A ) ( x2 + 5 ) . ( x2 - 25 ) = 0
B ) ( x2 - 5 ) . ( x2 - 25 < 0
nhanh nha
K
Khách
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1 tháng 7 2021
a)4x2+4x+1-x2-10x-25=0
`<=>(2x+1)^2-(x+5)^2=0`
`<=>(2x+1-x-5)(2x+1+x+5)=0`
`<=>(x-4)(3x+6)=0`
`<=>(x-4)(x+2)=0`
`<=>` \(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\)
b)(x^2+x+7)(x^2+x-7)=(x2+x)2-7x
`<=>(x^2+x)^2-7^2=(x^2+x)^2-7x`
`<=>-7^2=-7x`
`<=>-49=-7x`
`<=>x=7`
Vậy x=7
21 tháng 2 2018
ta có : \(\left(x^2+5\right)\left(x^2-25\right)< 0\)
vì \(x^2+5\ge5>0\forall x\) \(\Rightarrow\left(x^2+5\right)\left(x^2-25\right)< 0\) \(\Leftrightarrow x^2-25< 0\)
\(\Leftrightarrow x^2< 25\Leftrightarrow-5< x< 5\)
vậy \(-5< x< 5\)
HN
2
27 tháng 10 2021
\(a,\Leftrightarrow x\left(x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\\ b,\Leftrightarrow\left(x+4-4\right)\left(x+4+4\right)=0\\ \Leftrightarrow x\left(x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\\ c,\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\)
27 tháng 10 2021
a) \(\Leftrightarrow x\left(x+9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\)
b) \(\Leftrightarrow x\left(x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)
c) \(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) \(\Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x=5\)
Y
2
5 tháng 2 2021
\(a,\left(-31\right).\left(x+7\right)=0\\ \Rightarrow x+7=0\\ \Rightarrow x=-7\\ b,\left(8-x\right).\left(x+13\right)=0\\ \Rightarrow\left[{}\begin{matrix}8-x=0\\x+13=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-13\end{matrix}\right.\\ c,\left(x^2-25\right)\left(3-x\right)=0\\ \Rightarrow\left(x-5\right)\left(x+5\right)\left(3-x\right)=0\\\Rightarrow \left[{}\begin{matrix}x-5=0\\x+5=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\\x=3\end{matrix}\right.\\ d,\left(x-3\right)\left(x^2+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x^2=-4\left(loại\right)\end{matrix}\right.\\ \Rightarrow x=3\)
11 tháng 1 2021
a) Ta có: \(7x^2-28=0\)
\(\Leftrightarrow7\left(x^2-4\right)=0\)
\(\Leftrightarrow7\left(x-2\right)\left(x+2\right)=0\)
mà 7>0
nên (x-2)(x+2)=0
hay \(\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-2\right\}\)
b) Ta có: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)
\(\Leftrightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)
mà \(\dfrac{2}{3}>0\)
nên x(x-2)(x+2)=0
hay \(\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-2;2\right\}\)
c) Ta có: \(2x\left(3x-5\right)-\left(5-3x\right)=0\)
\(\Leftrightarrow2x\left(3x-5\right)+\left(3x-5\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=5\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{5}{3};-\dfrac{1}{2}\right\}\)
d) Ta có: \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-1-5\right)\left(2x-1+5\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-2\right\}\)
23 tháng 2 2023
a: =>(2x-5x-1)(2x+5x+1)=0
=>(-3x-1)(7x+1)=0
=>x=-1/3 hoặc x=-1/7
b: =>(5x-5)^2-(x+2)^2=0
=>(5x-5-x-2)(5x-5+x+2)=0
=>(4x-7)(6x-3)=0
=>x=1/2 hoặc x=7/4
c: =>(x^2+4x-1-x^2+3x-2)(x^2+4x-1+x^2-3x+2)=0
=>(7x-3)(2x^2+x+1)=0
=>7x-3=0
=>x=3/7
Bài 1.
PT
2
a, ( x2 + 5 ) . ( x2 - 25 ) = 0
Vì \(x^2+5\ne0\forall x\Rightarrow\left(x^2+5\right)\left(x^2-25\right)=0\Leftrightarrow x^2-25=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x-5=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
Vậy x = -5; 5