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a; \(\dfrac{-x}{4}\) = \(\dfrac{-2}{x}\)
-\(x.x\) = -2.4
-\(x^2\) = -8
\(x^2\) = 8
\(\left[{}\begin{matrix}x=-\sqrt{8}\\x=\sqrt{8}\end{matrix}\right.\)
Vậy \(x\in\) {-\(\sqrt{8}\); \(\sqrt{8}\)}
34 + (9-21)=3417-(x+341)
22 = 3417-(x+341)
3417-(x+341)=22
x+341=3417-22
x+341=3395
x=3395-341 ; x =3054
Xin lỗi Quảng Bùi Quỳnh Trang, tớ lỡ bấm vào sai rồi
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1a/ \(\left(15-x\right)+\left(x-12\right)=7-\left(-5+x\right)\)
=> \(\left(15-x\right)+\left(x-12\right)+\left(-5+x\right)=7\)
=> \(15-x+x-12-5+x=7\)
=> \(\left(15-12-5\right)-\left(x+x+x\right)=7\)
=> \(\left(15-12-5\right)-7=3x\)
=> \(3x=-2-7\)
=> \(3x=-9\)
=> \(x=\frac{-9}{3}=-3\)
b/ \(x-\left\{57-\left[42+\left(-23-x\right)\right]\right\}=13-\left\{47+\left[25-\left(32-x\right)\right]\right\}\)
=> \(x-57-42-23-x=13-47+25-32+x\)
=> \(x-x+x=13-47+25-32+57+42+23\)
=> \(x=\left(13+23\right)-\left(47+57\right)+\left(25+57\right)-\left(32+42\right)\)
=> \(x=36-104+82-74\)
=> \(x=-60\)
d/ \(\left(x-3\right)\left(2y+1\right)=7\)
Vì 7 là số nguyên tố nên ta có 2 trường hợp:
TH1: \(\hept{\begin{cases}x-3=1\\2y+1=7\end{cases}}\)=> \(\hept{\begin{cases}x=4\\y=3\end{cases}}\).
TH2: \(\hept{\begin{cases}x-3=7\\2y+1=1\end{cases}}\)=> \(\hept{\begin{cases}x=10\\y=0\end{cases}}\).
Các cặp (x, y) thoả mãn điều kiện: \(\left(4;3\right),\left(10;0\right)\).
1) x-2=-6
x=-6+2=-4
2)-5x-(-3)=13
-5x =13+(-3)=10
x=10:(-5)=-2
3) 15-(x-7)=-21
x-7=15-(-21)=36
x=7+36=43
4) 3x+17=2
3x=2-17=-15
x=-15:3=-5
1) x-2 = 6
<=> x = 6+2 =8
2) -5x-(-3)=13
<=> -5x+3 =13
<=> -5x=13-3=10
<=> x = 10:(-5)
<=> x=-2
3) 15-(x-7)=-21
<=> x-7 = 15 -(-21) =36
<=> x = 36 +7 = 43
4) 3x+17 = 2
<=> 3x = 2-17 =-15
<=> x=-15 :3 = -5
a) Ta có:+) \(\frac{12}{16}=\frac{-x}{4}\) <=> 12.4 = 16.(-x)
<=> 48 = -16x
<=> x = 48 : (-16) = -3
+) \(\frac{12}{16}=\frac{21}{y}\) <=> 12y = 21.16
<=> 12y = 336
<=> y = 336 : 12 = 28
+) \(\frac{12}{16}=\frac{z}{-80}\) <=> 12. (-80) = 16z
<=> -960 = 16z
<=> z = -960 : 16 = -60
b) Ta có: \(\frac{x+3}{7+y}=\frac{3}{7}\) <=> (x + 3).7 = 3(7 + y)
<=> 7x + 21 = 21 + 3y
<=> 7x = 3y
<=> \(\frac{x}{3}=\frac{y}{7}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{3}=\frac{y}{7}=\frac{x+y}{3+7}=\frac{20}{10}=2\)
=> \(\hept{\begin{cases}\frac{x}{3}=2\\\frac{y}{7}=2\end{cases}}\) => \(\hept{\begin{cases}x=2.3=6\\y=2.7=14\end{cases}}\)
Vậy ...
b)
\(4\frac{5}{9}:2\frac{5}{18}-7< x< \left(3\frac{1}{5}:3,2+4,5.1\frac{31}{45}\right):\left(21.\frac{1}{2}\right)\)
\(\Rightarrow\frac{41}{9}:\frac{41}{18}-7< x< \left(\frac{16}{5}:\frac{16}{5}+\frac{9}{2}.\frac{76}{45}\right):\frac{21}{2}\)
\(\Rightarrow2-7< x< \left(1+\frac{38}{5}\right):\frac{21}{2}\)
\(\Rightarrow-5< x< \frac{43}{5}:\frac{21}{2}\)
\(\Rightarrow-5< x< \frac{86}{105}\)
Vì \(x\in Z\left(gt\right)\)
\(\Rightarrow x\in\left\{-4;-3;-2;-1;0\right\}.\)
Vậy \(x\in\left\{-4;-3;-2;-1;0\right\}.\)
a) \(\left(x+3\right):4=12\)
\(x+3=12\cdot4\)
\(x+3=48\)
\(\Rightarrow x=45\)
b) 52015 : x = 52013
=> x = 52015 : 52013
=> x = 52
c) 21 + 3 . ( x - 7 ) = 72
3 . ( x - 7 ) = 72 - 21
3 . ( x - 7 ) = 51
x - 7 = 51 : 3
x - 7 = 17
=> x = 24
a. (x+3)-21=-21
=> x+3=-21+21
=> x+3=0
=> x=0-3
Vậy x=-3.
b. |x|-5=7
=> |x|=7+5
=> |x|=12
=> x = 12 hoặc x=-12.
Vậy x \(\in\){-12; 12}
c. |3-x|+2=13
=> |3-x|=13-2
=> |3-x|=11
+) 3-x=11
=> x=3-11
=> x=-8
+) 3-x=-11
=> x=3-(-11)
=> x=3+11
=> x=14
Vậy x \(\in\) {-8; 14}.
a , (x+3) -21=-21 b ,|x|-5=7 c, |3-x|+2=13
x+3=0 x= -3 |x|=12 x=12 |3-x|=11
=-12 3-x=11 x= -8
3-x=-11 =14