1) \(\dfrac{11}{12}\times\dfrac{28}{13}-\dfrac{11}{12}\times\dfrac{15}{13}=\dfrac{11}{12}\times\left(\dfrac{28}{13}-\dfrac{15}{13}\right)=\dfrac{11}{12}\times\dfrac{13}{13}=\dfrac{11}{12}\times1=\dfrac{11}{12}\)

Vậy biểu thức trên có kết quả là : \(\dfrac{11}{12}\)

2) \(x+653=87\times11\)

    \(x+653=957\) 

    \(x=957-653\)

    \(x=304\)

Vậy `x = 304 `

3) \(\text{70 000 + 800 + 20 + 9}=70829\)

a: \(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}\)

\(=\dfrac{3x+9\sqrt{x}}{x-9}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)

b: Khi x=11+6 căn 2 thì \(M=\dfrac{3\left(3+\sqrt{2}\right)}{3+\sqrt{2}-3}=\dfrac{9+3\sqrt{2}}{\sqrt{2}}=\dfrac{9\sqrt{2}+6}{2}\)

c: M<1
=>\(\dfrac{3\sqrt{x}-\sqrt{x}+3}{\sqrt{x}-3}< 0\)

=>căn x-3<0

=>0<x<9

`a,` \(M=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{9-x}\) \(\left(x\ne\pm3;x>0\right)\)

\(M=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{x-9}-\dfrac{3+11\sqrt{x}}{x-9}\)

\(M=\dfrac{2x-6\sqrt{x}}{x-9}+\dfrac{x+3\sqrt{x}+\sqrt{x}+3}{x-9}-\dfrac{3+11\sqrt{x}}{x-9}\)

\(M=\dfrac{3x+9\sqrt{x}}{x-9}\)

\(M=\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)

\(M=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)

`b,`Ta có :

 \(M=\dfrac{3\sqrt{11+6\sqrt{2}}}{\sqrt{11+6\sqrt{2}}-3}\)

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

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

\(M=\dfrac{9+3\sqrt{2}}{\sqrt{2}}\)

\(M=\dfrac{6+9\sqrt{2}}{2}\)

`c,`  Để `M<1` Ta có :

 \(\dfrac{3\sqrt{x}}{\sqrt{x}-3}< 1\)

\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}-1< 0\)

\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-3}< 0\)

\(\dfrac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)

\(\sqrt{x}-3< 0\) ( vì \(2\sqrt{x}+3>0\) )

\(\sqrt{x}< 3\)

\(x< 9\)

Đối chiếu ĐKXĐ ta có : `0<x<9`

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

b) Thay x=0 vào A, ta được:

\(A=\dfrac{15\cdot\sqrt{0}-11}{0+2\sqrt{0}-3}-\dfrac{3\sqrt{0}-2}{\sqrt{0}-1}-\dfrac{2\sqrt{0}+3}{\sqrt{0}+3}\)

\(=\dfrac{-11}{-3}-\dfrac{-2}{-1}-\dfrac{3}{3}\)

\(=\dfrac{11}{3}-2-1\)

\(=\dfrac{11}{3}-\dfrac{9}{3}=\dfrac{2}{3}\)

1) Sửa đề: x=0,09

Thay x=0,09 vào A, ta được:

\(A=\dfrac{\sqrt{0.09}}{\sqrt{0.09}-1}=\dfrac{0.3}{0.3-1}=\dfrac{0.3}{-0.7}=\dfrac{-3}{7}\)

1:

\(A=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

3: A nguyên

=>-5căn x-15+17 chia hết cho căn x+3

=>căn x+3 thuộc Ư(17)

=>căn x+3=17

=>x=196

\(a,A=\dfrac{9-3x+x^2+10x+25-x^2+1}{\left(x-1\right)\left(x+5\right)}\\ A=\dfrac{7x+35}{\left(x-1\right)\left(x+5\right)}=\dfrac{7\left(x+5\right)}{\left(x-1\right)\left(x+5\right)}=\dfrac{7}{x-1}\\ b,A\in Z\\ \Leftrightarrow x-1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Leftrightarrow x\in\left\{-6;0;2;8\right\}\left(tm\right)\\ b,A< 0\Leftrightarrow x-1< 0\left(7>0\right)\\ \Leftrightarrow x< 1;x\ne-5\\ c,\left|A\right|=3\Leftrightarrow\dfrac{7}{\left|x-1\right|}=3\Leftrightarrow\left|x-1\right|=\dfrac{7}{3}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}+1=\dfrac{10}{3}\left(tm\right)\\x=-\dfrac{7}{3}+1=-\dfrac{4}{3}\left(tm\right)\end{matrix}\right.\)

A = (15/√x) - (11x + 2√x - 3) - (3√x - 2√x - 1) - (2√x + 3√x - 3)

Tiếp theo, kết hợp các thành phần tương tự:

A = 15/√x - 11x - 2√x + 3 + 3√x - 2√x + 1 - 2√x - 3√x + 3

Đơn giản hóa biểu thức:

A = -11x + 15/√x + 4

Để tìm giá trị lớn nhất của A, ta có thể tìm điểm đạt cực đại của hàm số A(x). Tuy nhiên, để làm điều này, cần biết thêm về giá trị của x.

Sửa đề: (3căn x-2)/căn x-1-(2căn x+3)/(căn x+3)\(A=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

\(A=\dfrac{-5\sqrt{x}-15+17}{\sqrt{x}+3}==-5+\dfrac{17}{\sqrt{x}+3}< =\dfrac{17}{3}-5=\dfrac{2}{3}\)

Dấu = xảy ra khi x=0

a: Ta có: \(A=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}\)

\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

\(=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)

b: Để \(A\ge0\) thì \(\sqrt{x}-3>0\)

hay x>9