a: \(M=\dfrac{2x+2+4x-4+7x+15}{\left(x-1\right)\left(x+1\right)}=\dfrac{13x+13}{\left(x-1\right)\left(x+1\right)}=\dfrac{13}{x-1}\)

`a,` Với `x=3`

\(B=\dfrac{x^2-x}{2x+1}\\ \Rightarrow\dfrac{3^2-3}{2\cdot3+1}\\ =\dfrac{9-3}{6+1}\\ =\dfrac{6}{7}\)

`b,` Ta có `M=A*B`

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

`c,` Để `M=1/2`

`=> x/(x+1)=1/3`

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

`<=> 3x=x+1`

`<=>3x-x=1`

`<=>2x=1`

`<=>x=1/2`

các học bá đâu rùi

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

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

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

b: Thay x=-1/2 vào M, ta được:

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

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

\(=\left(\dfrac{-1+x-3x-3}{2\left(x-1\right)\left(x+1\right)}\right):\dfrac{2}{x-1}=\dfrac{-2x-4}{2\left(x-1\right)\left(x+1\right)}:\dfrac{2}{x-1}=\dfrac{-\left(x+2\right)}{2\left(x+1\right)}\)

b, Thay x  =-1/2 vào ta đc 

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

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

\(=\dfrac{-2x+2}{2\left(x-1\right)\left(x+1\right)}:\dfrac{2}{x-1}=\dfrac{-2\left(x-1\right)^2}{4\left(x-1\right)\left(x+1\right)}=\dfrac{-\left(x-1\right)}{2\left(x+1\right)}\)

b, Thay x = -1/2 vào ta đc 

\(M=\dfrac{\dfrac{1}{2}+1}{2\left(\dfrac{-1}{2}+1\right)}=\dfrac{\dfrac{3}{2}}{\dfrac{2.1}{2}}=\dfrac{3}{2}\)

MN ƠI GIÚP EM VS 15PHÚT NX EM PK NỘP R =(((

1.\(x=49\Rightarrow B=\dfrac{49-\sqrt{49}}{2\sqrt{49}+1}=\dfrac{14}{5}\)

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

\(\Rightarrow M=A.B=\left(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)

\(\Rightarrow M=A.B=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)

\(\Rightarrow M=A.B=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)

3,\(M=\dfrac{1}{3}\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{1}{3}\\ \Rightarrow3\sqrt{x}=\sqrt{x}+1\\ \Rightarrow2\sqrt{x}=1\\ \Rightarrow\sqrt{x}=\dfrac{1}{2}\\ \Rightarrow x=\dfrac{1}{4}\)

a)B =  \(\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{7x+3}{9-x^2}\left(ĐK:x\ne\pm3\right)\)

= \(\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}-\dfrac{7x+3}{x^2-9}\)

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

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

b) \(\left|2x+1\right|=7< =>\left[{}\begin{matrix}2x+1=7< =>x=3\left(L\right)\\2x+1=-7< =>x=-4\left(C\right)\end{matrix}\right.\)

Thay x = -4 vào B, ta có:

B = \(\dfrac{-4.3}{-4+3}=12\)

c) Để B = \(\dfrac{-3}{5}\)

<=> \(\dfrac{3x}{x+3}=\dfrac{-3}{5}< =>\dfrac{3x}{x+3}+\dfrac{3}{5}=0\)

<=> \(\dfrac{15x+3x+9}{5\left(x+3\right)}=0< =>x=\dfrac{-1}{2}\left(TM\right)\)

d) Để B nguyên <=> \(\dfrac{3x}{x+3}\) nguyên

<=> \(3-\dfrac{9}{x+3}\) nguyên <=> \(9⋮x+3\)

a: \(M=1:\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{3x}{2\left(x-4\right)}+\dfrac{1}{2\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{4-2\sqrt{x}}{1}\)

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

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

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

b: M=20

=>\(-4\left(x-4\right)\left(\sqrt{x}-2\right)=-60x+60\sqrt{x}-40\)

=>\(x\sqrt{x}-2x-4\sqrt{x}+8=-15x+15\sqrt{x}-10\)

=>\(x\sqrt{x}+13x-19\sqrt{x}+18=0\)

=>\(x\in\varnothing\)

\(M=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)

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

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

2) Thay x=9 vào M đã rút gọn ta được:

\(M=\dfrac{\sqrt{9}-1}{9+\sqrt{9}+1}=\dfrac{2}{13}\)

3) Có \(M=\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\)

\(\Leftrightarrow x.M+\sqrt{x}\left(M-1\right)+1+M=0\) (*)

Tại x=0 pt (*) <=> M=-1  (1)

Tại x khác 0, coi pt (*) là pt bậc 2 ẩn \(\sqrt{x}\)

Pt (*) có nghiệm không âm <=> \(\left\{{}\begin{matrix}\Delta\ge0\\S\ge0\\P\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-3M^2-6M+1\ge0\\\dfrac{1-M}{M}\ge0\\\dfrac{1+M}{M}\ge0\end{matrix}\right.\)

\(\Rightarrow0< M\le\dfrac{-3+2\sqrt{3}}{3}\) (2)

Từ (1) (2)=>  \(M_{min}=-1\) <=> x=0

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 5:

Em đánh trên word cho nó dễ đánh ạ!