1: \(=\dfrac{4\left(x-1\right)}{\left(x+1\right)^2}\cdot\dfrac{3\left(x+1\right)}{-20\left(x-1\right)}=\dfrac{-12}{20}\cdot\dfrac{1}{x+1}=\dfrac{-3}{5x+5}\)

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

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

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

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

=-2x

1. I'd rather you didn't ask me that question

2. I haven't seen Bob seen I was in HCM City

3. He would rather read books than watch TV

4. It took Peter three hours to repaint his house

5. He asked me if I knew to speak English

6. We haven't met each other for ten years

7. The film was so boring that she fell asleep

8. The furniture was too expensive for me to buy

9. The weather is so good that they are going for a picnic

10. The coffee is too hot for me to drink

20 strong enough to carry that heavy box.

21 isn't hot enough to boil the kettle.

22 wasn't warm enough for us to go swimming.

23 swims skillfully.

24 a very good English learner.

25 swims well.

26 a very fast runner.

27 more interesting than plays.

28 more difficult than English.

29 as intelligent as Jill is

30 to have a friend like you

31 to smoke 20 cigarettes a day, but now he doesn't smoke any more.

32 enough to lift that heavy table.

1: AD=8-2=6cm

AD/AB=6/8=3/4

AE/AC=9/12=3/4

=>AD/AB=AE/AC

2: Xét ΔADE và ΔABC có

AD/AB=AE/AC
góc A chung

=>ΔADE đồng dạng với ΔABC

3: AI là phân giác

=>IB/IC=AB/AC

=>IB/IC=AD/AE

=>IB*AE=AD*IC

1: Sửa đề: Qua N kẻ đường song song với PC cắt AB tại F

Xét tứ giác CNFP có NF//PC

nên CNFP là hình thang

vcl

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)

Bài `1`

\(a,A=a\left(a+b\right)-b\left(a+b\right)\\ =\left(a+b\right)\left(a-b\right)\)

Với `a=9;=10`

Ta có :

 \(\left(a+b\right)\left(a-b\right)\\=\left(9+10\right)\left(9-10\right)\\ =19.\left(-1\right)\\ =-19\)

\(b,B=\left(3x+2\right)^2+\left(3x-2\right)^2-2\left(3x+2\right)\left(3x-2\right)\\ =\left(3x+2\right)^2-2\left(3x+2\right)\left(3x-2\right)+\left(3x-2\right)^2\\ =\left[\left(3x+2\right)-\left(3x-2\right)\right]^2\)

Với `x=-4`

Ta có :

\(\left[\left(3x+2\right)-\left(3x-2\right)\right]^2\\ =\left(3.4+2-3.4+2\right)^2\\ =\left(12+2-12+2\right)^2\\ =4^2\\ =16\)

\(2,\\ x^3-6x^2+9x\\ =x\left(x^2-6x+9\right)\\ =x\left(x-3\right)^2\\ x^2-2x-4y^2-4y\\ \)

`->` có đúng đề ko cậu

2:

b; x^2-4y^2-2x-4y

=(x-2y)*(x+2y)-2(x+2y)

=(x+2y)(x-2y-2)

a: x^3-6x^2+9x

=x(x^2-6x+9)

=x(x-3)^2

- Dạng khí

- Vì giun đất hô hấp qua da, khi ngập nước sẽ lm ngập cơ thể chúng