\(\dfrac{3}{5×7}\) × \(\dfrac{3}{7×9_{ }}\)× .... × \(\dfrac{3}{59×61}\)
Những câu hỏi liên quan
\(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\)
Giải phương trinh trên
\(\Leftrightarrow x+66=0\)
hay x=-66
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\(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\\ \Rightarrow\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+3}{63}+1\right)=\left(\dfrac{x+5}{61}+1\right)+\left(\dfrac{x+7}{59}+1\right)\\ \Rightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}-\dfrac{x+66}{61}-\dfrac{x+66}{59}=0\\ \Rightarrow\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\\ \Rightarrow x+66=0\\ \Rightarrow x=-66\)
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\(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\\ \dfrac{x+66}{65}+\dfrac{x+66}{63}-\dfrac{x+66}{61}-\dfrac{x+66}{59}=0\\ \left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\\ x+66=0\\ x=-66\)
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\(\dfrac{x-1}{65}+\dfrac{x-3}{63}=\dfrac{x-5}{61}+\dfrac{x-7}{59}\)
\(\dfrac{x-1}{65}+\dfrac{x-3}{63}=\dfrac{x-5}{61}+\dfrac{x-7}{59}\)
\(\Leftrightarrow\dfrac{x-1}{65}-1+\dfrac{x-3}{63}-1=\dfrac{x-5}{61}-1+\dfrac{x-7}{59}-1\)
\(\Leftrightarrow\dfrac{x-66}{65}+\dfrac{x-66}{63}=\dfrac{x-66}{61}+\dfrac{x-66}{59}\)
\(\Leftrightarrow\left(x-66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\)
\(\Leftrightarrow x-66=0\)
\(\Leftrightarrow x=66\)
Vậy x=66.
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Giải phương trình sau: \(\dfrac{x+1}{65} + \dfrac{x+3}{63} = \dfrac{x+5}{61} + \dfrac{x+7}{59}\)
\(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\)
\(\Leftrightarrow\dfrac{x+1}{65}+1+\dfrac{x+3}{63}+1=\dfrac{x+5}{61}+1+\dfrac{x+7}{59}+1\)
\(\Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}=\dfrac{x+66}{61}+\dfrac{x+66}{59}\)
\(\Leftrightarrow\left(x+66\right)\cdot\left(\dfrac{1}{65}+\dfrac{1}{63}\right)=\left(x+66\right)\cdot\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\)
\(\Rightarrow x=-66\)
Vậy x = -66.
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\(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\)
\(\Leftrightarrow\dfrac{x+1}{65}+\dfrac{x+3}{63}-\dfrac{x+5}{61}-\dfrac{x+7}{59}=0\)
\(\Leftrightarrow\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+3}{63}+1\right)-\left(\dfrac{x+5}{61}+1\right)-\left(\dfrac{x+7}{59}+1\right)=0\)\(\Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}-\dfrac{x+66}{61}-\dfrac{x+66}{59}=0\)\(\Leftrightarrow\left(x+66\right)\left[\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\right]=0\)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 !!!
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\(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\\ \Leftrightarrow\dfrac{x+1}{65}+\dfrac{x+3}{63}+2=\dfrac{x+5}{61}+\dfrac{x+7}{59}+2\\ \Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}=\dfrac{x+66}{61}+\dfrac{x+66}{59}\\ \Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}-\dfrac{x+66}{61}-\dfrac{x+66}{59}=0\\ \Leftrightarrow\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\\ \Rightarrow x+66=0\Rightarrow x=-66\)
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Giá trị của x thỏa mãn
\(\dfrac{x+1}{65}\)+\(\dfrac{x+3}{63}\)=\(\dfrac{x+5}{61}\)+\(\dfrac{x+7}{59}\)
GIÚP MÌNH VS MÌNH ĐANG CẦN LẮM MN Ạ
Câu 1: Thực hiện phép tính
a, \(\dfrac{7}{9}-\dfrac{16}{9}\)
b, \(\dfrac{2}{-15}+\dfrac{7}{10}\)
c, \(\left(4\dfrac{2}{3}-4\dfrac{3}{4}\right):\dfrac{-5}{12}-\dfrac{4}{5}\)
d, \(\dfrac{-141}{157}.\dfrac{23}{59}-\dfrac{141}{157}.\dfrac{36}{59}+\dfrac{16}{-157}\)
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
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Giải phương trình sau: ( giúp mk nha)
a) \(\dfrac{2x-\dfrac{4-5x}{5}}{15}=\dfrac{7x-\dfrac{x-3}{2}}{5}-x+1\)
b) \(x-\dfrac{\dfrac{x}{2}-\dfrac{3+x}{4}}{2}=\dfrac{2x-\dfrac{10-7x}{3}}{2}-6x+1\)
c) \(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\)
giải các phương trình sau:
a,7+2x32-3z c, dfrac{x}{3}+dfrac{2x-6}{6}2-dfrac{x}{3}
b,dfrac{x-1}{x}+dfrac{1}{x+1}dfrac{2x-1}{x^2+x} d,dfrac{x+1}{65}+dfrac{x+3}{63}dfrac{x+5}{61}+dfrac{x+7}{59}
làm hộ mk nha mk đang cần gấp.thanks trc ạ
Đọc tiếp
giải các phương trình sau:
a,7+2x=32-3z c, \(\dfrac{x}{3}+\dfrac{2x-6}{6}=2-\dfrac{x}{3}\)
b,\(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\) d,\(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\)
làm hộ mk nha mk đang cần gấp.thanks trc ạ
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)
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giải các phương trình sau:
a,7+2x=32-3z
c,\(\dfrac{x}{3}+\dfrac{2x-6}{6}=2-\dfrac{x}{3}\)
b,\(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)
d,\(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\)
lm giúp mk đi thanks nhìu
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
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\(\dfrac{1}{7}\cdot2\dfrac{1}{3}+\dfrac{5}{2}\cdot\dfrac{3}{7}-\dfrac{59}{6}\cdot\dfrac{1}{7}\)
giup mink nha
\(\dfrac{1}{7}.2\dfrac{1}{3}+\dfrac{5}{2}.\dfrac{3}{7}-\dfrac{59}{6}.\dfrac{1}{7}\)
=\(\dfrac{1}{7}.\dfrac{7}{3}+\dfrac{5}{2}.\dfrac{3}{7}-\dfrac{59}{6}.\dfrac{1}{7}\)
=\(\dfrac{1}{7}.\left(\dfrac{7}{3}-\dfrac{59}{6}\right)+\dfrac{5}{2}.\dfrac{3}{7}\)
=\(\dfrac{1}{7}.\dfrac{-15}{2}+\dfrac{5}{2}.\dfrac{3}{7}\)
=\(\dfrac{-15}{14}+\dfrac{15}{14}\)
= 0
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