Bài 1:

a: \(A=\left(\sqrt{\frac23}+\sqrt{\frac{50}{3}}-\sqrt{24}\right)\cdot\sqrt6\)

\(=\left(\frac{2\sqrt6}{6}+\sqrt{\frac{100}{6}}-2\sqrt6\right)\cdot\sqrt6\)

\(=2+\sqrt{100}-2\cdot6=2+10-12=0\)

b: \(B=\left(\frac{\sqrt{14}-\sqrt7}{\sqrt2-1}+\frac{\sqrt{15}-\sqrt5}{\sqrt3-1}\right):\frac{1}{\sqrt7-\sqrt5}\)

\(=\left(\frac{\sqrt7\left(\sqrt2-1\right)}{\sqrt2-1}+\frac{\sqrt5\left(\sqrt3-1\right)}{\sqrt3-1}\right)\cdot\left(\sqrt7-\sqrt5\right)\)

\(=\left(\sqrt7+\sqrt5\right)\left(\sqrt7-\sqrt5\right)\)

=7-5

=2

Tham khảo thanh này để soạn đề chính xác hơn nha :vvv

a) Ta có: \(M=\left(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\right)\cdot\dfrac{x+3\sqrt{x}}{7-\sqrt{x}}\)

\(=\left(\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{7-\sqrt{x}}\)

\(=\dfrac{x-9-\left(x-2\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{7-\sqrt{x}}\)

\(=\dfrac{x-9-x+\sqrt{x}+2}{\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{-\left(\sqrt{x}-7\right)}\)

\(=\dfrac{\sqrt{x}-7}{\sqrt{x}-2}\cdot\dfrac{-1}{\sqrt{x}-7}\)

\(=\dfrac{-1}{\sqrt{x}-2}\)(1)

b) Ta có: \(x^2-4x=0\)

\(\Leftrightarrow x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=4\left(loại\right)\end{matrix}\right.\)

Thay x=0 vào biểu thức (1), ta được:

\(M=\dfrac{-1}{\sqrt{0}-2}=\dfrac{-1}{-2}=\dfrac{1}{2}\)

Vậy: Khi \(x^2-4x=0\) thì \(M=\dfrac{1}{2}\)

\(\dfrac{1}{\sqrt{5}}-\sqrt{3}-\dfrac{1}{\sqrt{5}}+\sqrt{3}\)

= \(\dfrac{1}{\sqrt{5}}-\dfrac{1}{\sqrt{5}}-\sqrt{3}+\sqrt{3}\)

=0

a: ĐKXĐ: x>=0; x<>1

b \(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\)

\(=\dfrac{x+2\sqrt{x}-x-\sqrt{x}-1}{x\sqrt{x}-1}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\)

\(=\dfrac{1}{\sqrt{x}+2}\)

c: Khi x=9-4 căn 5 thì \(A=\dfrac{1}{\sqrt{5}-2+2}=\dfrac{\sqrt{5}}{5}\)

d: căn x+2>=2

=>A<=1/2

Dấu = xảy ra khi x=0

gõ latex đi b=)

\(A=\sqrt{x}+1\) (đã thu gọn)

\(B=\dfrac{4\sqrt{x}}{x+4}\) (đã thu gọn)

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

\(A=\dfrac{3}{2\sqrt{x}}\) (đã thu gọn)

\(A=\dfrac{3}{\sqrt{x}+3}\) (đã thu gọn)

\(A=1-\sqrt{x}\) (đã thu gọn)

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

Bạn vui lòng viết đề bằng công thức toán để được hỗ trợ tốt hơn.

`a)A=sqrtx/(sqrtx+3)+(2sqrtx)/(sqrtx-3)-(3x+9)/(x-9)(x>=0,x ne 9)`

`=(sqrtx(sqrtx-3)+2sqrtx(sqrtx+3)-3x-9)/(x-9)`

`=(x-3sqrtx+2x+6sqrtx-3x-9)/(x-9)`

`=(3sqrtx-9)/(x-9)`

`=(3(sqrtx-3))/((sqrtx-3)(sqrtx+3))`

`=3/(sqrtx+3)`

`b)A=1/3`

`<=>3/(sqrtx+3)=1/3`

`<=>sqrtx+3=9`

`<=>sqrtx=6`

`<=>x=36(tm)`

`c)A=3/(sqrtx+3)`

`sqrtx+3>=3>0`

`=>A<=3/3=1`

Dấu "=" xảy ra khi `x=0`

\(R=\left(\dfrac{3\sqrt{x}}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{3x-5\sqrt{x}}{4-x}\right):\left(\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}-1\right)\left(ĐK:x\ge0,x\ne4\right)\\ =\left(\dfrac{3\sqrt{x}}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{3x-5\sqrt{x}}{\sqrt{x}^2-2^2}\right):\dfrac{2\sqrt{x}-1-\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\)

\(=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)+\sqrt{x}\left(\sqrt{x}+2\right)+3x-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}-2}{2\sqrt{x}-1-\sqrt{x}+2}\\ =\dfrac{3x-6\sqrt{x}+x+2\sqrt{x}+3x-5\sqrt{x}}{\sqrt{x}+2}.\dfrac{1}{\sqrt{x}+1}\)

\(=\dfrac{7x-9\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)

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