`1/9(x-3)^2-1/25(x+5)^2=0`

`<=>(1/3x-1)^2-(1/5x+1)^2=0`

`<=>(1/3x-1-1/5x-1)(1/3x-1+1/5x+1)=0`

`<=>(2/15x-2). 8/15x=0`

`<=>2/15x-2=0` hoặc `8/15x=0`

`<=>x=15`         hoặc `x=0`

Vậy `S=`{`15;0`}

Ta có : \(x^2+x+4=x^2+x+\frac{1}{4}+\frac{15}{4}=\left(x+\frac{1}{2}\right)^2+\frac{15}{4}>0\left(\forall x\right)\)

+) \(\left(x-1\right)\left(x^2+x+4\right)=0\)

\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

\(\left(x-1\right)\left(x^2+x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2+x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x^2+x=-4\end{cases}}\)

+) x² + x = - 4

<=> ( x + 1/2 )² = - 4 + 1/4 = -15/4

Mà ( x + 1/2 )² lớn hơn hoặc bằng 0 với mọi x

=> x² + x + 4 = 0 ktm

Vậy pt = 0 <=> x = 1

a) 2x²-4x-x+2=0

=> 2x(x-2)-(x-2)=0

=> (2x-1)(x-2)=0

=> \(\left[{}\begin{matrix}2x-1=0\\x-2=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

b) 3x²-12x+5x-20=0

=> 3x(x-4)+5.(x-4)=0

=> (x-4)(3x+5)=0

=> \(\left[{}\begin{matrix}x-4=0\\3x+5=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=4\\x=-\dfrac{5}{3}\end{matrix}\right.\)

c)x³+2x²-x²-2x+2x+4=0

=> x²(x+2)-x(x+2)+2(x+2)=0

=>(x²-x+2)(x+2)=0

=> x=-2( vi x²-x+2>0)

d) x³-x²-4x²+4x+4x-4=0

=> x²(x-1)-4x(x-1)+4(x-1)=0

=>(x-1)(x²-4x+4)=0

=> \(\left[{}\begin{matrix}x-1=0\\x^2-4x+4=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

2x²-5x+2=0

⇔2x²-x-4x+2=0

⇔x(2x-1)-2(2x-1)=0

⇔(x-2)(2x-1)=0

⇔\(\left[{}\begin{matrix}x-2=0\\2x-1=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=2\\2x=1\Leftrightarrow x=\dfrac{1}{2}\end{matrix}\right.\)

sậy S=\(\left\{2;\dfrac{1}{2}\right\}\)

x³+x²+4=0

⇔x³+2x²-x²-2x+2x+4=0

⇔(x³+2x²)-(x²+2x)+(2x+4)=0

⇔x²(x+2)-x(x+2)+2(x+2)=0

⇔(x+2)(x²-x+2)=0

⇔x+2=0 và x²-x+2=0

⇔x=-2 và \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\)(vô lý)

vậy S={-2}

\(x^4-2x^3+3x^2-2x+1=0\)

Chia cả hai vé cho \(x^2\)

\(\Leftrightarrow x^2-2x+3-\dfrac{2}{x}+\dfrac{1}{x^2}\)

\(\Leftrightarrow x^2+2+\dfrac{1}{x^2}-2\left(x+\dfrac{1}{x}\right)+1=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2-2\left(x+\dfrac{1}{x}\right)+1=0\)

Đặt x+1/x = a, ta có:

\(a^2-2a+1=0\)

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

\(\Leftrightarrow a=1\)

\(\Leftrightarrow x+\dfrac{1}{x}=1\)

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

\(\Leftrightarrow x^2-x+1=0\)

\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\)

Do \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+3>0\)

Do đó phương trình vô nghiệm

a) \(x^4+2x^3-2x^2+2x-3=0\)

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

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

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

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

Giải (1) : \(\left(x+3\right)\left(x^2+1\right)=0\)

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

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

Mà \(x^2\)>0

\(\Rightarrow\)pt vô nghiệm

Vậy \(x\in\left(-3;1\right)\)

\(\)

Câu 8 A

Câu 7 C

Câu 6D

5D

4D

2C

1A

e camon nhiều

\(a,=\left(x+8-x+2\right)^2=10^2=100\\ b,=x^2\left(x^2-16\right)-\left(x^4-1\right)=x^4-16x^2-x^4+1=1-16x^2\\ c,=x^3+1-x^3+1=2\)