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1.
a, \(x-14=3x+18\)
\(\Rightarrow x-3x=18+14\)
\(\Rightarrow-2x=32\Rightarrow x=\frac{32}{-2}=-16\)
b, \(\left(x+7\right).\left(x-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=9\end{cases}}}\)
c, \(\left|2x-5\right|-7=22\)
\(\Rightarrow\left|2x-5\right|=22+7\)
\(\Rightarrow\left|2x-5\right|=29\)
\(\Rightarrow\orbr{\begin{cases}2x+5=29\\2x-5=29\end{cases}}\Rightarrow\orbr{\begin{cases}2x=24\\2x=34\end{cases}\Rightarrow}\orbr{\begin{cases}x=12\\x=17\end{cases}}\)
d, \(\left(\left|2x\right|-5\right)-7=22\)
\(\Rightarrow\left(\left|2x\right|-5\right)=29\)
\(\Rightarrow\left|2x\right|=29+5\Rightarrow\left|2x\right|=34\Rightarrow x=\pm17\)
e, \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\)
Vì \(\left|x+3\right|\ge0;\left|x+9\right|\ge0;\left|x+5\right|\ge0;4x\ge0\)
Nên \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\ge0\)
\(\Rightarrow\left|x+3\right|>0\Rightarrow\left|x+3\right|=x+3\)
\(\left|x+9\right|>0\Rightarrow\left|x+9\right|=x+9\)
\(\left|x+5\right|>0\Rightarrow\left|x+5\right|=x+5\)
Ta có :
\(x+3+x+9+x+5=4x\)
\(\Rightarrow3x+\left(3+9+5\right)=4x\)
\(\Rightarrow4x-3x=17\)
\(\Rightarrow x=17\)
2. a , b sai đề bn
c, \(\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(\text{ }Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
x | 0 | -2/5 | 1/5 | -3/5 | 3/5 | -1 |
y | -3 | 5 | -1 | 3 | 0 | 2 |
d, \(5xy-5x+y=5\)
\(\Rightarrow\left(5xy-5x\right)+y=5\)
\(\Rightarrow5x.\left(y-1\right)+y=5\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
x | 0 | -2 | 1/5 | -3/5 | 3/5 | -1 |
y | -3 | 5 | -1 | 3 | 0 | 2 |
( 2 x y + 2/15 ) x 3 = 4/5
( 2 x y + 2/15 ) = 4/5 : 3
( 2 x y + 2/15 ) = 4/15
2 x y = 4/15 - 2/15
2 x y = 2/15
y = 2/15 :2
y = 1/15
(2 x y + 2/15) x 3 = 4/5
2 x y + 2/15) = 4/5 : 3
2 x y + 2/15 = 4/15
2 x y = 4/15 - 2/15
2 x y = 2/15
y = 2/15 : 2
y = 1/15
7/9 x (2 - 1/3 x y) = 14/15
(2 - 1/3 x y) = 14/15 : 7/9
(2 - 1/3 x y) = 6/5
2 - y = 6/5 x 1/3
2 - y = 2/5
y = 2/5 + 2
y = 12/5
4/21 + 5 x y - 8/7 = 1/3
4/21 + 5 x y = 1/3 + 8/7
4/21 + 5 x y = 31/21
5 x y = 31/21 - 4/21
5 x y = 9/7
y = 9/7 : 5
y = 9/35
7/12 x y - 3/12 x y = 5
y x (7/12 - 3/12) = 5
y x 1/3 = 5
y = 5 : 1/3
y = 15
\(\left(y+\frac{1}{5}\right).\frac{3}{7}=\frac{5}{14}\)
\(\left(y+\frac{1}{5}\right)=\frac{5}{14}:\frac{3}{7}\)
\(\left(y+\frac{1}{5}\right)=\frac{5}{6}\)
\(y=\frac{5}{6}-\frac{1}{5}\)
\(y=\frac{19}{30}\)
Vậy \(y=\frac{19}{30}\)
b)
ta có n + 1 chia hết cho n + 1
=> 2( n + 1 ) chia hết cho n + 1
=> 2n + 2 chia hết cho n + 1
mà 2n + 7 chia hết cho n + 1
=> 2n + 7 - ( 2n + 2 ) chia hết cho n + 1
=> 2n + 7 - 2n - 2 chia hết cho n + 1
=> 5 chia hết cho n + 1
=> n + 1 e Ư( 5 ) = { 1 ; 5 }
nếu n + 1 = 1 => n = 1 - 1 => n = 0 ( thỏa mãn )
nếu n + 1 = 5 => N = 5 - 1 => N = 4 ( thỏa mãn )
vậy n e { 0 ; 4 }
c) 728 và 5014
728 = ( 72 )14 = 4914
ta thấy 49 < 50 => 4914 < 5014
=> 728 < 5014
chúc bạn học giỏi ^^
mình biết làm câu b thôi hình như đây là dạng toán nâng cao hà
\(\left(y+\frac{1}{5}\right).\frac{3}{7}=\frac{5}{14}\)
\(\left(y+\frac{1}{5}\right)=\frac{5}{6}\)
\(y=\frac{5}{6}-\frac{1}{5}\)
\(y=\frac{19}{30}\)
Vậy \(y=\frac{19}{30}\)
\(\left(y+15\right).\frac{3}{7}=\frac{5}{14}\)
\(\Rightarrow y+15=\frac{5}{14}:\frac{3}{7}\)
\(\Rightarrow y+15=\frac{5}{6}\)
\(\Rightarrow y=\frac{5}{6}-15\)
\(\Rightarrow y=-\frac{85}{6}\)