Lời giải:

a) Để pt có 2 nghiệm phân biệt thì:

$\Delta'=16-4m^2>0\Leftrightarrow m^2< 4\Leftrightarrow -2< m< 2$

b) Để pt có 2 nghiệm phân biệt thì:

$\Delta'=(m-1)^2>0\Leftrightarrow m\neq 1$

\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11

e: Ta có: \(x^2-6x+y^2+4y+2=0\)

\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Dấu '=' xảy ra khi x=3 và y=-2

cho mik hỏi rằng là 3x2 + 4x = 0 hay  3x² + 4x = 0

ông ơi mấy bài này bấm máy tính là ra mà ông

a) \(3x^2+4x=0\Leftrightarrow\left(3x+4\right)x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+4=0\Leftrightarrow x=-\dfrac{4}{3}\end{matrix}\right.\)

   ➤\(x\in\left\{0;-\dfrac{4}{3}\right\}\)

b) \(-2x^2-8=0\Leftrightarrow-2x^2+\left(-2\right)\cdot4=0\)

                           \(\Leftrightarrow\left(x^2+4\right)\cdot\left(-2\right)=0\\ \Leftrightarrow x^2+4=0\\\Rightarrow x^2=\varnothing\Leftrightarrow x=\varnothing \) 

                          vì với mọi x, ta luôn đúng với: \(x^2\ge0\Leftrightarrow x^2+4\ge4>0\)

➤\(x=\varnothing\)

c)\(2x^2-7x^2+5=0\)

+) \(a+b+c=2+\left(-7\right)+5=7-7=0\)

Do đó, phương trình có 2 nghiệm sau:

\(x=1\) và \(x=\dfrac{5}{2}=2,5\)

➤\(x\in\left\{1;2,5\right\}\)

d) \(x^2-8x-48=0\)

+)\(\Delta=\left(-8\right)^2-4\cdot1\cdot\left(-48\right)=64+192=266>0\)

\(\Leftrightarrow\sqrt{\Delta}=\sqrt{266}\)

➢Do đó, ta có: \(\left[{}\begin{matrix}x=\dfrac{\sqrt{266}-\left(-8\right)}{2\cdot2}=\dfrac{\sqrt{266}+8}{4}\\x=\dfrac{-\sqrt{266}-\left(-8\right)}{2\cdot2}=\dfrac{8-\sqrt{266}}{4}\end{matrix}\right.\)

➤ \(x\in\left\{\dfrac{8+\sqrt{266}}{4};\dfrac{8-\sqrt{266}}{4}\right\}\)

a/ \(x^2-2.4x+16+y^2+2y+1+z^2=16\Leftrightarrow\left(x-4\right)^2+\left(y+1\right)^2+z^2=16\)

\(\Rightarrow\left\{{}\begin{matrix}I\left(4;-1;0\right)\\R=\sqrt{16}=4\end{matrix}\right.\)

b/ \(x^2+y^2+z^2+2x-y+5z-\dfrac{2}{3}=0\Leftrightarrow x^2+2x+1+y^2-2.\dfrac{1}{2}y+\dfrac{1}{4}+z^2+2.\dfrac{5}{2}z+\dfrac{25}{4}=\dfrac{2}{3}+1+\dfrac{1}{4}+\dfrac{25}{4}\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-\dfrac{1}{2}\right)^2+\left(z+\dfrac{5}{2}\right)^2=\dfrac{49}{6}\) \(\Rightarrow\left\{{}\begin{matrix}I\left(-1;\dfrac{1}{2};-\dfrac{5}{2}\right)\\R=\dfrac{7}{\sqrt{6}}\end{matrix}\right.\)

P/s: câu c bạn tự làm nốt ạ!

2a) pt <=> (x + 6)^2 = 0

<=> x = -6

b) pt <=> (4x - 1)^2 = 0

<=> x = 1/4

c) pt<=> (x + 1)^3 = 0

<=> x = -1

Bài 1:

a: Ta có: \(A=\left(4x+3y\right)^2+\left(4x-3y\right)^2\)

\(=16x^2+24xy+9y^2+16x^2-24xy+9y^2\)

\(=32x^2+18y^2\)

b: Ta có: \(B=\left(x-2\right)^3-\left(x+2\right)^3\)

\(=x^3-6x^2+12x-8-x^3-6x^2-12x-8\)

\(=-12x^2-24\)

Bài 2: 

a: Ta có: \(x^2+12x+36=0\)

\(\Leftrightarrow x+6=0\)

hay x=-6

b: Ta có: \(16x^2-8x+1=0\)

\(\Leftrightarrow4x-1=0\)

hay \(x=\dfrac{1}{4}\)

Bài 1: 

a: Ta có: \(A=\left(4x+3y\right)^2+\left(4x-3y\right)^2\)

\(=16x^2+24xy+9y^2+16x^2-24xy+9y^2\)

\(=32x^2+18y^2\)

b: Ta có: \(B=\left(x-2\right)^3-\left(x+2\right)^3\)

\(=x^3-6x^2+12x-8-x^3-6x^2-12x-8\)

\(=-12x^2-24\)

c: Ta có: \(C=\left(x+2y\right)^2+2\left(x+2y\right)\left(x-2y\right)+\left(x-2y\right)^2\)

\(=\left(x+2y+x-2y\right)^2\)

\(=4x^2\)

a) x² - 4 = 0

x² = 4

x = 2 hoặc x = -2

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

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

X + 5 = 0 hoặc 2x - 3 = 0

*) x + 5 = 0

x = -5

*) 2x - 3 = 0

2x = 3

x = 3/2

c) x³ - 6x² + 11x - 6 = 0

x³ - x² - 5x² + 5x + 6x - 6 = 0

(x³ - x²) - (5x² - 5x) + (6x - 6) = 0

x²(x - 1) - 5x(x - 1) + 6(x - 1) = 0

(x - 1)(x² - 5x + 6) = 0

(x - 1)(x² - 2x - 3x + 6) = 0

(x - 1)[(x² - 2x) - (3x - 6)] = 0

(x - 1)[x(x - 2) - 3(x - 2)] = 0

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

x - 1 = 0 hoặc x - 2 = 0 hoặc x - 3 = 0

*) x - 1 = 0

x = 1

*) x - 2 = 0

x = 2

*) x - 3 = 0

x = 3

Vậy x = 1; x = 2; x = 3

a) Ta có: \(x^3+x^2+x+1=0\)

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

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

mà \(x^2+1>0\forall x\)

nên x+1=0

hay x=-1

Vậy: S={-1}

b) Ta có: \(x^3-6x^2+11x-6=0\) 

\(\Leftrightarrow x^3-x^2-5x^2+5x+6x-6=0\)

\(\Leftrightarrow x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+6\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)

Vậy: S={1;2;3}

c) Ta có: \(x^3-x^2-21x+45=0\)

\(\Leftrightarrow x^3-3x^2+2x^2-6x-15x+45=0\)

\(\Leftrightarrow x^2\left(x-3\right)+2x\left(x-3\right)-15\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+2x-15\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+5x-3x-15\right)=0\)

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

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

Vậy: S={3;-5}

d) Ta có: \(x^4+2x^3-4x^2-5x-6=0\)

\(\Leftrightarrow x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6=0\)

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

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

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

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

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

mà \(x^2+x+1>0\forall x\)

nên (x-2)(x+3)=0

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

Vậy: S={2;-3}