\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)

\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)

\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)

\(=6x^2y\)

\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)

\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)

\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)

1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy

2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3

=6x^2y

3: =(x+y-x+y)^2=(2y)^2=4y^2

4: =(2x+3-2x-5)^2=(-2)^2=4

5: =18^8-18^8+1=1

042557

2006 x [43 x 10 - 2 x 43 x 5] +100

=2006x0+100

=0+100

=100

64x4+18x4+9x8

=256+72+72

=400

44x5+18x10+20x5

=220+180+100

=500

3x4+4x6+9x2+18

=12+24+18+18

=72

2x5+5x7+9x3

=10+35+27

=72

15:5+27:5+8:5

=[15+27+8]:5

=10

99:5-26:5-14:5

=[99-26-14]:5

=11.8

Câu cuối sai đề nha mà nếu đề như vậy thì đó là toán lớp 6

Cảm ơn bạn I LOVE YOU rất nhiều mình đang vội ,cảm ơn bạn nhé!!!

Trả lời:

\(2006\times\left(43\times10-2\times43\times5\right)+100\)

\(=2006\times\left(43\times10-43\times10\right)+100\)

\(=2006\times0+100\)

\(=0+100\)

\(=100\)

\(44\times5+18\times10+20\times5\)

\(=44\times5+18\times2\times5+20\times5\)

\(=44\times5+36\times5+20\times5\)

\(=5\times\left(44+36+20\right)\)

\(=5\times100\)

\(=500\)

\(3\times4+4\times6+9\times2+18\)

\(=12+24+18+18\)

\(=12+24+36\)

\(=72\)

\(2\times5+5\times7+9\times3\)

\(=5\times\left(2+7\right)+9\times3\)

\(=5\times9+9\times3\)

\(=9\times\left(5+3\right)\)

\(=9\times8\)

\(=72\)

\(15\div5+27\div5+8\div5\)

\(=\left(15+27+8\right)\div5\)

\(=50\div5\)

\(=10\)

\(99\div5-26\div5-14\div5\)

\(=\left(99-26-14\right)\div5\)

\(=59\div5\)

\(=\frac{59}{5}\)

a= 9

b=45

A=9

B=45

mai nah

Bài 46:

11: Ta có: \(-4\left|x-2\right|=-8\)

\(\Leftrightarrow\left|x-2\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)

Vậy: x∈{0;4}

12: Ta có: \(5\left|x+2\right|=-10\cdot\left(-2\right)\)

\(\Leftrightarrow5\left|x+2\right|=20\)

\(\Leftrightarrow\left|x+2\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

Vậy: x∈{-6;2}

13: Ta có: \(6\left|x-2\right|=18:\left(-3\right)\)

\(\Leftrightarrow6\left|x-2\right|=-6\)(1)

Ta có: \(\left|x-2\right|\ge0\forall x\)

\(\Rightarrow6\left|x-2\right|\ge0\forall x\)(2)

Ta có: -6<0(3)

Từ (1), (2) và (3) suy ra x∈∅

Vậy: x∈∅

14: Ta có:\(-7\left|x+4\right|=21:\left(-3\right)\)

\(\Leftrightarrow-7\left|x+4\right|=-7\)

\(\Leftrightarrow\left|x+4\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=1\\x+4=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

Vậy: x∈{-5;-3}

15: Ta có: \(4\left|x+1\right|=8\left(-2\right)-8\left(-5\right)\)

\(\Leftrightarrow4\left|x+1\right|=-16-\left(-40\right)\)

\(\Leftrightarrow4\left|x+1\right|=24\)

\(\Leftrightarrow\left|x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)

Vậy: x∈{-7;5}

16: Ta có: \(3\left|x+5\right|=-9\)(4)

Ta có: |x+5|≥0∀x

⇒3|x+5|≥0∀x(5)

Ta có: -9<0(6)

Từ (4), (5) và (6) suy ra x∈∅

Vậy: x∈∅

17: Ta có: \(-8\left|x-3\right|=24-16:2\)

\(\Leftrightarrow-8\left|x-3\right|=16\)

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

mà |x-3|≥0>-2∀x

nên x∈∅

Vậy: x∈∅

18: Ta có: \(-3\left|x+6\right|=6\cdot2-9\)

\(\Leftrightarrow-3\left|x+6\right|=3\)

\(\Leftrightarrow\left|x+6\right|=-1\)

mà |x+6|≥0>-1∀x

nên x∈∅

Vậy: x∈∅

19: Ta có: \(5-\left|x+7\right|=4\)

\(\Leftrightarrow\left|x+7\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+7=-1\\x+7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-6\end{matrix}\right.\)

Vậy: x∈{-8;-6}

20: Ta có: \(12-\left|x+8\right|=10\)

\(\Leftrightarrow\left|x+8\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=2\\x+8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-10\end{matrix}\right.\)

Vậy: x∈{-10;-6}

1: x=3/4-1/2=3/4-2/4=1/4

2: x-1/5=2/11

=>x=2/11+1/5=21/55

3: x-5/6=16/42-8/56

=>x-5/6=8/21-4/28=5/21

=>x=5/21+5/6=15/14

4: x/5=5/6-19/30

=>x/5=25/30-19/30=6/30=1/5

=>x=1

5: =>|x|=1/3+1/4=7/12

=>x=7/12 hoặc x=-7/12

6: x=-1/2+3/4

=>x=3/4-1/2=1/4

11: x-(-6/12)=9/48

=>x+1/2=3/16

=>x=3/16-1/2=-5/16

1)x= 1/4

2)x= 2/11+ 1/5

   x= 21/55

3)x - 5/6 = 5/21

   x         = 5/21+5/6

   x         = 15/14

4)x/5 = 5/6 + -19/30

   x:5 = 1/5

   x    = 1/5.5

   x    = 1

5) |x| - 1/4 = 6/18

    |x|           = 6/18 - 1/4

    |x|            =7/12

⇒x= 7/12 hoặc -7/12

6)x = -1/2 +3/4

   x= 1/4

7) x/15 = 3/5 + -2/3

   x:15  = -1/15

  x        = -1/15. 15

  x        = -1

8)11/8 + 13/6 = 85/x  

       85/24      = 85/x

  ⇒      x           = 24

9) x - 7/8 = 13/12

   x          = 13/12 + 7/8

   x          = 47/24

10)x - -6/15 = 4/27  

     x            = 4/27 + (-6/15)

    x             = -34/135

11) -(-6/12)+x = 9/48

                    x= 9/48 - 6/12

                    x = -5/16

12) x - 4/6 = 5/25 + -7/15

      x -4/6  =  -4/15

     x           = -4/15 + 4/6

    x             = 2/5

1. 2x(x - 5) + (x - 2)(x + 3)

= 2x² - 10x + x² + 3x - 2x - 6

= 3x² - 9x - 6

2;3 tương tự 1