d: =-76x10=-760

\(\dfrac{2x-3}{4}.\dfrac{6}{5}=\dfrac{21}{10}\\ \Leftrightarrow\dfrac{2x-3}{4}=\dfrac{7}{4}\\ \Leftrightarrow2x-3=7\\ \Leftrightarrow2x=10\\ \Leftrightarrow x=5\)

x = 5

x = 5

Lời giải:

a.

\(G=\frac{x^2-4}{x+1}+\frac{2}{x+1}:\frac{(2x-3)(x+1)-(2x+1)(x-1)}{(x-1)(x+1)}\)

\(=\frac{x^2-4}{x+1}+\frac{2}{x+1}:\frac{-2}{(x-1)(x+1)}=\frac{x^2-4}{x+1}+\frac{2}{x+1}.\frac{(x+1)(x-1)}{-2}\)

\(=\frac{x^2-4}{x+1}-(x-1)=\frac{x^2-4-(x^2-1)}{x+1}=\frac{-3}{x+1}\)

b.

Để $A\in\mathbb{Z}^+$ thì $x+1$ là ước âm của $-3$

$\Rightarrow x+1\in\left\{-1;-3\right\}$

$\Leftrightarrow x\in\left\{-2;-4\right\}$ (tm)

c.

$G< -1\Leftrightarrow \frac{-3}{x+1}+1< 0$

$\Leftrightarrow \frac{x-2}{x+1}< 0$

$\Leftrightarrow x-2<0< x+1$ hoặc $x-2>0>x+1$

$\Leftrightarrow -1< x< 2$ (chọn) hoặc $-1> x>2$ (loại)

Vậy $-1< x< 2$ và $x\neq 1$

Bài 8:

a: Ta có: \(G=\dfrac{x^2-4}{x+1}+\dfrac{2}{x+1}:\left(\dfrac{2x-3}{x-1}-\dfrac{2x+1}{x+1}\right)\)

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

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

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

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

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

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

c: Để G<-1 thì G+1<0

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

\(\Leftrightarrow\dfrac{x-2}{x+1}< 0\)

\(\Leftrightarrow-1< x\le2\)

Kết hợp ĐKXĐ, ta được:

\(\left\{{}\begin{matrix}-1< x\le2\\x\ne1\end{matrix}\right.\)

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

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

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

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

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

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

\(x^2-5x+6=0\\ \Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\\ \Rightarrow\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

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

Thay \(x=2\), ta có:

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

    \(=0\)

Thay \(x=3\), ta có:

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

    \(=-\dfrac{2}{5}\)

D nguyên âm \(\left[{}\begin{matrix}\left\{{}\begin{matrix}-2\left(x-2\right)< 0\\x+2>0\end{matrix}\right.\\\left\{{}\begin{matrix}-2\left(x-2\right)>0\\x+2< 0\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-2\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -2\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x>2\\x< -2\end{matrix}\right.\)

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

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

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

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

b: Ta có: \(x^2-5x+6=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

hay x=3

Thay x=3 vào D, ta được:

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

Còn số quãng đường là :

1-2/7-3/8=19/56(quãng đường)

ĐS:...

19/56 quãng đường

19/56 quãng đường

Đề yêu cầu gì vậy em? Rút gọn?

\(D=\dfrac{1}{3^2}+\dfrac{1}{3^4}+\dfrac{1}{3^6}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow9D=1+\dfrac{1}{3^2}+\dfrac{3}{3^4}+...+\dfrac{1}{3^{98}}\)

\(\Rightarrow9D-D=1-\dfrac{1}{3^{100}}\)

\(\Rightarrow8D=1-\dfrac{1}{3^{100}}\)

\(\Rightarrow D=\dfrac{1}{8}\left(1-\dfrac{1}{3^{100}}\right)\)

1.classmates → classmate

2.are → is

3.good → well

4.to ride → riding

5.in → around

6.Do → Will

7.doing → going

8.interesting → is interesting

9.interest → is interesting

10.played → will play

3. good → well

4. to ride  → riding

5. in  → to

6. Do  → Will

7. doing  → going

8. think  → find
9. interest  → is interesting

10. played  → will play

1

Với \(\left\{{}\begin{matrix}x\ne2\\x\ne-1\\x\ne\sqrt{\dfrac{1}{2}}\end{matrix}\right.\)

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

2

Để M = 0 thì \(\dfrac{x+1}{\left(2-x\right)\left(2x^2-1\right)}=0\Rightarrow x+1=0\Rightarrow x=-1\) (loại)

Vậy không có giá trị x thỏa mãn M = 0

1) \(M=\left(\dfrac{x-1}{2-x}-\dfrac{x^2}{x^2-x-2}\right)\cdot\dfrac{x^2+2x+1}{4x^4-4x^2+1}\) (ĐK: \(\left\{{}\begin{matrix}x\ne2\\x\ne-1\\x\ne\sqrt{\dfrac{1}{2}}\end{matrix}\right.\))

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

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

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

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

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

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

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

2) Ta có: \(M=0\)

\(\Rightarrow\dfrac{-\left(x+1\right)}{\left(x-2\right)\left(2x^2-1\right)}=0\)

\(\Leftrightarrow-\left(x+1\right)=0\)

\(\Leftrightarrow-x=1\)

\(\Leftrightarrow x=-1\left(ktm\right)\)

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

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

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

2: M=0

=>x+1=0

=>x=-1(loại)

K up hình lên đc :((