1,
A, 2017-|-26|×|-2³|+12
b M=26+|-x| .y -18 với x=12,y=6
2,tìm x
a x=|-32|
b 26-3 |x|=8
c |x-1|=6-|-6|
Giúp mk vs mk đag cần gấp.
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, -19 - x = -20
x = -19 - (-20)
x = -19 + 20
x = 1
b, 5x - 6 = 3x + 12
5x - 6 - 3x = 12
5x - 3x = 12 + 6
(5 - 3)x = 18
2x = 18
x = 18 : 2
x = 9
c, 15 - 3 (x - 1) = 8 - 2x
15 - 3 (x - 1) + 2x = 8
-3x - 3 - 2x = 8 - 15
-3x - 3 - 2x = -7
-3x - 2x - 3 = 7
-3x - 2x = 7 + 3
(-3 - 2) x = 10
-5x = 10
x = 10 : (-5)
x = -2
d, (5x - 6)2 = 16
(5x - 6)2 = 42
=> 5x - 6 = 4
5x = 4 + 6
5x = 10
x = 10 : 5
x = 2
f, 26 - | x + 9 | = 13
| x + 9 | = 26 - 13
=> | x + 9 | = 13
=> x + 9 = +- 13
* Với x + 9 = 13
x = 13 - 9
x = 4
* Với x + 9 = -13
x = -13 - 9
x = -22
Vậy x = {4;-22}
e, | 3 + x | = 19
=> 3 + x = +- 19
* Với 3 + x = 19
x = 19 - 3
x = 16
* Với 3 + x = -19
x = -19 - 3
x = -22
Vậy x = {16;-22}
a, X = -19+20=1
b, (5-3)X = 18
2X = 18
=> X = 9
c, 3X + 3 -2X = 7
X+3 =7
X = 4
f, |X+9| = 13
ta có 2 trường hợp:
TH1: X+9 = 13
=> X= 4
TH2 : X+9 = -13
=> X= -22
e, ta có 2 trường hợp:
TH1: 3+X = 19
=> X= 16
TH2: 3+X = -19
=> X= -22
a,ĐKXĐ:\(x\ge2\)
\(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\dfrac{x-2}{4}}=26\\ \Leftrightarrow4\sqrt{x-2}+3\sqrt{x-2}-\dfrac{\sqrt{x-2}}{2}=26\\ \Leftrightarrow8\sqrt{x-2}+6\sqrt{x-2}-\sqrt{x-2}=52\\ \Leftrightarrow13\sqrt{x-2}=52\\ \Leftrightarrow\sqrt{x-2}=4\\ \Leftrightarrow x-2=16\\ \Leftrightarrow x=18\left(tm\right)\)
b,ĐKXĐ:\(x\in R\)
\(3x+\sqrt{4x^2-8x+4}=1\\ \Leftrightarrow2\sqrt{x^2-2x+1}=1-3x\\ \Leftrightarrow\left|x-1\right|=\dfrac{1-3x}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1-3x}{2}\\x-1=\dfrac{3x-1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-2=1-3x\\2x-2=3x-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
c, ĐKXĐ:\(x\ge0\)
\(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)-2\left(2\sqrt{x}+1\right)=7\\ \Leftrightarrow2x+\sqrt{x}-4\sqrt{x}-2=7\\ \Leftrightarrow2x-3\sqrt{x}-9=0\\ \Leftrightarrow\left(2x+3\sqrt{x}\right)-\left(6\sqrt{x}+9\right)=0\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+3\right)-3\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left(\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\2\sqrt{x}=-3\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=9\left(tm\right)\)
\(\left(x+3\right).y=6\Rightarrow\left(x+3\right).y-6=0\)
\(\Rightarrow\hept{\begin{cases}x+3=0\\y-6=0\end{cases}\Rightarrow\hept{\begin{cases}x=-3\\y=6\end{cases}}}\)
\(\left(x+1\right).\left(y-2\right)=12\Rightarrow\left(x+1\right).\left(y-2\right)-12=0\)\(\Rightarrow\hept{\begin{cases}x+1=6\\y-2=2\end{cases}\Rightarrow\hept{\begin{cases}x=5\\y=4\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}x+1=3\\y-2=4\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=6\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x+1=1\\y-2=12\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\y=14\end{cases}}}\)
( x + 3 ) . y = 6
=> ( x + 3 ) . y = 1 . 6 = 6 . 1 = -1 . ( - 6 ) = -6 . ( -1 )
= 2 . 3 = 3 . 2 = - 2 . ( -3 ) = -3 . ( - 2 )
x + 3 | 1 | 6 | -1 | -6 | 2 | 3 | -2 | -3 |
y | 6 | 1 | -6 | -1 | 3 | 2 | -3 | -2 |
x | -2 | 3 | -4 | -9 | -1 | 0 | -5 | -6 |
y | 6 | 1 | -6 | -1 | 3 | 2 | -3 | -2 |
Vậy các cặp ( x,y ) thỏa mãn là : ( -2 , 6 ) ; ( 3 , 1 ) ; ( -4 , -6 ) ; ( -9 , -1 ) ; ( -1 ,3 ) ; ( 0 , 2 ) ; ( -5 , -3 ) ; ( -6 , -2 )
\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{4}\)và \(x+y-z=26\)
\(BCNN\left(3,5\right)=15\)
\(\Rightarrow\frac{x}{2}=\frac{y}{3}\)\(\Rightarrow\frac{x}{10}=\frac{y}{15}\)(1)
\(\frac{y}{5}=\frac{z}{4}\)\(\Rightarrow\frac{y}{15}=\frac{z}{12}\)(2)
Từ (1) và (2) \(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)
\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x+y-z}{10+15-12}=\frac{26}{13}=2\)
\(\Rightarrow x=2.10=20\)
\(y=2.15=30\)
\(z=2.12=24\)
Vậy x = 20 ; y = 30 ; z = 24