-5606066240

1,2 x y x 4 = 20

1,2 x y = 20 : 4

1,2 x y = 5

x = 5 : 1,2 

x = 4,166666667

1,2×y×4=20

1,2×y=20÷4

1,2×y=5

y=5÷1,2

y=25/6

1,2 x y x 4 = 20

=> 1,2 x y = 20 : 4 = 5

=> y = 5 : 1,2 

=> y = \(\frac{25}{6}\)

a) x-2006 = 1-1/2+1/2-1/3+1/3-.....-1/2006

=>x-2006= 1- 1/2006

=> x-2006 = 2005/2006

=> x = 2006 \(\frac{2005}{2006}\)

a, \(\Rightarrow\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{2006-2005}{2005.2006}=x-2006\)

\(\Rightarrow\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+...+\frac{2006}{2005.2006}-\frac{2005}{2005.2006}=x-2006\)

Giản ước tử cho mẫu của từng phân số ta được: 

Đề bài phần b không rõ lắm nên mình chưa làm

cảm ơn hai bn nha

Gọi A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{x.\left(x+1\right)}=\dfrac{19}{20}\)

\(\Rightarrow\) A = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)

\(\Rightarrow\) A = 1 - \(\dfrac{1}{x+1}\)

\(\Rightarrow\) 1 - \(\dfrac{1}{x+1}\) = \(\dfrac{19}{20}\)

\(\Rightarrow1-\dfrac{19}{20}=\dfrac{1}{x+1}\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{20}\)

\(\Rightarrow\) x + 1 = 20\(\Rightarrow\) x=19

\(\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\); \(\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\);.....; \(\frac{1}{x.\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)

=> \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=1-\frac{1}{x+1}=\frac{x}{x+1}\)

=> \(\frac{x}{x+1}=\frac{19}{20}\)=> 20x=19x+19 => x=19

ĐS: x=19

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{x\times\left(x+1\right)}=\frac{19}{20}\)\(\frac{19}{20}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{19}{20}\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{19}{20}\)

\(\Rightarrow\frac{x}{x+1}=\frac{19}{20}\)

\(\Rightarrow20x=19x+19\)\(\Rightarrow x=19\)

Vậy \(x=19\)

0,36 x 350 + 1,2 x 20 x 3 + 9 x 4 x 4,5

= 126  + 72 + 162

= 360

1/1.2+1/3.4+1/5.6+...+1/49.50

=1/1-1/2+1/3-1/4+...+1/49-1/50

=1/1+1/2+1/3+1/4+...+1/49+1/50-2(1/2+1/4+1/6+...+1/50)

=1/1+1/2+1/3+1/4+...+1/49+1/50-(1/1+1/2+1/3+1/4+...+1/25)

=1/26+1/27+...+1/50=1/26+1/27+...+1/50(đpcm)

b. 1/1-1/2+1/3-1/4+...+1/99-1/100=99/100

7/12=175/300; 5/6=10/12=250/300; 99/100=297/300

(hình như khúc này đề bài sai hả bạn) bạn tự tính ra nhé

bài 2: a.x+1/10+x/12+x/14+...x+1/20

(x+x+x...+x)+(1/10+1/12+...+1/20)

ko có kết quả sao tìm x được bạn:[

b.x+1/2000+x+2/1999=x+3/1998+x+4/1997

x+1/2000+x+2/1999=x+3/1998+x+4/1997

(x+1/2000+1)+(x+2/1999+1)=(x+3/1998+1)+(x+4/1997+1)

x+2002/2000+x+2002/1999=x+2002/1998+x+2002/1997

x+2002(1/2000+1/1999)=(x+2002)(1/1998+1/1997)

=>(1/2000+1/1999)=(1/1998+1/1997)

x+2002(1/2000+1/1999)-(x+2002)(1/1998+1/1997)=0

(x+2002)(1/2000+1/1999-1/1998-1/1997)=0

(x+2002).0=0

(x+2002)=0

x =0-2002=-2002

Chúc bạn học tốt.

help 

Bài 21: 

Ta có: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)

\(\Leftrightarrow\dfrac{x+1}{99}+1+\dfrac{x+2}{98}+1+\dfrac{x+3}{97}+1+\dfrac{x+4}{96}+1=0\)

\(\Leftrightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}+\dfrac{x+100}{97}+\dfrac{x+100}{96}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}\right)=0\)

mà \(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}>0\)

nên x+100=0

hay x=-100

Vậy: x=-100

`x :3*5 = 3/4 :(-5/6)`

`x :15 =3/4*(-6/5)=-9/10`

`x = -9/10 *15 =-27/2`

`x-1*2/2 = 8/x -1.2`

`x- 1*1 = 8/x -2`

`x-8/x = -2+1`

`x-8/x =-1`

`x^2 -8x =-x`

`x^2 -8x +x=0`

`x^2 -7x =0`

`x(x-7) =0`

`=>[(x=0),(x=7):}`

`a, x \div 15=-9/10`

`x=-9/10*14`

`x=-27/2`

`b, (x-1*2)/2=8/(x-1*2)`

\(\left(x-1\cdot2\right)\cdot\left(x-1\cdot2\right)=8\cdot2\)

`(x-1*2)^2=16`

`(x-1*2)^2=(+-4)^2`

\(\Rightarrow\left[{}\begin{matrix}x-1\cdot2=4\\x-1\cdot2=-4\end{matrix}\right.\)

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

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

`=>`\(\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)