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a: \(\left(x+5\right)^2>=0\forall x\)
\(\left(2y-8\right)^2>=0\forall y\)
Do đó: \(\left(x+5\right)^2+\left(2y-8\right)^2>=0\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+5=0\\2y-8=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-5\\y=4\end{matrix}\right.\)
b: \(\left(x+3\right)\left(2y-1\right)=5\)
=>\(\left(x+3\right)\left(2y-1\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)
=>\(\left(x+3;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(-2;3\right);\left(2;1\right);\left(-4;-2\right);\left(-8;0\right)\right\}\)
b: \(\left(2x+1\right)^2=25\)
=>\(\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
c: \(\left(1-3x\right)^3=64\)
=>\(\left(1-3x\right)^3=4^3\)
=>1-3x=4
=>3x=1-4=-3
=>x=-3/3=-1
d: \(\left(4-x\right)^3=-27\)
=>\(\left(4-x\right)^3=\left(-3\right)^3\)
=>4-x=-3
=>x=4+3=7
e: \(x^2-5x=0\)
=>\(x\left(x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
\(a,-4\left(2x+9\right)=\left(-8x+3\right)\)
\(\Rightarrow-8x-36=-8x+3\)
\(\Rightarrow-8x+8x=3+36\)
\(\Rightarrow0x=39\left(vô-lí\right)\)
\(b,1+x-2\left(5+3x\right)=4-5x\)
\(\Rightarrow1+x-10-6x=4-5x\)
\(\Rightarrow x-6x+5x=4+10-1\)
\(\Rightarrow0x=13\left(vô-lí\right)\)
\(c,3\left(2-x\right)+1=-3x+7\)
\(\Rightarrow6-3x+1=-3x+7\)
\(\Rightarrow-3x+3x=7-1-6\)
\(\Rightarrow0x=0\Rightarrow x=0\)
1) 3x - 6 = 5x + 2
=> 3x - 5x = 2 + 6
=> -2x = 8
=> x = -4
2) 15 - x = 4x - 5
=> 15 + 5 = 4x + x
=> 20 = 5x
=> x = 4
3) x - 15 = 6 + 4x
=> x - 4x = 6 + 15
=> -3x = 21
=> x = -7
4) -12 + x = 5x - 20
=> x - 5x = -20 + 12
=> -4x = -8
=> x = 2
5) 7x - 4 = 20 + 3x
=> 7x - 3x = 20 + 4
=> 4x = 24
=> x = 6
1) 3x- 6 = 5x + 2
5x - 3x = -6 - 2
2x = -8 => x = -4
Tương tự như trên
a) ( x - 1 )( x + 4 ) < 0
Xét hai trường hợp :
1. \(\hept{\begin{cases}x-1< 0\\x+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 1\\x>-4\end{cases}}\Rightarrow-4< x< 1\)
2. \(\hept{\begin{cases}x-1>0\\x+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>1\\x< -4\end{cases}}\)( loại )
Vậy với -4 < x < 1 thì ( x - 1 )( x + 4 ) < 0
b) 5x+2 - 5x-1 = 3100
<=> 5x( 52 - 5-1 ) = 3100
<=> 5x( 25 - 1/5 ) = 3100
<=> 5x.124/5 = 3100
<=> 5x = 125
<=> 5x = 53
<=> x = 3
c) 3x+1 - 3x-2 = 702
<=> 3x( 3 - 3-2 ) = 702
<=> 3x( 3 - 1/9 ) = 702
<=> 3x.26/9 = 702
<=> 3x = 243
<=> 3x = 35
<=> x = 5
a) (x - 1)(x + 4) < 0
Xét các trường hợp
TH1\(\hept{\begin{cases}x-1>0\\x+4< 0\end{cases}}\Rightarrow\hept{\begin{cases}x>1\\x< -4\end{cases}}\left(\text{loại}\right)\)
TH2\(\hept{\begin{cases}x-1< 0\\x+4>0\end{cases}}\Rightarrow\hept{\begin{cases}x< 1\\x>-4\end{cases}}\Rightarrow-4< x< 1\left(tm\right)\)
Vậy -4 < x < 1
b) 5x + 2 - 5x - 1 = 3100
=> 5x(52 - 1/5) = 3100
=> 5x.124/5 = 3100
=> 5x = 125
=> 5x = 53
=> x = 3
c) 3x + 1 - 3x - 2 = 702
=> \(3^x.3-3^x.\frac{1}{3^2}=702\)
=> 3x(3 - 1/9) = 702
=> 3x.26/9 = 702
=> 3x = 243
=> 3x = 35
=> x = 5
Vậy x = 5