khó vẽ

\(\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot...\cdot\frac{99\cdot101}{100\cdot100}\)

\(\frac{1\cdot101}{2\cdot100}\)

\(\frac{101}{200}\)

Lần sau bạn chú ý ghi đầy đủ yêu cầu đề.

Lời giải:

1. $9x^2+4+12x=(3x)^2+2.3x.2+2^2=(3x+2)^2$

2. Đề sai sai. Bạn xem lại 

3.

$(16x^2-4xy+y^2)(4x+y)=(4x+y)[(4x)^2-4x.y+y^2]$

$=(4x)^3+y^3=64x^3+y^3$

4.

$(5x-7)(25x^2+35x+49)=(5x-7)[(5x)^2+5x.7+7^2]$

$=(5x)^3-7^3=125x^3-343$

\(9x^2+12x+4=\left(3x+2\right)^2\)

\(\left(16x^2-4xy+y^2\right)\cdot\left(4x+y\right)=64x^3+y^3\)

\(\left(5x-7\right)\left(25x^2+35x+49\right)=125x^3-343\)

\(2,\\ a,\sqrt{4x-4}+\sqrt{9x-9}-\sqrt{25x-25}=7\left(x\ge1\right)\\ \Leftrightarrow2\sqrt{x-1}+3\sqrt{x-1}-5\sqrt{x-1}=7\\ \Leftrightarrow0\sqrt{x-1}=7\Leftrightarrow x\in\varnothing\\ b,\sqrt{2x^2-3}=4\left(x\le-\dfrac{\sqrt{6}}{2};\dfrac{\sqrt{6}}{2}\le x\right)\\ \Leftrightarrow2x^2-3=16\\ \Leftrightarrow x^2=\dfrac{19}{2}\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{19}{2}}\left(tm\right)\\x=-\sqrt{\dfrac{19}{2}}\left(tm\right)\end{matrix}\right.\)

\(1,\\ A=\sqrt{5+4x}+\sqrt{7-3x}\\ ĐKXĐ:\left\{{}\begin{matrix}5+4x\ge0\\7-3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{5}{4}\\x\le\dfrac{7}{3}\end{matrix}\right.\)

Bài 2:

a) \(\sqrt{4x-4}+\sqrt{9x-9}-\sqrt{25x-25}=7\left(đk:x\ge1\right)\)

\(\Leftrightarrow2\sqrt{x-1}+3\sqrt{x-2}-5\sqrt{x-1}=7\)

\(\Leftrightarrow0=7\left(VLý\right)\)

Vậy \(S=\varnothing\)

b) \(\sqrt{2x^2-3}=4\left(đk:-\sqrt{\dfrac{3}{2}}\ge x\ge\sqrt{\dfrac{3}{2}}\right)\)

\(\Leftrightarrow2x^2-3=16\)

\(\Leftrightarrow2x^2=19\Leftrightarrow x^2=\dfrac{19}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{19}{2}}\left(tm\right)\\x=-\sqrt{\dfrac{19}{2}}\left(tm\right)\end{matrix}\right.\)

1. -x+20 = -(-15)-8+13

=> -x=15-8+13-20

=> -x=0

=> x=0

2. -(-10)+x=-13+(-9)+(-6)

=> 10+x=-13-9-6

=> x = -13-9-6-10

=> x = -38

3. 8-(-12)+10=-(-14)-x

=> 8+12+10=14-x

=> x = 14-8-12-10

=> x = -16

4. -(+12)+(-x)-(-3)=5-(-7)

=> -12-x+3=5+7

=> -x=5+7+12-3

=> -x=21

=> x=-21

5. 14-x+(-10)=-(-9)+(+15)

=> 14-x-10=9+15

=> -x=9+15-14+10

=> -x=20

=> x=-20

6. 12-(-17)+(-3)=-5+x

=> 12+17-3+5=x

=> x=31

7. x-(-19)-(+32)=14-(+16)

=> x+19-32=14-16

=> x=14-16+32-19

=> x=11

8. x-|-15|-|7|=-(-9)+|-5|

=> x-15-7=9+5

=> x=9+5+7+15

=> x=36

9. 15-x+17=13-(-21)

=> 15-x+17=13+21

=> -x=13+21-15-17

=> -x=2

=> x=-2

10. -|-5|-(-x)+4=3-(-25)

=> -5+x+4=3+25

=> x=3+25-4+5

=> x=29

Bạn ơi bạn làm đc bao nhiêu thì làm phụ mình nhé 

bài1

a, 

5 - 7 + 3 +(-8)

= -2 + 3 + ( - 8)

=     1    + (-8)= -7

a; - (-5) - (+7) + (+3) + (-8)

  = 5 - 7 + 3  - 8

= (5 + 3 - 8) - 7

= (8 - 8) - 7

= 0 - 7

= - 7

\(\Leftrightarrow\left(\dfrac{x-5}{1990}-1\right)+\left(\dfrac{x-15}{1980}-1\right)+\left(\dfrac{x-25}{1970}-1\right)\\ +\left(\dfrac{x-1990}{5}-1\right)+\left(\dfrac{x-1980}{15}-1\right)+\left(\dfrac{x-1970}{25}-1\right)=0\\ \Leftrightarrow\dfrac{x-1995}{1990}+\dfrac{x-1995}{1980}+\dfrac{x-1995}{1970}+\dfrac{x-1995}{5}\\ +\dfrac{n-1995}{15}+\dfrac{n-1995}{25}=0\\ \Rightarrow\left(x-1995\right)\left(\dfrac{1}{1990}+\dfrac{1}{1980}+\dfrac{1}{1970}+\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}\right)=0\) 

\(\Rightarrow x-1995=0\\ \Rightarrow x=1995\)

a) Ta có: \(\sqrt{25x+75}+2\sqrt{9x+27}=5\sqrt{x+3}+18\)

\(\Leftrightarrow5\sqrt{x+3}+6\sqrt{x+3}-5\sqrt{x+3}=18\)

\(\Leftrightarrow\sqrt{x+3}=3\)

\(\Leftrightarrow x+3=9\)

hay x=6

b) Ta có: \(\sqrt{4x-8}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)

\(\Leftrightarrow2\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=8\)

\(\Leftrightarrow-3\sqrt{x-2}=8\)(Vô lý)

\(x^2-2x+3=\left(x^2-2x+1\right)+2=\left(x-1\right)^2+2\ge2\forall x\in R\)