Làm cho bạn 1 con thôi dài quá trôi hết màn hình:

c) có vẻ khó nhất (con khác tương tự)

đặt 2x+2=t=> x+1=t/2

\(\left(t-1\right).\left(\frac{t}{2}\right)^{^2}.\left(t+1\right)=18\Leftrightarrow\left(t^2-1\right)t^2=4.18\)

\(t^4-t^2=4.18\Leftrightarrow y^2-2.\frac{1}{2}y+\frac{1}{4}=4.18+\frac{1}{4}=\frac{16.18+1}{4}=\left(\frac{17}{2}\right)^2\)

<=> \(\left(y-\frac{1}{2}\right)^{^2}=\left(\frac{17}{2}\right)^2\Rightarrow\left[\begin{matrix}y=\frac{1}{2}-\frac{17}{2}=-8\\y=\frac{1}{2}+\frac{17}{2}=9\end{matrix}\right.\Rightarrow\left[\begin{matrix}2x+2=-8\Rightarrow x=-5\\2x+2=9\Rightarrow x=\frac{7}{2}\end{matrix}\right.\)

CÓ Sai ĐỀ KO BẠN

Ko sai ban oi

\(\left(12x+7\right)^2\left(3x+2\right)\left(2x+1\right)=3\Leftrightarrow\)\(\left(144x^2+168x+49\right)\left(6x^2+7x+2\right)=3\)

Đặt \(6x^2+7x+2=a\Rightarrow144x^2+168x+49=24a+1\)

Phương trình tương đương \(a\left(24a+1\right)=3\)\(\Leftrightarrow24a^2+a-3=0\)

tự giải tiếp

1/ \(\Leftrightarrow\left(4x+1\right)\left(3x+2\right)\left(12x-1\right)\left(x+1\right)=28\)

\(\Leftrightarrow\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)=28\)

Đặt \(12x^2+11x+2=t\)

\(\Rightarrow12x^2+11x-1=t-3\)

\(\Rightarrow t\left(t-3\right)=28\Leftrightarrow t^2-3t-28=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=7\\t=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}12x^2+11x+2=7\\12x^2+11x+2=-4\end{matrix}\right.\)

Bạn tự giải nốt và kl

b/ \(\Leftrightarrow\left(x^2+2x+1\right)\left(4x^2+8x+3\right)=18\)

\(\Leftrightarrow\left(4x^2+8x+4\right)\left(4x^2+8x+3\right)=72\)

Đặt \(4x^2+8x+3=t\Rightarrow t+1=4x^2+8x+4\)

\(\Rightarrow t\left(t+1\right)=72\Leftrightarrow t^2+t-72=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=8\\t=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4x^2+8x+3=8\\4x^2+8x+3=-9\end{matrix}\right.\)

Bạn tự giải và kết luận

Đặt \(2x+2=a\Rightarrow\left\{{}\begin{matrix}2x+1=a-1\\2x+3=a+1\\x+1=\frac{a}{2}\end{matrix}\right.\)

khi đó ta có phương trình:

\(\left(\frac{a}{2}\right)^2\left(a-1\right)\left(a+1\right)=18\Leftrightarrow\frac{a^2}{4}\left(a^2-1\right)=18\)

\(\Leftrightarrow\left(a^2-\frac{1}{2}\right)^2=\frac{289}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}a^2-\frac{1}{2}=\frac{17}{2}\\a^2-\frac{1}{2}=\frac{-17}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a^2=9\\a^2=-8\left(vôlý\right)́\end{matrix}\right.\)

<=> x=3 hoặc x=-3

(12x+7)²(3x+2)(2x+1)=3

<=> (144x²+168x+49)(6x²+7x+2)=3

<=>(144x²+168x+49)(144x+168+48)=72

Đặt 144x²+168x+48=t

=> 144x²+168x+49=t+1(*)

Do đó phương trình đã cho là

(t+1)t=72

<=> t²+t-72=0

<=> (t-8)(t+9)=0

<=>\(\left[{}\begin{matrix}t-8=0\\t+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=8\\t=-9\end{matrix}\right.\)

Bạn tự thay t vào (*) rồi tìm x nha

\(\left(12x+7\right)^2\left(3x+2\right)\left(2x+1\right)=3\)

\(\Leftrightarrow\left(12x+7\right)^2\cdot4\left(3x+2\right)\cdot6\left(2x+1\right)=3\cdot4\cdot6\)

\(\Leftrightarrow\left(12x+7\right)^2\left(12x+8\right)\left(12x+6\right)=72\) (1)

Đặt 12x + 7 = a

(1) \(\Leftrightarrow a^2\left(a+1\right)\left(a-1\right)=72\)

\(\Leftrightarrow a^2\left(a^2-1\right)=72\) (2)

Đặt \(a^2=b\)

(2) \(\Leftrightarrow b\left(b-1\right)=72\)

\(\Leftrightarrow b^2-b-72=0\)

\(\Leftrightarrow b^2+8b-9b-72=0\)

\(\Leftrightarrow b\left(b+8\right)-9\left(b+8\right)=0\)

\(\Leftrightarrow\left(b-9\right)\left(b+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}b-9=0\\b+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}b=9\\b=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a^2=9\Leftrightarrow a=\pm3\\a^2=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}12x+7=3\\12x+7=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}12x=-4\\12x=-10\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-\dfrac{5}{6}\end{matrix}\right.\)

\(\left(12x+7\right)^2\left(3x+2\right)\left(2x+1\right)=3\)

\(\Leftrightarrow\left(144x^2+168x+49\right)\left(6x^2+7x+2\right)=3\)

\(\Leftrightarrow\left(144x^2+168x+49\right)\left(144x^2+168+48\right)=72\)

Đặt \(144x^2+168x+48=u\)

\(\Rightarrow144x^2+168x+49=u+1\left(1\right)\)

Do đó: \(u\left(u+1\right)=72\Leftrightarrow u^2+u-72=0\)

\(\Leftrightarrow\left(u-8\right)\left(u+9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}u-8=0\\u+9=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}u=8\\u=-9\end{matrix}\right.\)

Với \(u=8;u=-9\) bạn thay vào (1) và tìm x nha.