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

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

1 .      \(\sqrt{x^4-2x^2+1}=x-1\)

<=>  \(\sqrt{\left(x^2-1\right)^2}=x-1\)

<=> \(x^2-1=x-1\)

<=> \(x^2-x=0\)(vậy pt vô nghiệm)

1,\(\sqrt{\left(x^2-1\right)^2}=x-1\)

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

<=>\(\orbr{\begin{cases}x1=0\\x2=1\end{cases}}\)

1,\(\sqrt{\left(x^2+4\right)}=5-\sqrt{\left(x^2+10\right)}\)

<=>\(x^2+4=25-10\sqrt{x^2+10}+x^2+10\)

<=>x^2 = -0.39 vô lý  => vô nhiệm 

a: ĐKXĐ: \(\left\{{}\begin{matrix}-2< =x< =2\\x< >0\end{matrix}\right.\)

c: \(f\left(-x\right)=\dfrac{\sqrt{2-\left(-x\right)}-\sqrt{2+\left(-x\right)}}{-x}=\dfrac{\sqrt{2+x}-\sqrt{2-x}}{-x}=\dfrac{\sqrt{2-x}-\sqrt{2+x}}{x}=f\left(x\right)\)

a: f(1)=-1,5

f(2)=-6

f(3)=-13,5

=>f(1)>f(2)>f(3)

b: \(f\left(-3\right)=-1,5\cdot9=-13,5\)

f(-2)=-1,5x4=-6

f(-1)=-1,5x1=-1,5

=>f(-3)<f(-2)<f(-1)

c: Hàm số này đồng biến khi x<0 và nghịch biến khi x>0