\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+6\right)=168x\\ \Leftrightarrow\left(x+7\sqrt{x}+6\right)\left(x+5\sqrt{x}+6\right)-168x=0\\ \Leftrightarrow\left(x+6\sqrt{x}+6\right)^2-\left(13\sqrt{x}\right)^2=0\\ \left(x-7\sqrt{x}+6\right)\left(x+19\sqrt{x}+6\right)=0 \\ \left(\sqrt{x}-1\right)\left(\sqrt{x}-6\right)=0\)

a) ĐKXĐ: \(x\ge0\)

Ta có: \(\left(x+3\sqrt{x}+2\right)\left(x+9\sqrt{x}+18\right)=168x\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+6\right)=168x\)

\(\Leftrightarrow\left(x+6\right)^2+12\sqrt{x}\left(x+6\right)-133=0\)

\(\Leftrightarrow\left(x+6\right)^2+19\sqrt{x}\left(x+6\right)-7\sqrt{x}\left(x+6\right)-133=0\)

\(\Leftrightarrow\left(x+6\right)\left(x+19\sqrt{x}+6\right)-7\sqrt{x}\left(x+19\sqrt{x}+6\right)=0\)

\(\Leftrightarrow\left(x-7\sqrt{x}+6\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=36\end{matrix}\right.\)

Dòng thứ 2 qua dòng thứ 3 anh làm chậm lại được không ạ, tại tắt quá e không hiểu

Dễ thấy \(x=0\) không là nghiệm của phương trình. Ta có "

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=168x^2\Leftrightarrow\left(x^2+7x+6\right)\left(x^2+5x+6\right)=168x^2\)

                                                          \(\Leftrightarrow\left(x+\frac{6}{x}+7\right)\left(x+\frac{6}{x}+5\right)=168\)

Đặt \(t=x+\frac{6}{x}\) ta được :

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=168x^2\Leftrightarrow\left(t+7\right)\left(t+5\right)=168\)

                                                          \(\Leftrightarrow t^2+12t-133=0\Leftrightarrow\left[\begin{array}{nghiempt}t=7\\t=-19\end{array}\right.\)

Do vậy :

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=168x^2\Leftrightarrow\begin{cases}x+\frac{6}{x}=7\\x+\frac{6}{x}=-19\end{cases}\)

                                                          \(\Leftrightarrow\begin{cases}x^2-7x+6=0\\x^2+19x+6=0\end{cases}\)

                                                          \(\Leftrightarrow\begin{cases}x=1\\x=6\\x=\frac{-19\pm\sqrt{337}}{2}\end{cases}\)

Vậy phương trình đã cho có tập nghiệm :

\(\left\{1;6;\frac{-19-\sqrt{337}}{2};\frac{-19+\sqrt{337}}{2}\right\}\)

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=168x^2\)

<=>\(\left(x+1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)=168x^2\)

<=>\(\left(x^2+7x+6\right)\left(x^2+5x+6\right)=168x^2\)(1)

Đặt t=x²+5x+6

PT (1) trở thành: (t+2x)t=168x²

<=>t²+2tx-168x²=0

<=>t²-12tx+14tx-168x²=0

<=>t.(t-12x)+14x.(t-12x)=0

<=>(t-12x)(t+14x)=0

<=>t-12x=0 hoặc t+14x=0

*t-12x=0 (thích giải denta cũng được)

<=>x²-7x+6=0

<=>x²-x-6x+6=0

<=>x.(x-1)-6.(x-1)=0

<=>(x-1)(x-6)=0

<=>x=1 hoặc x=6

*t+14x=0

<=>x²+19x+6=0

Giải denta là vừa tại số lớn lắm tự làm típ ..............

a. Đề bài sai, phương trình không giải được

b.

ĐKXĐ: \(x\ge-\dfrac{2}{3}\)

\(\left(2x+10\right)\left(\dfrac{1-\left(3+2x\right)}{1+\sqrt{3+2x}}\right)^2=4\left(x+1\right)^2\)

\(\Leftrightarrow\dfrac{\left(2x+10\right)4.\left(x+1\right)^2}{\left(1+\sqrt{3+2x}\right)^2}=4\left(x+1\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}4\left(x+1\right)^2=0\Rightarrow x=-1\\2x+10=\left(1+\sqrt{3+2x}\right)^2\left(1\right)\end{matrix}\right.\)

Xét (1)

\(\Leftrightarrow2x+10=2x+4+2\sqrt{2x+3}\)

\(\Leftrightarrow\sqrt{2x+3}=3\)

\(\Leftrightarrow x=3\)

cho em hỏi , em thấy câu a có nghiệm mà

ĐKXĐ: 5-x>=0 và x+8>=0

=>-8<=x<=5

Ta có: \(13\sqrt{5-x}+18\sqrt{x+8}=61+x+3\sqrt{\left(5-x\right)\left(x+8\right)}\)

=>\(13\sqrt{5-x}-26+18\sqrt{x+8}-54=x-19+3\sqrt{\left(5-x\right)\left(x+8\right)}\)

=>\(13\cdot\left(\sqrt{5-x}-2\right)+18\left(\sqrt{x+8}-3\right)=x-1-18+3\sqrt{\left(5-x\right)\left(x+8\right)}\)

=>\(13\cdot\frac{5-x-4}{\sqrt{5-x}+2}+18\cdot\frac{x+8-9}{\sqrt{x+8}+3}=x-1+3\left(\sqrt{\left(5-x\right)\left(x+8\right)}-6\right)\)

=>\(13\cdot\frac{1-x}{\sqrt{5-x}+2}+18\cdot\frac{x-1}{\sqrt{x+8}+3}=x-1+3\left(\sqrt{5x+40-x^2-8x}-6\right)\)

=>\(-13\cdot\frac{\left(x-1\right)}{\sqrt{5-x}+2}+18\cdot\frac{x-1}{\sqrt{x+8}+3}=x-1+3\left(\sqrt{-x^2-3x+40}-6\right)\)

=>\(-13\cdot\frac{\left(x-1\right)}{\sqrt{5-x}+2}+18\cdot\frac{x-1}{\sqrt{x+8}+3}=x-1+3\cdot\frac{-x^2-3x+40-36}{\sqrt{-x^2-3x+40}+6}\)

=>(x-1)\(\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}\right)=x-1+3\cdot\frac{-x^2-3x+4}{\sqrt{-x^2-3x+40}+6}\)

=>\(\left(x-1\right)\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}\right)=x-1+3\cdot\frac{-x^2-4x+x+4}{\sqrt{-x^2-3x+40}+6}\)

=>\(\left(x-1\right)\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}\right)=x-1+3\cdot\frac{\left(x+4\right)\left(-x+1\right)}{\sqrt{-x^2-3x+40}+6}\)

=>\(\left(x-1\right)\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}\right)=x-1-3\cdot\frac{\left(x+4\right)\left(x-1\right)}{\sqrt{-x^2-3x+40}+6}\)

=>\(\left(x-1\right)\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}-1+3\cdot\frac{x+4}{\sqrt{-x^2-3x+40}+6}\right)=0\)

=>x-1=0

=>x=1(nhận)