Câu 2: 

\(\dfrac{7}{9}\cdot\dfrac{3}{35}=\dfrac{1}{5}\cdot\dfrac{1}{3}=\dfrac{1}{15}\)

\(\dfrac{9}{22}\cdot55=\dfrac{9\cdot55}{22}=\dfrac{9\cdot5}{2}=\dfrac{45}{2}\)

12 : 3 = 4 x 4 = 16 + 16 = 32 + 17 + 17 + 17 = 83

3/14 × 5/17 + 11/14 × 5/17 + 12/17 × 5/16 + 12/17 × 11/16

= 5/17 × ( 3/14 + 11/14) + 12/17 × ( 5/16 + 11/16)

= 5/17 × 1 + 12/17 × 1

= 5/17 + 12/17

= 1

3/14 × 5/17 + 11/14 × 5/17 + 12/17 × 5/16 + 12/17 × 11/16

= 5/17 × ( 3/14 + 11/14) + 12/17 × ( 5/16 + 11/16)

= 5/17 × 1 + 12/17 × 1

= 5/17 + 12/17

= 1

\(x:x-54\cdot17\cdot\frac{6}{17}\cdot3\cdot4=12\)

\(x:x=12+\left(54\cdot17\cdot\frac{6}{17}\cdot3\cdot4\right)\)

x:x = ?

còn lại lm giúp mk nhá +0

( Mik làm mấy phần mà bạn dưới chưa làm)

11) xy+x+y=9

\(\Leftrightarrow\) xy+x+y+1=9+1

\(\Leftrightarrow\left(xy+x\right)+\left(y+1\right)\)=10

\(\Leftrightarrow x\left(y+1\right)+\left(y+1\right)=10\)

\(\Leftrightarrow\) (x+1)(y+1)=10=1.10=10.1=-1.-10=-10.-1=2.5=5.2=-2.-5=-5.-2

\(\Rightarrow\) TH1: x+1=1 ; y+1=10

\(\Leftrightarrow x=0;y=9\)

TH2: x+1=10;y+1=1

\(\Leftrightarrow\)x=9;y=0

TH3: x+1=-1;y+1=-10

\(\Leftrightarrow\) x=-2;y=-11

...........

Vậy:........

( Bạn tự làm nốt chứ dài quá, mik chỉ hướng dẫn cách làm bài thôi)

1) -x = -7

=> x = 7

2) - x = 17

=> x = - 17

3) |x| = 17

=> x = ±17

4) -(-x) = |-17|

=> x = 17

5) - 19 - x = 17

=> - x = 17 + 19

=> x = - 36

6) - 19 - x = - 17

=> - x = - 17 + 19

=> -x = 2

=> x = - 2

7) - 5 - (10 - x) = 7

=> - 5 - 10 + x = 7

=> - 15 + x = 7

=> x = 7 + 15

=> x = 22

8) |x + 3| + 7 = 12

=> |x + 3| = 12 - 7

=> |x + 3| = 5

=> x + 3 = 5 hoặc x + 3 =- 5

=> x = 2 hoặc x = - 8

9) 2 - |x - 2| = x

=> - |x - 2| - x = - 2

TH1: x >= 2

- (x - 2) - x = - 2

=> - x + 2 - x =- 2

=> - 2x = - 4

=> x = 2 (nhận)

TH2: x < 2

-[-(x - 2)] - x = - 2

=> x - 2 - x = - 2

=> 0x = 0 (vô số nghiệm)

viết sai đầu bài rùi

chú ý viết đề bài cẩn thận

có đúng ko

\(x^3=3+2\sqrt{2}+3-2\sqrt{2}+3\cdot\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\\ \Leftrightarrow x^3=6+3x\sqrt[3]{1}\\ \Leftrightarrow x^3-3x=6\)

\(y^3=17+12\sqrt{2}+17-12\sqrt{2}+3\sqrt[3]{\left(17-12\sqrt{2}\right)\left(17+12\sqrt{2}\right)}\left(\sqrt[3]{17-12\sqrt{2}}+\sqrt[3]{17+12\sqrt{2}}\right)\\ \Leftrightarrow y^3=34+3x\sqrt[3]{1}\\ \Leftrightarrow y^3-3y=34\)

Thay vào P, ta được

\(P=x^3+y^3-3x-3y+1979\\ P=\left(x^3-3x\right)+\left(y^3-3y\right)+1979\\ P=6+34+1979=2019\)

\(x^3=6+3\sqrt[3]{\left(3+2\sqrt[]{2}\right)\left(3-2\sqrt[]{2}\right)}\left(\sqrt[3]{3+2\sqrt[]{2}}+\sqrt[3]{3-2\sqrt[]{2}}\right)\)

\(\Rightarrow x^3=6+3x\)

\(\Rightarrow x^3-3x=6\)

Tương tự:

\(y^3=34+3\sqrt[3]{\left(17+12\sqrt[]{2}\right)\left(17-12\sqrt[]{2}\right)}\left(\sqrt[3]{17+12\sqrt[]{2}}+\sqrt[3]{17-12\sqrt[]{2}}\right)\)

\(\Rightarrow y^3=34+3y\)

\(\Rightarrow y^3-3y=34\)

Do đó:

\(P=\left(x^3-3x\right)+\left(y^3-3y\right)+1979=6+34+1979=...\)

\(x=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)

\(\Rightarrow x^3=3+2\sqrt{2}+3-2\sqrt{2}+3\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\)

\(=6+3\sqrt[3]{9-8}.x=6+3x\)

\(\Rightarrow x^3-3x=6\)

\(y=\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}}\)

\(\Rightarrow y^3=17+12\sqrt{2}+17-12\sqrt{2}+3\sqrt[3]{\left(17+12\sqrt{2}\right)\left(17-12\sqrt{2}\right)}\left(\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}}\right)\)

\(=34+3\sqrt[3]{289-288}.y=34+3y\)

\(\Rightarrow y^3-3y=34\)

\(P=x^3+y^3-3\left(x+y\right)+2009=\left(x^3-3x\right)+\left(y^3-3y\right)+2009\)

\(=6+34+2009=2049\)

Có \(x^3=3+2\sqrt{2}-3\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\right)-\left(3-2\sqrt{2}\right)\)

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

Có \(y^3=17+12\sqrt{2}-3\sqrt[3]{\left(17+12\sqrt{2}\right)\left(17-12\sqrt{2}\right)}\left(\sqrt[3]{17+12\sqrt{2}}-\sqrt[3]{17-12\sqrt{2}}\right)-\left(17-12\sqrt{2}\right)\)

\(\Leftrightarrow y^3=24\sqrt{2}-3y\) \(\Leftrightarrow y^3+3y=24\sqrt{2}\) (2)

Từ (1) (2)\(\Rightarrow x^3+3x-y^3-3y=-20\sqrt{2}\)

Có \(M=\left(x-y\right)^3+3\left(x-y\right)\left(xy+1\right)=\left(x-y\right)\left[\left(x-y\right)^2+3\left(xy+1\right)\right]\)

\(=\left(x-y\right)\left(x^2+xy+y^2+3\right)=x^3-y^3+3\left(x-y\right)=-20\sqrt{2}\)

Vậy \(M=-20\sqrt{2}\)

theo bài ra

\(x=\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\)

\(=>x^3=\left(\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\right)^3\)

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

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

\(x^3=4\sqrt{2}-3.\left[\sqrt[3]{9-\left(2\sqrt{2}\right)^2}\right]x\)

\(x^3=4\sqrt{2}-3.1x\)

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

\(< =>x^3+3x-4\sqrt{2}=0\)

rồi làm y tương tự rồi thế vào M là ra