a:

ĐKXĐ: x>=0; x<>1

Ta có: \(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\)

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

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

Ta có: \(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

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

\(=\frac{x+\sqrt{x}-2+3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

Ta có: \(P=\left(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\right):\left(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)

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

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

b: Thay \(x=6-2\sqrt5=\left(\sqrt5-1\right)^2\) vào P, ta được:

\(P=\frac{2\cdot\sqrt{\left(\sqrt5-1\right)^2}-1}{\sqrt{\left(\sqrt5-1\right)^2}+1}\)

\(=\frac{2\left(\sqrt5-1\right)-1}{\sqrt5-1+1}=\frac{2\sqrt5-3}{\sqrt5}=2-\frac{3}{\sqrt5}=2-\frac{3\sqrt5}{5}=\frac{10-3\sqrt5}{5}\)

c: \(P=\frac{1}{\sqrt{x}}\)

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

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

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

=>\(x-\sqrt{x}-\frac12=0\)

=>\(x-\sqrt{x}+\frac14-\frac34=0\)

=>\(\left(\sqrt{x}-\frac12\right)^2=\frac34\)

=>\(\left[\begin{array}{l}\sqrt{x}-\frac12=\frac{\sqrt3}{2}\\ \sqrt{x}-\frac12=-\frac{\sqrt3}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{x}=\frac{\sqrt3+1}{2}\\ \sqrt{x}=\frac{-\sqrt3+1}{2}\left(loại\right)\end{array}\right.\)

=>\(\sqrt{x}=\frac{\sqrt3+1}{2}\)

=>\(x=\left(\frac{\sqrt3+1}{2}\right)^2=\frac{4+2\sqrt3}{4}=\frac{2+\sqrt3}{2}\)

d: Để P là số nguyên thì \(2\sqrt{x}-1\) ⋮\(\sqrt{x}+1\)

=>\(2\sqrt{x}+2-3\) ⋮\(\sqrt{x}+1\)

=>-3⋮\(\sqrt{x}+1\)

=>\(\sqrt{x}+1\in\left\lbrace1;3\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace0;2\right\rbrace\)

=>x∈{0;4}

e: \(P<1-\sqrt{x}\)

=>\(\frac{2\sqrt{x}-1}{\sqrt{x}+1}<1-\sqrt{x}\)

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

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

=>\(x+2\sqrt{x}+1-3<0\)

=>\(\left(\sqrt{x}+1\right)^2<3\)

=>\(\sqrt{x}+1<\sqrt3\)

=>\(\sqrt{x}<\sqrt3-1\)

=>\(x<4-2\sqrt3\)

Kết hợp ĐKXĐ, ta được: 0<=x<\(4-2\sqrt3\)

a:

ĐKXĐ: x>=0; x<>1

Ta có: \(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\)

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

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

Ta có: \(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

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

\(=\frac{x+\sqrt{x}-2+3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

Ta có: \(P=\left(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\right):\left(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)

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

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

b: Thay \(x=6-2\sqrt5=\left(\sqrt5-1\right)^2\) vào P, ta được:

\(P=\frac{2\cdot\sqrt{\left(\sqrt5-1\right)^2}-1}{\sqrt{\left(\sqrt5-1\right)^2}+1}\)

\(=\frac{2\left(\sqrt5-1\right)-1}{\sqrt5-1+1}=\frac{2\sqrt5-3}{\sqrt5}=2-\frac{3}{\sqrt5}=2-\frac{3\sqrt5}{5}=\frac{10-3\sqrt5}{5}\)

c: \(P=\frac{1}{\sqrt{x}}\)

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

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

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

=>\(x-\sqrt{x}-\frac12=0\)

=>\(x-\sqrt{x}+\frac14-\frac34=0\)

=>\(\left(\sqrt{x}-\frac12\right)^2=\frac34\)

=>\(\left[\begin{array}{l}\sqrt{x}-\frac12=\frac{\sqrt3}{2}\\ \sqrt{x}-\frac12=-\frac{\sqrt3}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{x}=\frac{\sqrt3+1}{2}\\ \sqrt{x}=\frac{-\sqrt3+1}{2}\left(loại\right)\end{array}\right.\)

=>\(\sqrt{x}=\frac{\sqrt3+1}{2}\)

=>\(x=\left(\frac{\sqrt3+1}{2}\right)^2=\frac{4+2\sqrt3}{4}=\frac{2+\sqrt3}{2}\)

d: Để P là số nguyên thì \(2\sqrt{x}-1\) ⋮\(\sqrt{x}+1\)

=>\(2\sqrt{x}+2-3\) ⋮\(\sqrt{x}+1\)

=>-3⋮\(\sqrt{x}+1\)

=>\(\sqrt{x}+1\in\left\lbrace1;3\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace0;2\right\rbrace\)

=>x∈{0;4}

e: \(P<1-\sqrt{x}\)

=>\(\frac{2\sqrt{x}-1}{\sqrt{x}+1}<1-\sqrt{x}\)

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

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

=>\(x+2\sqrt{x}+1-3<0\)

=>\(\left(\sqrt{x}+1\right)^2<3\)

=>\(\sqrt{x}+1<\sqrt3\)

=>\(\sqrt{x}<\sqrt3-1\)

=>\(x<4-2\sqrt3\)

Kết hợp ĐKXĐ, ta được: 0<=x<\(4-2\sqrt3\)

???

\(B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)+\dfrac{2-2\sqrt{x}}{\sqrt{x}}(x \geq 0,x \neq 1\)
`=((2x+1-x+\sqrtx)/(x\sqrtx-1))(((\sqrtx+1)(x-\sqrtx+1))/(\sqrtx+1)-\sqrtx)+(2-2sqrtx)/sqrtx`
`=((x-\sqrtx+1)/((\sqrtx-1))(x+sqrtx+1)))(x-2\sqrtx+1)-(2\sqrtx-2)/sqrtx`
`=(1/(\sqrtx-1))(\sqrtx-1)^2-(2(\sqrtx-1))/sqrtx`
`=\sqrtx-1-(2(\sqrtx-1))/sqrtx`
`=(x-\sqrtx-2\sqrtx+2)/sqrtx`
`=(x-3sqrtx+2)/sqrtx`

\(\dfrac{3\sqrt{10}+\sqrt{20}-3\sqrt{6}-\sqrt{12}}{\sqrt{5}+\sqrt{3}}\)

\(=\dfrac{3\sqrt{2}.\sqrt{5}+2\sqrt{5}-3\sqrt{2}.\sqrt{3}-2\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)

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

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

Bạn coi lại xem dưới mẫu đúng dấu ''+'' không á, phải dấu ''-'' mới rút với tử ở trên được nha.

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

b: Thay x=1 vào P, ta được:

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

Sửa đề; \(A=\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\)

a: \(A=\dfrac{\sqrt{x}-1+\sqrt{x}+1-2}{x-1}=\dfrac{2\sqrt{x}-2}{x-1}=\dfrac{2}{\sqrt{x}+1}\)

b: Khi x=3+2căn 2 thì \(A=\dfrac{2}{\sqrt{2}+1+1}=\dfrac{2}{\sqrt{2}+2}=2-\sqrt{2}\)

1 quy đồng lên ra được

2 \(A=\dfrac{1}{x-2\sqrt{x-5}+3}\le\dfrac{1}{5-2.0+3}=\dfrac{1}{8}\)

dấu"=" xảy ra<=>x=5

ở câu 1 mình làm cách quy đồng rồi nhưng nó ko ra, bạn có cách khác ko?

Câu 1: 

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

a) Thay x=16 vào B, ta được:

\(B=\dfrac{1}{\sqrt{16}-3}=\dfrac{1}{4-3}=1\)

Vậy: Khi x=16 thì B=1

b) Ta có: M=A-B

\(=\dfrac{x+3}{x-9}+\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\)

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

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

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

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

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

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

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

c) Để \(M=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) thì \(\dfrac{\sqrt{x}-2}{\sqrt{x}-3}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)=\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)\)

\(\Leftrightarrow x-4=x-2\sqrt{x}-3\)

\(\Leftrightarrow-2\sqrt{x}-3=-4\)

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

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

hay \(x=\dfrac{1}{4}\)(thỏa ĐK)

Vậy: Để \(M=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) thì \(x=\dfrac{1}{4}\)

Câu 2: 

b) Gọi thời gian tổ 1 hoàn thành công việc khi làm một mình là x(giờ)

thời gian tổ 2 hoàn thành công việc khi làm một mình là y(giờ)

(Điều kiện: x>12; y>12)

Trong 1 giờ, tổ 1 làm được: \(\dfrac{1}{x}\)(công việc)

Trong 1 giờ, tổ 2 làm được: \(\dfrac{1}{y}\)(công việc)

Trong 1 giờ, hai tổ làm được: \(\dfrac{1}{12}\)(công việc)

Do đó, ta có phương trình: \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\)(1)

Vì khi tổ 1 làm trong 2 giờ, tổ 2 làm trong 7 giờ thì hai tổ hoàn thành được một nửa công việc nên ta có phương trình: \(\dfrac{2}{x}+\dfrac{7}{y}=\dfrac{1}{2}\)(2)

Từ (1) và (2) ta lập được hệ phương trình:

\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{2}{x}+\dfrac{7}{y}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{2}{y}=\dfrac{1}{6}\\\dfrac{2}{x}+\dfrac{7}{y}=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-5}{y}=\dfrac{-1}{3}\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=15\\\dfrac{1}{x}+\dfrac{1}{15}=\dfrac{1}{12}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{60}\\y=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=60\\y=15\end{matrix}\right.\)(thỏa ĐK)

Vậy: Tổ 1 cần 60 giờ để hoàn thành công việc khi làm một mình

Tổ 2 cần 15 giờ để hoàn thành công việc khi làm một mình

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

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

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

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

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