\(\dfrac{3}{5×7}\) × \(\dfrac{3}{7×9_{ }}\)× .... × \(\dfrac{3}{59×61}\)
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PL
1
12 tháng 9 2017
\(\dfrac{x-1}{65}+\dfrac{x-3}{63}=\dfrac{x-5}{61}+\dfrac{x-7}{59}\)
\(\Leftrightarrow\dfrac{x-1}{65}-1+\dfrac{x-3}{63}-1=\dfrac{x-5}{61}-1+\dfrac{x-7}{59}-1\)
\(\Leftrightarrow\dfrac{x-66}{65}+\dfrac{x-66}{63}=\dfrac{x-66}{61}+\dfrac{x-66}{59}\)
\(\Leftrightarrow\left(x-66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\)
\(\Leftrightarrow x-66=0\)
\(\Leftrightarrow x=66\)
Vậy x=66.
TD
4
DM
6 tháng 8 2017
\(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\)
\(\Leftrightarrow\dfrac{x+1}{65}+1+\dfrac{x+3}{63}+1=\dfrac{x+5}{61}+1+\dfrac{x+7}{59}+1\)
\(\Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}=\dfrac{x+66}{61}+\dfrac{x+66}{59}\)
\(\Leftrightarrow\left(x+66\right)\cdot\left(\dfrac{1}{65}+\dfrac{1}{63}\right)=\left(x+66\right)\cdot\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\)
\(\Rightarrow x=-66\)
Vậy x = -66.
N
6 tháng 8 2017
\(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\)
\(\Leftrightarrow\dfrac{x+1}{65}+\dfrac{x+3}{63}-\dfrac{x+5}{61}-\dfrac{x+7}{59}=0\)
\(\Leftrightarrow\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+3}{63}+1\right)-\left(\dfrac{x+5}{61}+1\right)-\left(\dfrac{x+7}{59}+1\right)=0\)\(\Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}-\dfrac{x+66}{61}-\dfrac{x+66}{59}=0\)\(\Leftrightarrow\left(x+66\right)\left[\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\right]=0\)Nhận xét : Do \(\dfrac{1}{65}< \dfrac{1}{63}< \dfrac{1}{61}< \dfrac{1}{59}\)
\(\Rightarrow\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)< 0\)
Vậy để \(\left(x+66\right)\left[\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\right]=0\)\(\Leftrightarrow x+66=0\Leftrightarrow x=-66\)
Vậy....
tik mik nha !!!
21 tháng 4 2022
a. 7/9 - 16/9 = -9/9 = -1
b. 2/-15 + 7/10 = 17/30
c. (4 2/3 - 4 3/4) : -5/12 - 4/5
= (14/3 - 19/4) : (-5/12) - 4/5
= -1/12 : (-5/12) - 4/5
= 1/5 - 4/5
= -3/5
22 tháng 6 2022
c: \(\Leftrightarrow2x+2x-6=12-2x\)
=>4x-6=12-2x
=>6x=18
hay x=3
b: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)+x=2x-1\)
\(\Leftrightarrow x^2-1+x=2x-1\)
=>x2-x=0
=>x(x-1)=0
=>x=0(loại) hoặc x=1(nhận)
28 tháng 2 2018
lớp 9 gì như lớp 6 thế
a) đề sai
c) <=>x/3 +x/3 -1 =2-x/3
<=>3.x/3 =3 => x=3
b) x<> 0; -2 <=>
x^2 -1 +x =2x-1
<=>x^2 -x =0 => x =0 (l) x =1 nhận
d ; <=> (x+1)/65+1 +(x+3)/63 +1 =(x+5)/61+1 +(x+7)/59+1
<=>(x+66) [1/65+1/63-1/61-1/59] =0
[...] khác 0
x=-66
2 tháng 5 2017
\(\dfrac{1}{7}.2\dfrac{1}{3}+\dfrac{5}{2}.\dfrac{3}{7}-\dfrac{59}{6}.\dfrac{1}{7}\)
=\(\dfrac{1}{7}.\dfrac{7}{3}+\dfrac{5}{2}.\dfrac{3}{7}-\dfrac{59}{6}.\dfrac{1}{7}\)
=\(\dfrac{1}{7}.\left(\dfrac{7}{3}-\dfrac{59}{6}\right)+\dfrac{5}{2}.\dfrac{3}{7}\)
=\(\dfrac{1}{7}.\dfrac{-15}{2}+\dfrac{5}{2}.\dfrac{3}{7}\)
=\(\dfrac{-15}{14}+\dfrac{15}{14}\)
= 0
\(\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+...+\dfrac{3}{59\cdot61}\)(đã sửa)
Đặt \(A=\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+....+\dfrac{3}{59\cdot61}\)
\(\dfrac{2}{3}A=\dfrac{2}{3}\left(\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+....+\dfrac{3}{59\cdot61}\right)\)
\(\dfrac{2}{3}A=\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{59\cdot61}\)
\(\dfrac{2}{3}A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\)
\(\dfrac{2}{3}A=\dfrac{1}{5}-\dfrac{1}{61}\)
\(A=\dfrac{56}{305}:\dfrac{2}{3}=\dfrac{56}{305}\cdot\dfrac{3}{2}=\dfrac{84}{305}\)
ST trả lời đúng rồi đó!!!