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TT
1
NB
5 tháng 8 2020
\(\Leftrightarrow2x^8-2x^5+2x^2-2x+2=0\\ \Leftrightarrow x^8-2x^5+x^2+x^2-2x+1+x^2+1=0\\ \Leftrightarrow\left(x^4\right)^2-2x^4x+x^2+\left(x-1\right)^2+\left(x^4\right)^2+1=0\)
\(\Leftrightarrow\left(x^4-x\right)^2+\left(x-1\right)^2+\left(x^4\right)^2+1=0\) vô lí
⇒ vô nghiệm
LT
5 tháng 8 2020
Bài 9 : Tìm x, biết :
a, (x - 2)(x - 3) + (x - 2) - 1 = 0
\(\Leftrightarrow\left(x-2\right)\left(x-3+1\right)-1=0\)
\(\Leftrightarrow\left(x-2\right)^2-1=0\)
\(\Leftrightarrow\left(x-2+1\right)\left(x-2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy x ={1; 3}
b, (x + 2)2 - 2x(2x + 3) = (x + 1)2
\(\Leftrightarrow\left(x+2\right)^2-\left(x+1\right)^2-2x\left(2x+3\right)=0\)
\(\Leftrightarrow\left(x+2+x+1\right)\left(x+2-x-1\right)-2x\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3-2x\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(1-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\1-2x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy \(x=\left\{-\frac{3}{2};\frac{1}{2}\right\}\)
c, 6x3 + x2 = 2x
\(\Leftrightarrow6x^3+x^2-2x=0\)
\(\Leftrightarrow x\left(6x^2+x-2\right)=0\)
\(\Leftrightarrow x\left(6x^2+4x-3x-2\right)=0\)
\(\Leftrightarrow x\left[2x\left(3x+2\right)-\left(3x+2\right)\right]=0\)
\(\Leftrightarrow x\left(3x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+2=0\\2x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\frac{2}{3}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy \(x=\left\{0;-\frac{2}{3};\frac{1}{2}\right\}\)
LT
3 tháng 8 2020
Bài 1 : Tìm x,biết :
a, x2(x + 5) - 9x = 45
⇔ x2(x + 5) - 9x - 45 = 0
⇔ x2(x + 5) - 9(x + 5) = 0
⇔ (x + 5)(x2 - 9) = 0
⇔ (x + 5)(x - 3)(x + 3) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-3=0\\x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=3\\x=-3\end{matrix}\right.\)
Vậy x ={-5; 3; -3}
b, 9(5 - x) + x2 - 10x = -25
⇔ 45 - 9x + x2 - 10x + 25 = 0
⇔ x2 - 19x + 70 = 0
⇔ x2 - 14x - 5x + 70 = 0
⇔ (x2 - 5x) - (14x - 70) = 0
⇔ x(x - 5) - 14(x - 5) = 0
⇔ (x - 5)(x - 14) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-14=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=14\end{matrix}\right.\)
Vậy x ={5; 14}
XU
3 tháng 8 2020
a, x2( x+5 ) - 9x = 45
x3 + 5x2 - 9x - 45 = 0
x2( x+5 ) - 9( x+5) = 0
(x2 - 9)(x + 5) = 0
(x + 3)(x - 3)(x + 5) = 0
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=3\\x=-5\end{matrix}\right.\)
b, 9( 5-x ) + x2 -10x = -25
45 - 9x + x2 - 10x + 25 = 0
x2 - 19x + 70 = 0
x2 - 14x - 5x + 70 = 0
x( x-14 ) - 5( x-14) = 0
(x - 5)(x - 14) = 0
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=14\end{matrix}\right.\)
5 tháng 8 2020
a, Ta có : \(x^5+x+1\)
\(=x^5-x^2+x^2+x+1\)
\(=x^2\left(x^3-1\right)+x^2+x+1\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^2\left(x-1\right)+1\right)\)
b, Ta có : \(x^8+x^4+1\)
\(=\left(x^4\right)^2+2x^4+1-x^4\)
\(=\left(x^4+1\right)^2-\left(x^2\right)^2\)
\(=\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\)
17 tháng 8 2021
\(a)\)
\(\frac{1}{x+1}-\frac{x-1}{x}=\frac{3x+1}{x\left(x+1\right)}\)
\(\Leftrightarrow x-x^2+1=3x+1\)
\(\Leftrightarrow x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(b)\)
\(\frac{\left(x+2\right)^2}{2x-3}-\frac{1}{1}=\frac{x^2+10}{2x-3}\)
\(\Leftrightarrow x^2+4x+4-2x-3=x^2+10\)
\(\Leftrightarrow x^2+2x+1=x^2+10\)
\(\Leftrightarrow2x-9=0\)
\(\Leftrightarrow2x=9\)
\(\Leftrightarrow x=\frac{2}{9}\)
Ta có: \(\dfrac{x}{2\cdot3}+\dfrac{x}{3\cdot4}+...+\dfrac{x}{99\cdot100}=-1\)
\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=-1\)
\(\Leftrightarrow x\cdot\dfrac{49}{100}=-1\)
hay \(x=-\dfrac{100}{49}\)
$\dfrac{x}{2.3}+\dfrac{x}{3.4}+...+\dfrac{x}{99.100}=1$
`<=>x/2 - x/3 +x/3-x/4+...+x/(99)-x/(100)=1`
`<=>x/2-x/(100)=1`
`<=>(50x)/(100)-x/(100)=(100)/(100)`
`<=>50x-x=100`
`<=>49x=100`
`<=>x=(100)/(49)`
Vậy `x=(100)/(49)`