8. Bpt 2x -1 >= 0 tương đương với bột nào?
A. 2x -1 + 1/x-3 >= 1/x -3
B. 2x -1 + 1/x-3 >=1/x-3
C. (2x-1).✓x -2018 >= ✓ x-2018
D. 2x -1/ ✓ x -2018 >= 1/✓ x -2018
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a) x( x + 2018 ) - 2x - 4036 = 0
<=> x( x + 2018 ) - 2( x + 2018 ) = 0
<=> ( x + 2018 )( x - 2 ) = 0
<=> \(\orbr{\begin{cases}x+2018=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2018\\x=2\end{cases}}\)
b) x + 5 = 2( x + 5 )2
<=> x + 5 = 2( x2 + 10x + 25 )
<=> x + 5 = 2x2 + 20x + 50
<=> 2x2 + 20x + 50 - x - 5 = 0
<=> 2x2 + 19x + 45 = 0
<=> 2x2 + 10x + 9x + 45 = 0
<=> 2x( x + 5 ) + 9( x + 5 ) = 0
<=> ( x + 5 )( 2x + 9 ) = 0
<=> \(\orbr{\begin{cases}x+5=0\\2x+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=-\frac{9}{2}\end{cases}}\)
c) ( x2 + 1 )( 2x - 1 ) + 2x = 1
<=> 2x3 - x2 + 4x - 1 - 1 = 0
<=> 2x3 - x2 + 4x - 2 = 0
<=> x2( 2x - 1 ) + 2( 2x - 1 ) = 0
<=> ( 2x - 1 )( x2 + 2 ) = 0
<=> \(\orbr{\begin{cases}2x-1=0\\x^2+2=0\end{cases}\Leftrightarrow}x=\frac{1}{2}\)( vì x2 + 2 ≥ 2 > 0 ∀ x )
d) \(\frac{x}{3}-\frac{x^2}{4}=0\)
\(\Leftrightarrow\frac{4x}{12}-\frac{3x^2}{12}=0\)
\(\Leftrightarrow\frac{4x-3x^2}{12}=0\)
\(\Leftrightarrow4x-3x^2=0\)
\(\Leftrightarrow x\left(4-3x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\4-3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)
a, Vì -2018 khác 0
=> x-11=0
=> x=11
b, Vì -2018 < 0
=> x+13 > 0
=> x > -13
c, Vì 2018 > 0 => 2x-10 > 0
=> 2x > 10
=> x > 5
d, => x-3=0 hoặc 3x-9=0
=> x=3
e, Vì x-1 < x+5
=> x-1 < 0 và x+5 > 0
=> x < 1 và x > -5
=> -5 < x < 1
Tk mk nha
Các đề bài trên khi chuyển vế đều bị mất đi x nên không có x thỏa mãn
b,2x.(x-5)-x.(3+2x)=26
2x2 - 10x - 3x - 2x2 = 26
-13x = 26
x = -2
c, (x+7)2-x.(x-3)=12
x2 +14x +49 - x2 + 3x = 12
17x + 49 = 12
17x = - 37
x = \(\dfrac{-37}{17}\)
d, 9( x -2018) - x+ 2018 =0
9( x -2018) - (x -2018) = 0
( 9-1)(x -2018) = 0
8( x -2018) = 0
x -2018 = 0
x = 2018
a: =>2x+10-x^2-5=0
=>-x^2+2x+5=0
=>\(x\in\left\{1+\sqrt{6};1-\sqrt{6}\right\}\)
e: =>4x^2+4x+9x^2-4=15
=>13x^2+4x-19=0
=>\(x\in\left\{\dfrac{-2+\sqrt{251}}{13};\dfrac{-2-\sqrt{251}}{13}\right\}\)
a) \(22-x\left(1-4x\right)=\left(2x+3\right)^3\)
\(\Leftrightarrow22-x+4x^2=8x^3+36x^2+54x+27\)
\(\Leftrightarrow-x-54x+4x^2-36x^2-8x^3=-22+27\)
\(\Leftrightarrow-8x^3-32x^2-55x=5\Leftrightarrow-8x^3-32x^2-55x-5=0\)
Bn tự làm tiếp nhé
b) \(\frac{2x}{3}+\frac{2x-1}{6}=\frac{4-x}{3}\Leftrightarrow\frac{2.2x}{6}+\frac{2x-1}{6}=\frac{2\left(4-x\right)}{6}\)
\(\Leftrightarrow2.2x+2x-1=2\left(4-x\right)\Leftrightarrow4x+2x-1=8-2x\)
\(\Leftrightarrow6x-1=8-2x\Leftrightarrow8x=9\Leftrightarrow x=\frac{9}{8}\)
Vậy phương trình có tập nghiệm S ={9/8}
c) \(\frac{x-1}{2019}+\frac{x-2}{2018}=\frac{x-3}{2017}+\frac{x-4}{2016}\)
\(\Leftrightarrow\left(\frac{x-1}{2019}-1\right)+\left(\frac{x-2}{2018}-1\right)=\left(\frac{x-3}{2017}-1\right)+\left(\frac{x-4}{2016}-1\right)\)
\(\Leftrightarrow\frac{x-2020}{2019}+\frac{x-2020}{2018}-\frac{x-2020}{2017}-\frac{x-2020}{2016}=0\)
\(\Leftrightarrow\left(x-2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\right)=0\)
Do \(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}>0\)
Nên \(x-2020=0\Leftrightarrow x=2020\)
a) \(\left(x+2018\right)\left(\frac{1}{2}+\frac{2}{7}\right)=\left(x+2018\right)\left(\frac{1}{5}+\frac{1}{6}\right)\)
\(\Leftrightarrow\) \(\left(x+2018\right)\left(\frac{1}{2}+\frac{2}{7}\right)-\left(x+2018\right)\left(\frac{1}{5}+\frac{1}{6}\right)\) = 0
\(\Leftrightarrow\left(x+2018\right)\left(\frac{1}{2}+\frac{2}{7}-\frac{1}{5}-\frac{1}{6}\right)=0\)
\(\Leftrightarrow x+2018=0\)
\(\Leftrightarrow x=-2018\)
b) \(7\left(x-1\right)+2x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(7+2x\right)=0\)
\(\Leftrightarrow\) x - 1 = 0 hoặc 7 + 2x = 0
1) x - 1 = 0 \(\Leftrightarrow\) x = 1
2) 7 + 2x = 0 \(\Leftrightarrow\) -3,5
Vậy: x = 1; -3,5
b) \(7\left(x-1\right)+2x\left(x-1\right)=0\)
=> \(\left(x-1\right).\left(7+2x\right)=0\)
=> \(\left\{{}\begin{matrix}x-1=0\\7+2x=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=0+1\\2x=0-7=-7\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=1\\x=\left(-7\right):2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=1\\x=-\frac{7}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{1;-\frac{7}{2}\right\}.\)
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