\(4x^2+8x-5=0\)

\(\Leftrightarrow4x^2-2x+10x-5=0\)

\(\Leftrightarrow2x\left(2x-1\right)+5\left(2x-1\right)=0\)

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

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

Vậy: Phương trình có tập nghiệm \(S=\left\{\dfrac{1}{2};-\dfrac{5}{2}\right\}\)

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

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

(4x - 3)² - (2x + 1)² = 0

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

\(\Leftrightarrow\) (2x - 4)(6x - 2) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\)

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

Vậy ...

3x - 12 - 5x(x - 4) = 0

\(\Leftrightarrow\) 3x - 12 - 5x² + 20x = 0

\(\Leftrightarrow\) -5x² + 23x - 12 = 0

\(\Leftrightarrow\) 5x² - 23x + 12 = 0

\(\Leftrightarrow\) 5x² - 20x - 3x + 12 = 0

\(\Leftrightarrow\) 5x(x - 4) - 3(x - 4) = 0

\(\Leftrightarrow\) (x - 4)(5x - 3) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-4=0\\5x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy ...

(8x + 2)(x² + 5)(x² - 4) = 0

\(\Leftrightarrow\) (8x + 2)(x² + 5)(x - 2)(x + 2) = 0

Vì x² \(\ge\) 0 \(\forall\) x nên x² + 5 > 0 \(\forall\) x

\(\Rightarrow\) (8x + 2)(x - 2)(x + 2) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}8x+2=0\\x-2=0\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=2\\x=-2\end{matrix}\right.\)

Vậy ...

Chúc bn học tốt!

a) Ta có: \(\left(4x-3\right)^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left(4x-3-2x-1\right)\left(4x-3+2x+1\right)=0\)

\(\Leftrightarrow\left(2x-4\right)\left(6x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{2;\dfrac{1}{3}\right\}\)

b) Ta có: \(3x-12-5x\left(x-4\right)=0\)

\(\Leftrightarrow3\left(x-4\right)-5x\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(3-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{4;\dfrac{3}{5}\right\}\)

c) Ta có: \(\left(8x+2\right)\left(x^2+5\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow2\left(4x+1\right)\left(x^2+5\right)\left(x-2\right)\left(x+2\right)=0\)

mà \(2>0\)

và \(x^2+5>0\forall x\)

nên \(\left(4x+1\right)\left(x-2\right)\left(x+2\right)=0\)

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

Vậy: \(S=\left\{-\dfrac{1}{4};2;-2\right\}\)

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

\(\Leftrightarrow x^2\left(x^2-4\right)+8\left(x+2\right)=0\)

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x^3-2x^2+8=0\end{array}\right.\)

Tới đây tự giải nhé :)

Đầu tiên ta phân tích : \(x^4+4=\left(x^4+4x^2+4\right)-4x^2=\left(x^2+2\right)^2-\left(2x\right)^2\)

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

Suy ra pt : \(\left(x^2-2x+2\right)\left(x^2+2x+2\right)-4x\left(x-2\right)=0\)

Nhận thấy x = 0 không là nghiệm của pt, do đó chia cả hai vế của pt cho \(x^4\ne0\) được : 

\(\left(1-\frac{2}{x}+\frac{2}{x^2}\right)\left(1+\frac{2}{x}+\frac{2}{x^2}\right)-4\left(\frac{1}{x^2}-\frac{2}{x^3}\right)=0\)

Đặt \(t=\frac{2}{x}\) , pt trở thành : \(\left(1-2t+2t^2\right)\left(1+2t+2t^2\right)-4\left(t^2-2t^3\right)=0\)

Tới đây thử giải pt với ẩn t xem có đc k

chia 2 vế cho x² đi xem nào

Lời giải:

$-4x^2-8x+11=0$

$\Leftrightarrow 4x^2+8x-11=0$

$\Leftrightarrow (2x)^2-2.2x.2+2^2-15=0$

$\Leftrightarrow (2x-2)^2-15=0$

$\Leftrightarrow (2x-2-\sqrt{15})(2x-2+\sqrt{15})=0$

\(\Rightarrow \left[\begin{matrix} 2x-2-\sqrt{15}=0\\ 2x-2+\sqrt{15}=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{2+\sqrt{15}}{2}\\ x=\frac{2-\sqrt{15}}{2}\end{matrix}\right.\)

\(\left(x^3-x^2\right)-4x^2+8x-4=0\)

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

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

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

\(\Leftrightarrow x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-4x-4\right)=0\)

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

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

Vậy......

\(\left(x^3-x^2\right)-ã^2+8x-4=0\)

\(< =>x^3-x^2-4x^2+8x-4\)

\(< =>x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

\(< =>\left(x-1\right)\left(x^2-4x+4=0\right)\)

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

Lời giải:

1. 

PT $\Leftrightarrow (x^2+5x)^2+2(x^2+5x)-24=0$

$\Leftrightarrow t^2+2t-24=0$ (đặt $x^2+5x=t$)

$\Leftrightarrow (t-4)(t+6)=0$

$\Rightarrow t-4=0$ hoặc $t+6=0$

Nếu $t-4=0\Leftrightarrow x^2+5x-4=0$

$\Leftrightarrow x=\frac{-5\pm \sqrt{41}}{2}$

Nếu $t+6=0$

$\Leftrightarrow x^2+5x+6=0$

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

2.

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

$\Leftrightarrow t^2+2t-3=0$ (đặt $x^2-4x+1=t$)

$\Leftrightarrow (t-1)(t+3)=0$

$\Rightarrow t-1=0$ hoặc $t+3=0$

Nếu $t-1=0\Leftrightarrow x^2-4x=0\Leftrightarrow x(x-4)=0$

$\Rightarrow x=0$ hoặc $x=4$

Nếu $t+3=0\Leftrightarrow x^2-4x+4=0$

$\Leftrightarrow (x-2)^2=0\Leftrightarrow x=2$

\(a,x^2-10x-39=0\)

\(\Leftrightarrow x^2-10x-39+64=64\)

\(\Leftrightarrow x^2-10x+25=64\)

\(\Leftrightarrow\left(x-5\right)^2=64\)

làm nốt

\(x^2-10x-39=0\Leftrightarrow x^2-13x+3x-39=0\Leftrightarrow x\left(x-13\right)+3\left(x-13\right)=0\)

\(\Leftrightarrow\left(x-13\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=13\\x=-3\end{cases}}\)

\(b,\frac{x^2}{x^3-9}=\frac{1}{x+3}\)

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

\(\Leftrightarrow x^3+3x^2=x^3-9\)

\(\Leftrightarrow3x^2=-9\left(VL\right)\)

Đặt \(\sqrt{x^2+4x+7}=t>0\), ta có pt sau:

\(2\left(t^2+3\right)-7t=0\)

⇔ \(t^2-7t+6=0\Leftrightarrow\left(t-2\right)\left(2t-3\right)=0\)

⇔\(\left[{}\begin{matrix}t=2\\t=\frac{3}{2}\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x^2+4x+7=4\\x^2+4x+7=\frac{9}{4}\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\\x=\frac{\pm\sqrt{79}-4}{2}\end{matrix}\right.\)

Vậy ...