bai 1

1 thay k=0 vao pt ta co 4x^2-25+0^2+4*0*x=0

<=>(2x)^2-5^2=0

<=>(2x+5)*(2x-5)=0

<=>2x+5=0 hoăc 2x-5 =0 tiếp tục giải ý 2 tương tự

Câu 1: 

a: x+2=0

nên x=-2

b: (x-3)(2x+8)=0

=>x-3=0 hoặc 2x+8=0

=>x=3 hoặc x=-4

a . 

x + 2 = 0

=> x = 0 - 2 = -2 

b ) .

<=> x - 3 = 0 ; 2x + 8 = 0

= > x = 3 ; x = -8/2 = -4 

c ) .

ĐKXĐ của pt : x - 5 khác 0 = > ddk : x khác 5

1)

a) \(x+2=0\)

\(\Leftrightarrow x=-2\)

Vậy S = {\(-2\)}

b) \(\left(x-3\right)\left(2x+8\right)=0\)

\(\Leftrightarrow x-3=0\) hoặc \(2x+8=0\)

*) \(x-3=0\)

\(\Leftrightarrow x=3\)

*) \(2x+8=0\)

\(\Leftrightarrow2x=-8\)

\(\Leftrightarrow x=-4\)

Vậy S = \(\left\{-4;3\right\}\)

2) ĐKXĐ:

\(x-5\ne0\Leftrightarrow x\ne5\)

b,

+ Với \(x=0\) \(\Rightarrow PTVN\)

+ Với \(x\ne0\), chia cả 2 vế cho \(x^2\) :

\(PT\Leftrightarrow x^2-16x+46+\frac{144}{x}+\frac{81}{x^2}=0\)

\(\Leftrightarrow\left(x^2+\frac{81}{x^2}\right)-16\left(x-\frac{9}{x}\right)+46=0\)

Đặt \(x-\frac{9}{x}=t\Rightarrow t^2=x^2+\frac{81}{x^2}-18\)

\(\Leftrightarrow t^2+18-16t+46=0\)

\(\Leftrightarrow t^2-16t+64=0\Rightarrow t=8\)

\(\Leftrightarrow x-\frac{9}{x}=8\Leftrightarrow x^2-8x-9=0\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=9\end{matrix}\right.\) (t/m)

a: =>7-x=0

hay x=7

b: \(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(x+5\right)\left(3x-8\right)=0\)

hay \(x\in\left\{\sqrt{2};-\sqrt{2};-5;\dfrac{8}{3}\right\}\)

a: =>-x+7=0

hay x=7

b: \(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(x+5\right)\left(3x-8\right)=0\)

hay \(x\in\left\{\sqrt{2};-\sqrt{2};-5;\dfrac{8}{3}\right\}\)

ta có : x^5+2x^4+3x^3+3x^2+2x+1=0

\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0

\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0

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

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

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

\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0

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

VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)

\(\Rightarrow\)x+1=0

\(\Rightarrow\)x=-1

CÒN CÂU B TỰ LÀM (02042006)

b: x^4+3x^3-2x^2+x-3=0

=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0

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

=>x-1=0

=>x=1

Bài 2 :

a, Ta có : \(\left(x+4\right)\left(x-1\right)=0\)

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

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

b, Ta có : \(\left(3x-2\right)\left(4x-7\right)=0\)

=> \(\left[{}\begin{matrix}3x-2=0\\4x-7=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}3x=2\\4x=7\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{7}{4}\end{matrix}\right.\)

c, Ta có : \(\left(x+5\right)\left(x^2+1\right)=0\)

=> \(\left[{}\begin{matrix}x+5=0\\x^2+1=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-5\\x^2+1=0\left(VL\right)\end{matrix}\right.\)

d, Ta có : \(x\left(x-1\right)\left(x^2+4\right)=0\)

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

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

e, Ta có : \(\left(3x+2\right)\left(x+\frac{1}{2}\right)=0\)

=> \(\left[{}\begin{matrix}3x+2=0\\x+\frac{1}{2}=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-\frac{2}{3}\\x=-\frac{1}{2}\end{matrix}\right.\)

f, Ta có : \(\left(x+2\right)\left(x+3\right)\left(x^2+7\right)=0\)

=> \(\left[{}\begin{matrix}x+2=0\\x-3=0\\x^2+7=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-2\\x=3\\x^2+7=0\left(VL\right)\end{matrix}\right.\)

Bài 1 :

a, Ta có : \(1-\frac{x+3}{4}-\frac{x-2}{6}=0\)

=> \(\frac{12}{12}-\frac{3\left(x+3\right)}{12}-\frac{2\left(x-2\right)}{12}=0\)

=> \(12-3\left(x+3\right)-2\left(x-2\right)=0\)

=> \(12-3x-9-2x+4=0\)

=> \(-5x=-7\)

=> \(x=\frac{7}{5}\)

3) \(x^2-7x+6=0\)

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

\(\Leftrightarrow x\left(x-6\right)-\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)

S=\(\left\{6;1\right\}\)

\(\)