\(48x\left(x+1\right)\left(x^3-4\right)=\left(x^4+8x+12\right)^2\)

\(\Leftrightarrow4\left(12x+12\right)\left(x^4-4x\right)=\left(x^4+8x+12\right)^2\)

Đặt \(\left\{{}\begin{matrix}x^4-4x=a\\12x+12=b\end{matrix}\right.\)

\(\Rightarrow4ab=\left(a+b\right)^2\)

\(\Leftrightarrow4ab=a^2+a^2+2ab\)

\(\Leftrightarrow\left(a-b\right)^2=0\)

\(\Leftrightarrow a-b=0\)

\(\Leftrightarrow x^4-16x-12=0\)

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

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

\(\Rightarrow x=1\pm\sqrt{3}\)

Giải:

\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-3\right)\left(x-2\right)}=\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-1\right)\left(x-4\right)}\)

ĐKXĐ: \(x\ne\left\{1;2;3;4\right\}\)

\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-3\right)\left(x-2\right)}=\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-1\right)\left(x-4\right)}\)

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

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

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

\(\Leftrightarrow0x=2\)

Vậy ...

Ta có:

\(\left|x-1\right|+\left|x-5\right|=\left|x-1\right|+\left|5-x\right|\)

≥ \(\left|x-1+5-x\right|=4\)

mà \(4-\left|x-3\right|\)≤ 4

Dấu "="⇔ \(x=3\)

bậc nhất môt ẩn đây ak

a)<=>\(x^4-4x^3-19x^2+106x-120=\left(x-4\right)\left(x-3\right)\left(x-2\right)\left(x+5\right)\)

=>TH1:x-4=0

=>x=4

=>TH2:x-3=0

=>x=3

=>TH3:x-2=0

=>x=2

và TH 5 : x+5=0

=>x=-5

b)<=>\(\left(\text{x+1)(x+2)(x+4)(x+5}\right)-40=x\left(x+6\right)\left(x^2+6x+13\right)\)

=>TH1:x=0

=>TH2:x+6=0

=>x=-6

=>\(x^2+6x+13=0\)

=>có biệt thức \(6^2-4\left(1.13\right)=-16\)

=>D<0

=>PT ko có nghiệm

=>x=-6 hoặc 0

a)x=-5;2;3;4( có 4 Trường hợp )

b)x=-6;0( có 2 trường hợp)

\(x^4-4x^3-19x^2+76x+30x-120=0\)

<=>\(x^3\left(x-4\right)-19x\left(x-4\right)+30\left(x-4\right)=\left(x-4\right)\left(x^3-19x+30\right)\)=0

=>x=4

ĐKXĐ: \(-4\le x\le1\)

Đặt \(\sqrt{x+4}-\sqrt{1-x}=t\)

\(\Rightarrow t^2=5-2\sqrt{\left(x+4\right)\left(1-x\right)}\Rightarrow\sqrt{\left(x+4\right)\left(1-x\right)}=\frac{5-t^2}{2}\)

Pt trở thành:

\(t\left(1+\frac{5-t^2}{2}\right)=3\Leftrightarrow t\left(7-t^2\right)=6\)

\(\Leftrightarrow t^3-7t+6=0\Leftrightarrow\left(t+3\right)\left(t-1\right)\left(t-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}t=-3\\t=1\\t=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x+4}-\sqrt{1-x}=-3\\\sqrt{x+4}-\sqrt{1-x}=1\\\sqrt{x+4}-\sqrt{1-x}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+4}+3=\sqrt{1-x}\left(vn\right)\\\sqrt{x+4}=1+\sqrt{1-x}\\\sqrt{x+4}=2+\sqrt{1-x}\end{matrix}\right.\) (1 vô nghiệm do \(VT\ge3;VP\le\sqrt{5}< 3\))

\(\Leftrightarrow\left[{}\begin{matrix}x+4=2-x+2\sqrt{1-x}\\x+4=5-x+4\sqrt{1-x}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1=1-x\\4x^2-4x+1=16-16x\end{matrix}\right.\) \(\Leftrightarrow...\)

ĐK: \(x\in R\backslash\left\{-4,-3,-2,-1\right\}\)

PT ban đầu

\(\Leftrightarrow\frac{x+2-x-1}{\left(x+1\right)\left(x+2\right)}+\frac{x+3-x-2}{\left(x+2\right)\left(x+3\right)}+\frac{x+4-x-3}{\left(x+3\right)\left(x+4\right)}+\frac{x+5-x-4}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+1}-403\\ \Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}=\frac{1}{x+1}-403\\ \Leftrightarrow\frac{1}{x+5}=403\\ \Leftrightarrow x+5=\frac{1}{403}\Leftrightarrow x=\frac{-2014}{403}\)

Chúc bạn học tốt nha.

Xửa đề:

\(\left(x+1\right)\left(x+4\right)+3\left(x+4\right)\sqrt{\frac{x+1}{x+4}}-18=0\)

Xet \(x+4>0\)

\(\Rightarrow\left(x+1\right)\left(x+4\right)+3\sqrt{\left(x+1\right)\left(x+4\right)}-18=0\)

Đặt \(\sqrt{\left(x+1\right)\left(x+3\right)}=a\)

\(\Rightarrow a^2+3a-18=0\)

Trường hợp \(x+4< 0\)

Làm tương tự