\(\Leftrightarrow\left(3x^2+7x+4\right)\left(36x^2+84x+49\right)=6\)

Đặt \(3x^2+7x=a\Rightarrow36x^2+84x=12a\)

\(\left(a+4\right)\left(12a+49\right)-6=0\)

\(\Leftrightarrow12a^2+97a+190=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=-\frac{10}{3}\\a=-\frac{19}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x^2+7x+\frac{10}{3}=0\\3x^2+7x+\frac{19}{4}=0\end{matrix}\right.\) \(\Leftrightarrow...\)

\(ĐK:x\in R\)

Đặt \(x^2-2x=a\), PTTT:

\(-a+\sqrt{6a+7}=0\\ \Leftrightarrow\sqrt{6a+7}=a\\ \Leftrightarrow a^2-6a-7=0\\ \Leftrightarrow\left[{}\begin{matrix}a=7\\a=-1\left(loại.do.a=\sqrt{6a+7}\ge0\right)\end{matrix}\right.\\ \Leftrightarrow a=7\\ \Leftrightarrow x^2-2x-7=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1+2\sqrt{2}\\x=1-2\sqrt{2}\end{matrix}\right.\)

Nhận thấy \(x=0\) ko phải nghiệm

Với \(x\ne0\) chia 2 vế của pt cho \(x^2\) ta được:

\(6\left(x^2+\dfrac{1}{x^2}\right)-5\left(x+\dfrac{1}{x}\right)-38=0\)

Đặt \(x+\dfrac{1}{x}=t\Rightarrow x^2+\dfrac{1}{x^2}=t^2-2\)

\(\Rightarrow6\left(t^2-2\right)-5t-38=0\)

\(\Leftrightarrow6t^2-5t-50=0\Rightarrow\left[{}\begin{matrix}t=\dfrac{10}{3}\\t=-\dfrac{5}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=\dfrac{10}{3}\\x+\dfrac{1}{x}=-\dfrac{5}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}3x^2-10x+3=0\\2x^2+5x+2=0\end{matrix}\right.\)

\(\Rightarrow x=\left\{-2;-\dfrac{1}{2};\dfrac{1}{3};3\right\}\)

Ta có:

(6x+8)(6x+6)(6x+7)² = 72

Đặt \(6x+7=a\)

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

\(\Leftrightarrow a^4-a^2-72=0\)

\(\Leftrightarrow\left(a^4+8a^2\right)+\left(-9a^2-72\right)=0\)

\(\Leftrightarrow\left(a^2+8\right)\left(a^2-9\right)=0\)

Đễ thấy \(a^2+8>0\)

\(\Rightarrow a^2-9=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=3\\a=-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}6x+7=3\\6x+7=-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=\frac{-5}{3}\end{cases}}\)

(36x^2+84x+48)(36x^2+84x+49)=72 
dat 36x^2+84x+48=a 
phuong trinh da cho co dang 
a(a+1)=72 
a^2+a-72=0 
a=8 hoac a=-9 
a=8=>36x^2+84x+48=8 
=>x=-2/3 hoac x=-5/3 
a=-9=>36x^2+84x+48=-9(vo nghiem)

Bang 9 nha Do Thi Hai Yen

Xét thấy x = 0 không thỏa mãn pt

Ta có : \(6x^4+7x^3-36x^2+7x+6=0\)

\(\Leftrightarrow x^2\left(6x^2+7x-36+\frac{7}{x}+\frac{6}{x^2}\right)=0\)

\(\Leftrightarrow6x^2+7x-36+\frac{7}{x}+\frac{6}{x^2}=0\)

\(\Leftrightarrow6\left(x^2+\frac{1}{x^2}\right)+7\left(x+\frac{1}{x}\right)-36=0\)

\(\Leftrightarrow6\left(x+\frac{1}{x}\right)^2-7\left(x+\frac{1}{x}\right)-36-12=0\)

\(\Leftrightarrow6\left(x+\frac{1}{x}\right)^2-7\left(x+\frac{1}{x}\right)-48=0\)

Đặt \(x+\frac{1}{x}=a\)

\(pt\Leftrightarrow6a^2-7a-48=0\)

\(\Leftrightarrow6\left(a^2-\frac{7}{6}a-8\right)=0\)

\(\Leftrightarrow a^2-\frac{7}{6}a-8=0\)

\(\Leftrightarrow a^2-2\cdot a\cdot\frac{7}{12}+\frac{49}{144}-\frac{1201}{144}=0\)

\(\Leftrightarrow\left(a-\frac{7}{12}\right)^2=\left(\frac{\pm\sqrt{1201}}{12}\right)^2\)

\(\Leftrightarrow a=\frac{\pm\sqrt{1201}+7}{12}\)

\(\Leftrightarrow x+\frac{1}{x}=\frac{\pm\sqrt{1201}+7}{12}\)

Giải nốt nha bạn. Nghiệm hơi xấu

\(\frac{3x-2}{x+7}=\frac{6x+1}{2x-3}\) (Đkxđ: \(x\ne-7;x\ne\frac{3}{2}\))

\(\Rightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)

\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)

\(\Leftrightarrow6x^2-9x-4x-6x^2-42x-x=7-6\)

\(\Leftrightarrow-56x=1\)

\(\Leftrightarrow x=-\frac{1}{56}\) (t/m đkxđ)

Vậy \(S=\left\{-\frac{1}{56}\right\}\)

ĐKXĐ: x khác -7 và 3/2

Từ đề bài <=> (3x-2)(2x-3) = (6x+1)(x+7)

<=> 6x^2-4x-9x+6 = 6x^2+x+42x+7

<=> -13x+6 = 43x+7

<=> 6-7 = 43x+13x

<=> 56x = -1

<=> x = -1/56 (TM)

Vậy ...

ĐKXĐ:x khác -7;x khác 1,5

=>(3x-2)(2x-3)=(6x+1)(x+7)

=>6x²-4x-9x+6=6x²+x+42x+7

=>6x²-13x+6=6x²+43x+7

=>6x²-6x²-13x-43x+6-7=0

=>-56x-1=0

=>-56x=1

=>x=\(\frac{-1}{56}\)

Ta có

\(\left\{{}\begin{matrix}\sqrt{2x^2-4x+3}=\sqrt{2\left(x-1\right)^2+1}\ge\sqrt{1}=1\\\sqrt{3x^2-6x+7}=\sqrt{3\left(x-1\right)^2+4}\ge\sqrt{4}=2\end{matrix}\right.\) \(\forall x\)

\(\Rightarrow VT=\sqrt{2x^2-4x+3}+\sqrt{3x^2-6x+7}\ge3\) \(\forall x\)

Lại có \(VP=2-x^2+2x=3-\left(x-1\right)^2\le3\) \(\forall x\)

\(\Rightarrow\sqrt{2x^2-4x+3}+\sqrt{3x^2-6x+7}=2-x^2+2x\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2\left(x-1\right)^2+1}=1\\\sqrt{3\left(x-1\right)^2+4}=2\\3-\left(x-1\right)^2=3\end{matrix}\right.\)

\(\Leftrightarrow\left(x-1\right)^2=0\Rightarrow x=1\)

Vậy pt có nghiệm duy nhất \(x=1\)