\(\left(\dfrac{\dfrac{x}{x+1}}{\dfrac{x^2}{x^2+x+1}}-\dfrac{2x+1}{x^2+x}\right)\dfrac{x^2-1}{x-1}\)ĐK : \(x\ne\pm1\)

\(=\left(\dfrac{x}{x+1}.\dfrac{x^2+x+1}{x^2}-\dfrac{2x+1}{x\left(x+1\right)}\right)\left(x+1\right)=\left(\dfrac{x^2+x-1}{x^2+x}-\dfrac{2x+1}{x\left(x+1\right)}\right)\left(x+1\right)\)

\(=\left(\dfrac{x^2+x-1-2x-1}{x\left(x+1\right)}\right)\left(x+1\right)=\dfrac{x^2-3x-2}{x}\)

à xin lỗi mình nhầm dòng cuối 

\(=\dfrac{x^2-x-2}{x}=\dfrac{\left(x+1\right)\left(x-2\right)}{x}\)

Để biểu thức trên nhận giá trị dương khi 

\(\dfrac{\left(x+1\right)\left(x-2\right)}{x}>0\)bạn tự xét TH cả tử và mẫu nhé, mình đánh trên này bị lỗi 

a) Ta có: \(\dfrac{x^2}{y^2}:\sqrt{\dfrac{x^2}{y^4}}\)

\(=\dfrac{x^2}{y^2}:\dfrac{x}{y^2}\)

=x

b) Ta có: \(\sqrt{\dfrac{27\left(x-1\right)^2}{12}}+\dfrac{3}{2}-\left(x-2\right)\sqrt{\dfrac{50x^2}{8\left(x-2\right)^2}}\)

\(=\sqrt{\dfrac{9}{4}}\cdot\sqrt{\left(x-1\right)^2}+\dfrac{3}{2}-\left(x-2\right)\cdot\sqrt{\dfrac{25}{4}}\cdot\sqrt{\dfrac{x^2}{\left(x-2\right)^2}}\)

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

\(=\dfrac{3}{2}x-\dfrac{3}{2}+\dfrac{3}{2}-\dfrac{5}{2}\left(x-2\right)\cdot\dfrac{-x}{x-2}\)

\(=\dfrac{3}{2}x+\dfrac{5}{2}\cdot\left(x\right)\)

=4x

\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

Có 

Câu 1:

Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

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

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

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

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

Câu 3: 

Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)

\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)

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

\(=\sqrt{a}\left(\sqrt{a}-2\right)\)

\(=a-2\sqrt{a}\)

1,

\(A=\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{4x^2+x-2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x^2-4}{\left(x-2\right)\left(x+2\right)}\)

\(x=4\Rightarrow A=\dfrac{4.x^2-4}{\left(4-2\right)\left(4+2\right)}=...\)

2.

\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3-5x}{\left(x-1\right)\left(x+1\right)}\)

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

\(=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)

3.

Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)

4.

\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)

\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)

5.

\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)

\(=\dfrac{x^2+2x+2\left(x+2\right)}{x\left(x+2\right)}=\dfrac{x^2+4x+4}{x\left(x+2\right)}\)

\(=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x+2}{x}\)

\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)

1: Ta có: \(P=\dfrac{x-\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\)

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

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

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

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)

2)

a) Thay \(x=\dfrac{9}{4}\) vào P, ta được:

\(P=\left(\dfrac{3}{2}+2\right):\left(\dfrac{3}{2}+3\right)=\dfrac{7}{2}:\dfrac{11}{2}=\dfrac{7}{11}\)

b) Ta có: \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

\(=5+\sqrt{2}-4-\sqrt{2}\)

=1

Thay x=1 vào P, ta được:

\(P=\dfrac{1+2}{1+3}=\dfrac{3}{4}\)

ĐKXĐ: \(x\ne\pm3\)

\(P=\left[\dfrac{x\left(x+3\right)}{x^2\left(x+3\right)+9\left(x+3\right)}+\dfrac{3}{x^2+9}\right]:\left[\dfrac{1}{x-3}-\dfrac{6x}{x^2\left(x-3\right)+9\left(x-3\right)}\right]\)

\(=\left[\dfrac{x\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}+\dfrac{3}{x^2+9}\right]:\left[\dfrac{1}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\)

\(=\dfrac{x+3}{x^2+9}:\dfrac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}=\dfrac{x+3}{x^2+9}.\dfrac{\left(x-3\right)\left(x^2+9\right)}{\left(x-3\right)^2}=\dfrac{x+3}{x-3}\)

Ý 2 mình k hiểu ý bạn lắm

\(P=\dfrac{x+3}{x-3}=\dfrac{x-3+6}{x-3}=1+\dfrac{6}{x-3}\in Z\)

\(\Leftrightarrow\left(x-3\right)\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

Kết hợp vs ĐKXĐ \(\Rightarrow x\in\left\{0;1;2;4;5;6;9\right\}\)

1.\(\sqrt{27}+\sqrt{48}-\sqrt{108}-\sqrt{12}=3\sqrt{3}+4\sqrt{3}-6\sqrt{3}-2\sqrt{3}=-\sqrt{3}\)

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

\(P=\left(\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)

\(P=\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)

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

\(a,\sqrt{27}+\sqrt{48}-\sqrt{108}-\sqrt{12}\\ =3\sqrt{3}+4\sqrt{3}-6\sqrt{3}-2\sqrt{3}\\ =-\sqrt{3}\)

\(b,P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{x-2\sqrt{x}+1}{2}\\ =\dfrac{\left(\sqrt{x}+1\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{2}\\ =\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{2}\\ =\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{2}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)