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\(\left|x+1\right|+\left|x+4\right|=3x\)
\(\Rightarrow1+x+4+x=3x\)
\(\Rightarrow5+2x=3x\)
\(\Rightarrow5=3x-2x\)
\(\Rightarrow5=x\)
x + y + xy = 2
x + y(x + 1) = 2
x + 1 + y(x + 1) = 3
(x + 1)(y + 1) = 3
=> x + 1 và y + 1 thuộc ước của 3
Ư(3) = { - 3; - 1; 1; 3 }
Ta có bảng sau :
| x + 1 | - 3 | - 1 | 3 | 1 |
| y + 1 | - 1 | - 3 | 1 | 3 |
| x | - 4 | - 2 | 2 | 0 |
| y | - 2 | - 4 | 0 | 2 |
Vậy ( x;y ) = { ( -4;-2 );( -2;-4 );( 2;0 );( 0;2 ) }
\(2.\)
\(2x^3-6x\)
\(\Leftrightarrow2x^3-6x=0\)
\(\Leftrightarrow2x\left(x^2-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x^2-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{3}\end{cases}}\)
\(x+\frac{5}{2015}+x+\frac{6}{2014}+x+\frac{3}{2017}=-3\)
\(\Rightarrow3x+\left(\frac{1}{403}+\frac{3}{1007}+\frac{3}{2017}\right)=-3\)
\(\Rightarrow\frac{1}{403}+\frac{3}{1007}+\frac{3}{2017}=-3-3x=-3.\left(1-x\right)\)
\(\Rightarrow\frac{\frac{1}{403}+\frac{3}{1007}+\frac{3}{2017}}{-3}-1=-x\)
a, ( 44 - x ) / 3 = ( x - 12 ) / 5
=> 5 ( 44 - x ) = 3 ( x - 12 )
220 - 5x = 3x - 36
- 5x - 3x = - 36 - 220
- 8 x = - 256
x = 32
b , ( 3 - x ) / 4 = ( 2x + 7 ) / 5
=> 5 ( 3 - x ) = 4 ( 2x + 7 )
15 - 5x = 8 x + 28
- 5 x - 8 x = 28 - 15
- 13 x = 13
x = -1
a, \(\frac{\left(44-x\right)}{3}=\frac{\left(x-12\right)}{5}\)
=> (44 - x) . 5 = (x - 12) . 3
=> 44 - x . 5 = x - 12 .3
=> 44 - x . 5 = x - 36
=> x5 + x = - 36 - 44
=> x5 + x = - 80
=> x . (5 + 1) = - 80
=> x . 6 = - 80
=> x = - 80 : 6
=> x = - 13,3
b, \(\frac{\left(3-x\right)}{4}=\frac{\left(2x+7\right)}{5}\)
=> (3 - x) . 5 = (2x + 7) . 4
=> 3 - x . 5 = 2x + 7 . 4
=> 3 - x . 5 = 2x + 28
=> -x . 5 + 2x = 28 - 3
=> -x . 5 + 2x = 25
=> x . 5 + 2x = 25
=> x . (5 + 2) = 25
=> x . 7 = 25
=> x = 25 : 7
=> x = 3,57
\(\frac{x-2}{x-1}=\frac{x+4}{x-7}\) Đk : x \(\ne\)1 ; 7
=> ( x - 2 ) . ( x - 7 ) = ( x - 1 ) . ( x + 4 )
=> x ( x - 7 ) - 2 ( x - 7 ) = x ( x - 1 ) + 4 ( x - 1 )
=> x 2 - 7x - 2x + 14 = x 2 - x + 4x - 4
=> - 7x - 2x + x - 4x = - 4 - 14
=> - 12 x = - 18
=> x = \(\frac{3}{2}\)
Bài 1:
Giải:
Ta có: \(3\left(x-1\right)=2\left(y-2\right)=3\left(z-3\right)\)
\(\Rightarrow\frac{x-1}{\frac{1}{3}}=\frac{y-2}{\frac{1}{2}}=\frac{z-3}{\frac{1}{3}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-1}{\frac{1}{3}}=\frac{y-2}{\frac{1}{2}}=\frac{z-3}{\frac{1}{3}}=\frac{2x-2}{\frac{2}{3}}=\frac{3y-6}{\frac{3}{2}}=\frac{z-3}{\frac{1}{3}}=\frac{2x-2+3y-6+z-3}{\frac{2}{3}+\frac{3}{2}+\frac{1}{3}}=\frac{\left(2x+3y+z\right)-\left(2+6+3\right)}{\frac{5}{2}}\)
\(=\frac{50-11}{\frac{5}{2}}=\frac{39}{\frac{5}{2}}=39.\frac{2}{5}=15,6\)
+) \(\frac{x-1}{\frac{1}{3}}=15,6\Rightarrow x-1=5,2\Rightarrow x=6,2\)
+) \(\frac{y-2}{\frac{1}{2}}=15,6\Rightarrow y-2=7,8\Rightarrow y=9,8\)
+) \(\frac{z-3}{\frac{1}{3}}=15,6\Rightarrow z-3=5,2\Rightarrow z=8,2\)
Vậy bộ số \(\left(x;y;z\right)\) là \(\left(6,2;9,8;8,2\right)\)
Ta có:
x+2x+3x+...+100x=50
x.(1+2+3+...+100)=50
x.5050=50
x=\(\frac{1}{101}\)
\(x+2x+3x+...+100x=50\)
\(\Rightarrow5050x=50\)
\(\Rightarrow x=\frac{50}{5050}\)
\(\Rightarrow x=\frac{1}{101}\)