Ta có: 

\(x^2+9x+2x=\left(x+4\right)\left(x+5\right)\)

\(x^2+11x+30=\left(x+6\right)\left(x+5\right)\)

\(x^2+13x+42=\left(x+6\right)\left(x+7\right)\)

ĐK: \(\left\{{}\begin{matrix}x\ne-4\\x\ne-5\\x\ne-6\\x\ne-7\end{matrix}\right.\)

pt \(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{18\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}-\dfrac{18\left(x+4\right)}{18\left(x+4\right)\left(x+7\right)}=\dfrac{\left(x+4\right)\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}\)

\(\Rightarrow18\left(x+7\right)-18\left(x+4\right)=\left(x+4\right)\left(x+7\right)\)

\(\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+13=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=2\end{matrix}\right.\)  (tm)

De bai sai ha ban ?

\(\Leftrightarrow\left|x-1,5\right|=-\left|2,5-x\right|.\)(1)

VT >=0; VP <=0. Để đẳng thức 1 xảy ra thì VT = VP = 0.

Nhưng vì VP = 0 =>x= 2,5 thì VT = 1 nên PT vô nghiệm.

Đặt \(\left\{{}\begin{matrix}x\sqrt{y}=a\\y\sqrt{x}=b\end{matrix}\right.\)

Hpt \(\Leftrightarrow\left\{{}\begin{matrix}a+b=6\\a^2+b^2=20\end{matrix}\right.\)

=> Hệ đối xứng loại 1 => EZ

Đặt \(\left\{{}\begin{matrix}a=x\sqrt{y}\\b=\sqrt{x}.y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=6\\a^2+b^2=20\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=6-a\\a^2+\left(6-a\right)^2=20\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=6-a\\2a^2-12a+16=0\end{matrix}\right.\)

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

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

Trường hợp \(\left\{{}\begin{matrix}a=4\\b=2\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x\sqrt{y}=2\sqrt{x}.y\\y\sqrt{x}=2\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{xy}\left(\sqrt{x}-2\sqrt{y}\right)=0\\y\sqrt{x}=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=2\sqrt{y}\\\sqrt{x}.y=2\end{matrix}\right.\)( vì \(\sqrt{xy}\ne0\) )

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=4y\\y\sqrt{y}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=4\end{matrix}\right.\)

TRường hợp 2 tương tự nha

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=40\\ \Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]=40\\ \Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=40\\ \Leftrightarrow\left(x^2+5x+4\right)\left[\left(x^2+5x+4\right)+2\right]=40\\ \Leftrightarrow\left(x^2+5x+4\right)^2+2\left(x^2+5x+4\right)-40=0\)

Mình thấy nghiệm xấu lắm, bạn xem có đúng đề ko

(x² + 5x + 6)(x² + 9x + 20) = 24

<=> (x + 2)(x + 3)(x + 4)(x + 5) - 24 = 0

<=> (x² + 7x + 10)(x² + 7x + 12) - 24 = 0 (1)

Đặt x² + 7x + 11 = t, ta có:

(1) <=> (t - 1)(t + 1) - 24 = 0

<=> t² - 1 - 24 = 0

<=> (t - 5)(t + 5) = 0

\(\Leftrightarrow\left[{}\begin{matrix}t-5=0\\t+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+7x+11-5=0\\x^2+7x+11+5=0\end{matrix}\right.\)

<=> (x + 1)(x + 6) = 0 (vì \(x^2+7x+16\ge\dfrac{15}{4}>0\))

<=> x = - 1 hoặc x = - 6

~ ~ ~ ~ ~

x⁴ - 24x = 32

<=> x⁴ - 24x - 32 = 0

<=> (x² - 2x - 4)(x² + 2x + 8) = 0

<=> \(\left(x-1-\sqrt{5}\right)\left(x-1+\sqrt{5}\right)=0\) (vì \(x^2+2x+8\ge7>0\))

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

\(\frac{x}{40}-\frac{x}{45}=\frac{3}{2}\)

\(\Leftrightarrow\frac{9x-8x}{360}=\frac{3}{2}\)

\(\Leftrightarrow2x=3.360\)

\(\Leftrightarrow2x=1080\)

\(\Leftrightarrow x=540\)

\(\frac{x}{40}-\frac{x}{45}=\frac{3}{2}\)

\(\Leftrightarrow\frac{18x}{720}-\frac{16x}{720}=\frac{1080}{720}\)

\(\Rightarrow18x-16x=1080\)(KHỬ MẪU)

\(\Leftrightarrow2x=1080\)

\(\Leftrightarrow x=\frac{1080}{2}\)

\(\Leftrightarrow x=540\)

Vậy tập nghiệm của phương trình là \(S=\left\{540\right\}\)

Đặt \(x+1=t\)

PT\(\Leftrightarrow\left(t-2\right)^4+\left(t+2\right)^4=40\)

\(\Leftrightarrow\left[\left(t-2\right)^2\right]^2+\left[\left(t+2\right)^2\right]^2=40\)

\(\Leftrightarrow\left[\left(t-2\right)^2+\left(t+2\right)^2\right]^2-2\left(t-2\right)^2\left(t-2\right)^2=40\)

\(\Leftrightarrow\left(t^2-4t+4+t^2+4t+4\right)^2-2\left(t^2-4\right)^2=40\)

​\(\Leftrightarrow\left(2t^2+8\right)^2-2\left(t^2-4\right)^2=40\)​

\(\Leftrightarrow...\)

X nặng 80 lần là gì hả bạn?