Phương trình tích

a) Ta có: \(\left(2x+7\right)^2=\left(x+3\right)^2\)

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

\(\Leftrightarrow\left(2x+7-x-3\right)\left(2x+7+x+3\right)=0\)

\(\Leftrightarrow\left(x+4\right)\cdot\left(3x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\3x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\3x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-\dfrac{10}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-4;-\dfrac{10}{3}\right\}\)

b) Ta có: \(\left(4x+14\right)^2=\left(7x+2\right)^2\)

\(\Leftrightarrow\left(4x+14\right)^2-\left(7x+2\right)^2=0\)

\(\Leftrightarrow\left(4x+14-7x-2\right)\left(4x+14+7x+2\right)=0\)

\(\Leftrightarrow\left(-3x+12\right)\left(11x+16\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-3x+12=0\\11x+16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=-12\\11x=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{16}{11}\end{matrix}\right.\)Vậy: \(S=\left\{4;-\dfrac{16}{11}\right\}\)

(2x+7)²=(x+3)²

=>(2x+7)²-(x+3)²=0

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

=>(x-4)(3x+10)=0

=>x-4=0 hoặc 3x+10=0

TH1:x-4=0=>x=4

TH2:3x+10=0=>x=-10/3

(4x+14)²=(7x+2)²

(4x+14)²-(7x+2)²=0

(4x+14-7x-2)(4x+14+7x+2)=0

(-3x+12)(11x+16)=0

TH1:-3x+12=0=>x=4

TH2:11x+16=0=>x=-16/11

Đề bài là gì bạn nhỉ ?

\(PT\Leftrightarrow\left(x^2+2\right)^2+5\left|x^2-2\right|+4=0\).

VT luôn lớn hơn 0 nên pt vô nghiệm.

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

\(\Leftrightarrow x^2-\frac{1}{3}x-\frac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\frac{1}{6}+\frac{1}{36}-\frac{13}{36}=0\)

\(\Leftrightarrow\left(x-\frac{1}{6}\right)^2=\frac{13}{36}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{6}=\frac{\sqrt{13}}{6}\\x-\frac{1}{6}=-\frac{\sqrt{13}}{6}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1+\sqrt{13}}{6}\\x=\frac{1-\sqrt{13}}{6}\end{matrix}\right.\)

Vậy phương trình có nghiệm \(x=\frac{1\pm\sqrt{13}}{6}\)

Lời giải:

Lần sau bạn chú ý ghi đầy đủ đề bài.

Đặt $x^2+3=a$ thì PT trở thành:

$(a-3x)(a-2x)=2x^2$

$\Leftrightarrow a^2-5ax+4x^2=0$

$\Leftrightarrow (a^2-ax)-(4ax-4x^2)=0$

$\Leftrightarrow a(a-x)-4x(a-x)=0$

$\Leftrightarrow (a-x)(a-4x)=0$

Nếu $a-x=0$

$\Leftrightarrow x^2-x+3=0$

$\Leftrightarrow (x-\frac{1}{2})^2=\frac{-11}{4}< 0$ (vô lý- loại)

Nếu $a-4x=0$

$\Leftrightarrow x^2-4x+3=0$

$\Leftrightarrow (x-1)(x-3)=0$

$\Rightarrow x=1$ hoặc $x=3$

Vậy.......

a/\(\left(4x-1\right)\left(x+5\right)=x^2-25\Leftrightarrow4x^2+20x-x-5=x^2-25\Leftrightarrow3x^2+19x+20\)\(\Leftrightarrow\left[{}\begin{matrix}\frac{-4}{3}\\-5\end{matrix}\right.\)

b/

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

c/\(x\left(x+3\right)^3-\frac{x}{4}\left(x+3\right)=0\Leftrightarrow\left(x+3\right)\left[\left(x+3\right)^2x-\frac{x}{4}\right]=0\Leftrightarrow\left(x+3\right)\left[\left(x^2+6x+9\right)x-\frac{x}{4}\right]=0\Leftrightarrow\left(x+3\right)\left(x^3+6x^2+9x-\frac{x}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^3+6x^2+\frac{35}{4}x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{5}{2}\\x=-\frac{7}{2}\end{matrix}\right.\)

d/\(\left(x-1\right)^2=\left(2x+5\right)^2\Leftrightarrow\left(x-1\right)^2-\left(2x+5\right)^2=0\Leftrightarrow\left(x-1+2x+5\right)\left(x-1-2x-5\right)=0\Leftrightarrow\left(3x+4\right)\left(-x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}3x+4=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{-4}{3}\\0\\-6\end{matrix}\right.\)