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

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

Giúp vs m.n ơi mai mình kt òi

a) Với m=0

=> pt <=> \(x^2+5x=0\)

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

b) \(x^2+5x+3m=0\)

\(\Delta=25-12m\)

Để phương trình có 2 nghiệm phân biệt 

\(\Leftrightarrow\Delta>0\)

\(\Leftrightarrow25-12m>0\)

\(\Leftrightarrow m< \dfrac{25}{12}\)

2x³-7x²+5x=0

<=>x*(2x²-7x+5)=0

<=>x*(2x²-2x-5x+5)=0

<=>x*[(2x²-2x)-(5x-5)]=0

<=>x*[2x*(x-1)-5*(x-1)]=0

<=>x*(x-1)*(2x-5)=0

1, Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=-5\\x_1x_2=-6\end{matrix}\right.\)

\(A=\left(x_1-2x_2\right)\left(2x_1-x_2\right)\\ =2x_1^2-4x_1x_2-x_1x_2+2x_1^2\\ =2\left(x_1^2+x_2^2\right)-5x_1x_2\\ =2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-5x_1x_2\\ =2\left(-5\right)^2-4.\left(-6\right)-5.\left(-6\right)\\ =104\)

2, Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=5\\x_1x_2=-3\end{matrix}\right.\)

\(B=x_1^3x_2+x_1x_2^3\\ =x_1x_2\left(x_1^2+x_2^2\right)\\ =\left(-3\right)\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\\ =\left(-3\right)\left[5^2-2\left(-3\right)\right]\\ =-93\)

Đặt `x^2=t (t>=0)`, có:

`2t^2+5t+2=0`

`\Delta = 5^2-4.2.2=9>0`

`=>` PT có 2 nghiệm:

`t_1=-1/2 (KTM)`

`t_2=-2 (KTM)`

Vậy PTVN.

x₁=\(\dfrac{1}{2}\)

x₂=-3

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

\(\Leftrightarrow2x^2-x+6x-3=0\)

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

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

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

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

\(2x^2+5x-3=0\\ \Leftrightarrow2x^2+6x-x-3=0\\ \Leftrightarrow2x\left(x+3\right)-\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(x^2-5x+7=0\\ \Leftrightarrow\left(x^2-2.\dfrac{5}{2}x+\dfrac{25}{4}\right)+\dfrac{3}{4}=0\\ \Leftrightarrow\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\)

\(\Leftrightarrow x\left(5x^2-7x+5x-7\right)=0\\ \Leftrightarrow x\left[5x\left(x+1\right)+7\left(x+1\right)\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\5x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=\dfrac{7}{5}\end{matrix}\right.\)

\(\Leftrightarrow5x^3+5x^2-7x^2-7x=0\)

\(\Leftrightarrow5x^2\left(x+1\right)-7x\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(5x^2-7x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\5x^2-7x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=\dfrac{7}{5}\end{matrix}\right.\)

6x2y + 2y3+ 35 = 0 (1)

5x2+ 5y2+ 2xy + 5x + 13y = 0 (2)
Lấy phương trình (1) cộng với 3 lần phương trình (2) theo vế ta được:(6y + 15)x2+ 3(2y + 5)x + 2y3+ 15y2+ 39y + 35 = 0⇔ (2y + 5)*(3(x +12)mũ 2+(y +5/2)mũ 2)

=0 ⇔[y = −5/2 ;x = −1/2, y = −5/2
       
.Lần lượt thế vào phương trình (1) ta được:(1/2;−5/2);(−1/2;−5/2)là nghiệm của hệ.

 

Có người làm rồi nhé

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

\(x^2-\frac{5}{2}x+1=0\)

\(x^2+2.\frac{5}{4}x+\frac{25}{16}-\frac{25}{16}+1=0\)

\(\left(x+\frac{5}{4}\right)^2-\frac{9}{16}=0\)

\(\left(x+\frac{5}{4}\right)^2-\left(\frac{3}{4}\right)^2=0\)

\(\left(x+\frac{5}{4}-\frac{3}{4}\right)\left(x+\frac{5}{4}+\frac{3}{4}\right)=0\)

\(\left(x+\frac{1}{2}\right)\left(x+2\right)=0\)

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

2x²-5x+2

\(\Leftrightarrow2x^2-4x-x+2\)

\(\Leftrightarrow\left(2x^2-4x\right)-\left(x-2\right)\)

\(\Leftrightarrow2x\left(x-2\right)-\left(x-2\right)\)

\(\Leftrightarrow\left(2x-1\right)\left(x-2\right)\)

\(\Leftrightarrow2x-1=0hoacx-2=0\)

Nếu 2x-1=0

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

Nếu x-2=0 thì 

\(\Leftrightarrow x=2\)

Vậy pt đãcho có tập nghiệm là:S=\(\hept{\begin{cases}\\\end{cases}2;\frac{1}{2}}\)