a) đặc \(x^2=t\left(t\ge0\right)\)

pt \(\Leftrightarrow\) \(t^2-8t-9=0\)

\(\Delta'=\left(-4\right)^2-1\left(-9\right)\) = \(16+9=25>0\)

\(\Rightarrow\) phương trình có 2 nghiệm phân biệt

\(t_1=\dfrac{4+\sqrt{25}}{1}=9\left(tmđk\right)\)

\(t_2=\dfrac{4-\sqrt{25}}{1}=-1\left(loại\right)\)

\(t=x^2=9\) \(\Leftrightarrow\) \(x=\pm9\)

vậy \(x=\pm9\)

a: x-2y=3

=>2y=x-3

=>\(y=\frac{x-3}{2}\)

Vậy: \(\begin{cases}x\in R\\ y=\frac{x-3}{2}\end{cases}\)

b: 5x(2x-3)=0

=>x(2x-3)=0

=>\(\left[\begin{array}{l}x=0\\ 2x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=\frac32\end{array}\right.\)

c: \(\frac{2}{x}=1\) (ĐKXĐ: x<>0)

=>\(x=\frac22=1\) (nhận)

d: 2x+1>0

=>2x>-1

=>\(x>-\frac12\)

không chắc nhé 

a) \(x^2-6x+6=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{6+2\sqrt{3}}{2}\\x=\frac{6-2\sqrt{3}}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=3+\sqrt{3}\\x=3-\sqrt{3}\end{cases}}\)

I my va li it so ceut

`Answer:`

a) \(\left(\sqrt{2}+1\right)x-\sqrt{2}=2\)

\(\Leftrightarrow\left(\sqrt{2}+1\right)x=2+\sqrt{2}\)

\(\Leftrightarrow x=\frac{2+\sqrt{2}}{\sqrt{2}+1}\)

\(\Leftrightarrow x=\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)

\(\Leftrightarrow x=\sqrt{2}\)

b) \(x^4+x^2-6=0\)

\(\Leftrightarrow x^4+3x^2-2x^2-6=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2-2=0\\x^2+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\pm\sqrt{2}\\x^2=-3\text{(Vô lý)}\end{cases}}}\)

374

a: \(5x^2-8x=0\)

=>x(5x-8)=0

=>\(\left[\begin{array}{l}x=0\\ 5x-8=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ 5x=8\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=\frac85\end{array}\right.\)

b: \(-3x^2-6x=0\)

=>-3x(x+2)=0

=>x(x+2)=0

=>\(\left[\begin{array}{l}x=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-2\end{array}\right.\)

c: 2x(x-3)=x-3

=>2x(x-3)-(x-3)=0

=>(x-3)(2x-1)=0

=>\(\left[\begin{array}{l}x-3=0\\ 2x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=\frac12\end{array}\right.\)

d: 2x(x-3)+5x-15=0

=>2x(x-3)+5(x-3)=0

=>(x-3)(2x+5)=0

=>\(\left[\begin{array}{l}x-3=0\\ 2x+5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-\frac52\end{array}\right.\)

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

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

=>2*2x=0

=>4x=0

=>x=0

f: \(\left(x-2\right)^2=\left(3x+5\right)^2\)

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

=>(3x+5+x-2)(3x+5-x+2)=0

=>(4x+3)(2x+7)=0

=>\(\left[\begin{array}{l}4x+3=0\\ 2x+7=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac34\\ x=-\frac72\end{array}\right.\)

g: \(\left(6-9x\right)^2=\left(5x-7\right)^2\)

=>\(\left(9x-6\right)^2-\left(5x-7\right)^2=0\)

=>(9x-6-5x+7)(9x-6+5x-7)=0

=>(4x+1)(14x-13)=0

=>\(\left[\begin{array}{l}4x+1=0\\ 14x-13=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac14\\ x=\frac{13}{14}\end{array}\right.\)

h: \(\left(x+1\right)^2\cdot\left(x+2\right)=0\)

=>\(\left[\begin{array}{l}x+1=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-1\\ x=-2\end{array}\right.\)

i: \(\left(3x-1\right)\cdot\left(3-x\right)^2=0\)

=>\(\left[\begin{array}{l}3x-1=0\\ \left(3-x\right)^2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=1\\ 3-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac13\\ x=3\end{array}\right.\)

\(x^2-6x+9=0\)     (1)

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

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

Vậy tập nghiệm của phương trình (1) là \(S=\left\{3\right\}\)

\(x^3-6x^2+11x-6=0\)

\(\Leftrightarrow\left(x^3-3x^2\right)-\left(3x^2-9x\right)+\left(2x-6\right)=0\)

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

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

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

\(\Leftrightarrow\)\(x=3\)

hoặc \(x=1\)

hoặc \(x=2\)

Vậy tập nghiệm của phương trình (2) là \(S=\left\{1;2;3\right\}\)

Mà 2 phương trình trên có 1 nghiệm chung

\(\Rightarrow\)Tập nghiệm của 2 phương trình là \(S=\left\{3\right\}\)

a) \(\sqrt{x^2-6x+9}=3\)

⇔ \(\sqrt{\left(x-3\right)^2}=3\)

⇔ \(\left|x-3\right|=3\)

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

b) \(\sqrt{x^2-8x+16}=x+2\)

⇔ \(\sqrt{\left(x-4\right)^2}=x+2\)

⇔ \(\left|x-4\right|=x+2\)

⇔ \(\orbr{\begin{cases}x-4=x+2\left(x\ge4\right)\\4-x=x+2\left(x< 4\right)\end{cases}\Leftrightarrow}x=1\)

c) \(\sqrt{x^2+6x+9}=3x-6\)

⇔ \(\sqrt{\left(x+3\right)^2}=3x-6\)

⇔ \(\left|x-3\right|=3x-6\)

⇔ \(\orbr{\begin{cases}x-3=3x-6\left(x\ge3\right)\\3-x=3x-6\left(x< 3\right)\end{cases}}\Leftrightarrow x=\frac{9}{4}\)

d) \(\sqrt{x^2-4x+4}-2x+5=0\)

⇔ \(\sqrt{\left(x-2\right)^2}-2x+5=0\)

⇔ \(\left|x-2\right|-2x+5=0\)

⇔ \(\orbr{\begin{cases}x-2-2x+5=0\left(x\ge2\right)\\2-x-2x+5=0\left(x< 2\right)\end{cases}}\Leftrightarrow x=3\)

Đặt: \(x^2=t\)

Sao đó giải như pt bậc 2 bình thường 

cops mạng đâu thế :((