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

\(=1:\left[\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x+1}{x^2+x+1}+\frac{-1}{x-1}\right]\)

\(=1:\left[\frac{\left(x^2+2\right)+\left(x+1\right)\left(x-1\right)+\left(-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)

\(=1:\left[\frac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)

\(=1:\left[\frac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left[\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)

\(=1:\frac{x}{x^2+x+1}=\frac{x^2+x+1}{x}\)

Giải các câu khác giúp mình với 

a) ĐKXĐ: \(x\ge0,x\ne1\)

\(P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

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

\(\Leftrightarrow-\sqrt{x}-1=\sqrt{x}-1\Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)

c) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\in Z\)

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

Kết hợp đk:

\(\Leftrightarrow x\in\left\{0\right\}\)

d) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\)

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

Do \(\sqrt{x}+1\ge1\Leftrightarrow-\dfrac{2}{\sqrt{x}+1}\ge-2\)

\(\Leftrightarrow P=1-\dfrac{2}{\sqrt{x}+1}\ge1-2=-1\)

\(minP=-1\Leftrightarrow x=0\)

\(a,P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\left(x\ge0;x\ne1\right)\\ P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ b,P=-1\Leftrightarrow\sqrt{x}-1=-\sqrt{x}-1\\ \Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\\ c,P\in Z\Leftrightarrow\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\in Z\\ \Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{1;2\right\}\left(\sqrt{x}+1\ge1\right)\\ \Leftrightarrow\sqrt{x}=0\left(x\ne1\right)\\ \Leftrightarrow x=0\)

\(d,P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\left(\dfrac{2}{\sqrt{x}+1}>0\right)\\ e,P=1-\dfrac{2}{\sqrt{x}+1}\\ \sqrt{x}+1\ge1\Leftrightarrow-\dfrac{2}{\sqrt{x}+1}\ge-\dfrac{2}{1}=-2\\ \Leftrightarrow P=1-\dfrac{2}{\sqrt{x}+1}\ge1-\left(-2\right)=3\)

Dấu \("="\Leftrightarrow x=0\)

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

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

\(=\dfrac{4}{x-1}\)

b: x=4 thì C=4/(4-1)=4/3

Khi x=-4 thì C=4/(-4-1)=-4/5

c: C>0

=>x-1>0

=>x>1

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

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

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

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

\(a,M=\dfrac{x+1-4x+4+7x-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{4\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{x-1}\\ b,x=-3\Rightarrow M=\dfrac{4}{-3-1}=-1\)

b.\(M=\dfrac{1}{-3-1}-\dfrac{4}{-3+1}+\dfrac{7\left(-3\right)-1}{\left(-3\right)^2-1}\)
\(M=\dfrac{-1}{4}-\left(-2\right)+\dfrac{11}{5}=3,95\)

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

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

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

b: Khi \(x=3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\) thì

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

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

c: M>0

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

mà \(\sqrt{x}>0\)

nên \(\sqrt{x}-2>0\)

=>\(\sqrt{x}>2\)

=>x>4

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

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!