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\(1,x-\dfrac{2}{7}=\dfrac{4}{3};x=\dfrac{4}{3}+\dfrac{2}{7}=\dfrac{34}{21}\)
\(2,x+\dfrac{3}{4}=\dfrac{6}{8};x=\dfrac{6}{8}-\dfrac{3}{4}=0\)
\(3,\left|x\right|-\dfrac{3}{4}=\dfrac{2}{3}+\dfrac{1}{4}=\dfrac{11}{12};\left|x\right|=\dfrac{11}{12}+\dfrac{3}{4};\left|x\right|=\dfrac{5}{3}\Rightarrow x=\left[{}\begin{matrix}\dfrac{-5}{3}\\\dfrac{5}{3}\end{matrix}\right.\)
\(4,\left(x+1\right).3=4.5=20;x+1=\dfrac{20}{3}\Leftrightarrow x=\dfrac{17}{3}\)
a ) 13/20
B)
C..........................................................
minh dang tính
Gọi ƯCLN(2n+1;6a+4)=d
2n+1 \(⋮\) d\(\Rightarrow\) 6n +3\(⋮\) d
6n+4\(⋮\)d
\(\Rightarrow\)(6n+4)-(6n+3)\(⋮\) d
\(\Rightarrow\)6n+4 - 6n-3\(⋮\) d
\(\Rightarrow1⋮d\Rightarrow d=1\)
Gọi d là ƯCLN (2a + 1; 6a + 4) Nên ta có :
2a + 1 ⋮ d và 6n + 4 ⋮ d
=> 3 ( 2a + 1 ) ⋮ d và 6n + 4 ⋮ d
=> 6a + 3 ⋮ d và 6a + 4 ⋮ d
=> (6a + 4) - (6a + 3) ⋮ d
=> 1 ⋮ d => d = 1
Vì ƯCLN (2a + 1; 6a + 4) = 1 => 2a + 1 và 6a + 4 là nguyên tố cùng nhau ( đpcm )
Cuối học kì I lớp 6 đề khó vậy !!
a ) 6 |x - 7| = 18 : (-3)
\(\Rightarrow6\left|x-7\right|=-6\)
\(\Rightarrow\left|x-7\right|=-1\) (1)
Mà \(\hept{\begin{cases}\left|x-7\right|\ge0\forall x\\-1< 0\end{cases}}\)
\(\Rightarrow\) | x - 7| = - 1 ( vô lí ) (2)
Từ (1) và (2) \(\Rightarrow\) \(x\in\varnothing\)
Vậy \(x\in\varnothing\)
Câu c tương tự nhé
b) -7 | x + 4| = 21 : (-3)
\(\Rightarrow-7\left|x+4\right|=-7\)
\(\Rightarrow\left|x+4\right|=1\)
\(\Rightarrow\orbr{\begin{cases}x+4=1\\x+4=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=-5\end{cases}}\)
Vậy \(x\in\left\{-3;-5\right\}\)
@@ Hc tốt @@
## Chiyuki Fujito
6 | x - 7 | = 18 : ( -3 )
6 | x - 7 | = ( -6 )
| x - 7 | = ( -6 ) : 6
| x - 7 | = ( -1 )
\(\Rightarrow\)x - 7 = ( -1 ) hoặc x - 7 = 1
x = ( -1 ) + 7 x = 1 + 7
x = 6 x = 8
\(\Rightarrow\) x = 6 ; x = 8
-7 | x + 4 | = 21 : ( -3 )
-7 | x + 4 | = ( -7 )
| x + 4 | = ( -7 ) : ( -7 )
| x + 4 | = 1
\(\Rightarrow\)x + 4 = 1 hoặc x + 4 = ( -1 )
x = 1 - 4 x = ( -1 ) - 4
x = -3 x = -5
\(\Rightarrow\) x = ( -3 ) ; x = ( -5 )
3 | x + 5 | = ( -9 )
| x + 5 | = ( -9 ) : 3
| x + 5 | = ( -3 )
\(\Rightarrow\)x + 5 = ( -3 ) hoặc x + 5 = 3
x = ( -3 ) - 5 x = 3 - 5
x = ( -8 ) x = ( -2 )
\(\Rightarrow\)x = ( -8 ) ; x = ( -2 )
@ Học tốt @
Nhớ k cho mình nha !!!!! Thank
( 3/8 - x ) + 2\(\dfrac{1}{3}\) : 4/3 = 75%
( 3/8 - x) + 7/3 x 3/4 = 75/100
( 3/8 - x) + 7/4 = 3/4
3/8 - x = 3/4 - 7/4
3/8 - x = -1
x = 3/8 + 1
x = 11/8
Hổng bt làm !!!!!!! Nhưng tui xin thông báo tui là Fan K-pop chân chính !!!!! Mong kết bạn !!!!!! Tui ko Anti bất kì nhóm nào
a. (-7) . (-13) + 8 . (-13) = (-7 + 8) . (-13) = 1 . (-13) = -13
b. (-5) . [-4 - (-14)] = (-5) . (-4) - (-5) . (-14) = (-5) . 10 = -50
2B:
a: \(A=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{4}+\dfrac{5}{8}=\dfrac{3}{8}+\dfrac{5}{8}=\dfrac{8}{8}=1\)
b: \(B=\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{3\left(\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}}{\dfrac{3}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}\right)}=\dfrac{1}{3}+\dfrac{2}{3}=1\)
a) \(A=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{4}+\dfrac{5}{8}\)
\(=\dfrac{3}{8}+\dfrac{5}{8}=1\)
b) \(B=\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{1}{2}-\dfrac{3}{10}}\)
\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{3\left(\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)}\)
\(=\dfrac{1}{3}+\dfrac{1}{\dfrac{3}{2}}=\dfrac{1}{3}+\dfrac{2}{3}=1\)