Tim so nguyen x biet
a) (x - 1)(x + 4) < 0
b) 5x + 2 - 5x - 1 = 3100
c) 3x + 1 - 3x - 2 = 702
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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: \(\left(x+3\right)\left(2y-1\right)=5\)
=>\(\left(x+3\right)\left(2y-1\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)
=>\(\left(x+3;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(-2;3\right);\left(2;1\right);\left(-4;-2\right);\left(-8;0\right)\right\}\)
a: Ta có: \(3x\left(3x-1\right)-\left(3x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow9x^2-3x-9x^2+1=0\)
\(\Leftrightarrow3x=1\)
hay \(x=\dfrac{1}{3}\)
b: Ta có: \(x^2-5x+25-5x=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x-5=0\)
hay x=5
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)\(-\frac{21}{x}+\frac{18}{x}=\frac{-21+18}{x}=\frac{-3}{x}\in Z\)
=>-3 chia hết x
=>x thuộc Ư(-3)
=>x thuộc {1;-1;3;-3}
b)\(\frac{2x-5}{x+1}=\frac{2\left(x+1\right)-7}{x+1}=\frac{2\left(x+1\right)}{x+1}-\frac{7}{x+1}=2-\frac{7}{x+1}\in Z\)
=>7 chia hết x+1
=>x+1 thuộc Ư(7)
=>x+1 thuộc {1;-1;7;-7}
=>x thuộc {0;-2;6;-8}
c)\(\frac{3x+2}{x-1}-\frac{x-5}{x-1}=\frac{3x+2-\left(x-5\right)}{x-1}=\frac{2x+7}{x-1}=\frac{2\left(x-1\right)+9}{x-1}=\frac{2\left(x-1\right)}{x-1}+\frac{9}{x-1}\)\(=2+\frac{9}{x-1}\in Z\)
=>9 chia hết x-1
=>x-1 thuộc Ư(9)
=>....
Còn lại bạn tự làm típ nha khi nào ko làm đc thì nhắn vs mk :)
a, 3x - 7 = 0
<=> 3x = 7
<=> x = 7/3
b, 8 - 5x = 0
<=> -5x = -8
<=> x = 8/5
c, 3x - 2 = 5x + 8
<=> -2x = 10
<=> x = -5
e) Ta có: \(\left(5x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=3\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{5};3\right\}\)
\(a,\Leftrightarrow\left(5x+1\right)\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x+1-x\right)=0\\ \Leftrightarrow5x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x^2-10x-2x^2-3x=26\\ \Leftrightarrow-13x=26\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^3+1-x^3+3x=15\\ \Leftrightarrow3x=14\\ \Leftrightarrow x=\dfrac{14}{3}\)
\(d,\Leftrightarrow x^3-5x+2x^2-10+5x-2x^2-17=0\\ \Leftrightarrow x^3-27=0\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x=3\)
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
Tuyet Anh Nguyen
1.a)(3x-2)(4x+5)=0
12x^2+7x-10=0>>x1=2/3,x2=-5/4
b)4x^3+2x^2+4x+2=0>>x=-1
c)0,23x^2-4,21x-13,8=0>>x1=21,14,x2=-2,8...
d)10x^3-13x^2-178x-35=0>>x1=5,x2=-1/5
b2/a)2x^3+5x^2-3x=0>>x1=1/2,x2=-3
b)(3x-1)(x^2-7x+12)=0>>x1=1/3,x2=4,x3=...
b3/
a)x^2+x-2=0>>x1=1,x2=-2
b)x1=-1,x2=-6
b4/a)0,5x^2-1,5x-1,5x^2+x+4,5x-3=0>>-x...
b)3x/7-1=3x/7-x>>x=1
c)2x^2-13x+15=0>>x1=5,x2=3/2
P/s: Tham khảo nha
\(a,\left(2x-3\right)\left(x^2-4\right)=0\\ \Leftrightarrow\left(2x-3\right)\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=-2\end{matrix}\right.\\ b,2x-\left(3-5x\right)=4\left(x+3\right)\\ \Leftrightarrow2x-3+5x=4x+12\\ \Leftrightarrow7x-3-4x-12=0\\ \Leftrightarrow3x-15=0\\ \Leftrightarrow x=5\)
\(c,ĐKXĐ:\left\{{}\begin{matrix}x\ne-1\\x\ne2\end{matrix}\right.\)
\(\dfrac{1}{x-2}-\dfrac{2}{x+1}=\dfrac{11-3x}{\left(x+1\right)\left(x-2\right)}\\ \Leftrightarrow\dfrac{x+1}{\left(x-2\right)\left(x+1\right)}-\dfrac{x-2}{\left(x+1\right)\left(x-2\right)}-\dfrac{11-3x}{\left(x+1\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x+1-x+2-11+3x}{\left(x+1\right)\left(x-2\right)}=0\\ \Rightarrow3x-8=0\\ \Leftrightarrow x=\dfrac{8}{3}\left(tm\right)\)
a: =>x-1=0 hoặc 3x-1=0
=>x=1 hoặc x=1/3
b: ĐKXĐ: x<>2; x<>-1
PT =>x-2-5(x+1)=15
=>x-2-5x-5=15
=>-4x-7=15
=>-4x=22
=>x=-11/2(nhận)
c: ĐKXĐ: x<>2; x<>-2
PT =>(x-1)(x-2)-x(x+2)=5x-2
=>x^2-3x+2-x^2-2x=5x-2
=>-5x+2=5x-2
=>-10x=-4
=>x=2/5(nhận)
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