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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

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 \(x=\left\{0;-\frac{2}{3};\frac{1}{2}\right\}\)

\(1,\)

\(2x\left(x-3\right)-\left(3-x\right)=0\)

\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)

\(2,\)

\(3x\left(x+5\right)-6\left(x+5\right)=0\)

\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)

\(3,\)

\(x^4-x^2=0\)

\(\Leftrightarrow x^2\left(x^2-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

\(4,\)

\(x^2-2x=0\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(5,\)

\(x\left(x+6\right)-10\left(x-6\right)=0\)

\(\Leftrightarrow x^2+6x-10x+60=0\)

\(\Leftrightarrow x^2-4x+60=0\)

\(\Leftrightarrow x^2-4x+4+56=0\)

\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)

=> Phương trình vô nghiệm

2 tháng 8 2020

a/ \(x^2\left(x-5\right)+5-x=0\)

\(\Leftrightarrow x^2\left(x-5\right)-\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=5\end{matrix}\right.\)

Vậy...

b/ \(3x^4-9x^3=-9x^2+27x\)

\(\Leftrightarrow3x^4-9x^3+9x^2-27x=0\)

\(\Leftrightarrow3x^3\left(x-3\right)+9x\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(3x^3+9x\right)=0\)

\(\Leftrightarrow3x\left(x-3\right)\left(x^2+3\right)=0\)

\(x^2+3>0\forall x\)

\(\Leftrightarrow3x\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Vậy..

c/ \(x^2\left(x+8\right)+x^2=-8x\)

\(\Leftrightarrow x^2\left(x+8\right)+x^2+8x=0\)

\(\Leftrightarrow x^2\left(x+8\right)+x\left(x+8\right)=0\)

\(\Leftrightarrow x\left(x+8\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+8=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\\x=-1\end{matrix}\right.\)

Vậy...

d/ \(\left(x+3\right)\left(x^2-3x+5\right)=x^2+3x\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+5\right)-x\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left[\left(x-2\right)^2+1\right]=0\)

\(\left(x-2\right)^2+1>0\forall x\)

\(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)

Vậy..

2 tháng 8 2020

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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}

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

\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=14\end{matrix}\right.\)

\(a)\)

\(\frac{1}{x+1}-\frac{x-1}{x}=\frac{3x+1}{x\left(x+1\right)}\)

\(\Leftrightarrow x-x^2+1=3x+1\)

\(\Leftrightarrow x^2-2x=0\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(b)\)

\(\frac{\left(x+2\right)^2}{2x-3}-\frac{1}{1}=\frac{x^2+10}{2x-3}\)

\(\Leftrightarrow x^2+4x+4-2x-3=x^2+10\)

\(\Leftrightarrow x^2+2x+1=x^2+10\)

\(\Leftrightarrow2x-9=0\)

\(\Leftrightarrow2x=9\)

\(\Leftrightarrow x=\frac{2}{9}\)

21 tháng 7 2017

\(x^3+x=0\)

\(\Leftrightarrow x\left(x^2+1\right)=0\)

thấy :x2+1>0 loại

suy ra x=0