Ta có:

\(\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+9}=\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2+4}\ge\sqrt{9}+\sqrt{4}=5\)

\(3-4x-2x^2=5-2\left(x+1\right)^2\le5\)

Đẳng thức xảy ra khi và chỉ khi:

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

Vậy pt có nghiệm duy nhất \(x=-1\)

a)    \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

\(\Leftrightarrow\)\(2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\)\(\left(3x+1\right)\left[2\left(3x+1\right)-\left(x-2\right)\right]=0\)

\(\Leftrightarrow\)\(\left(3x+1\right)\left(6x+2-x+2\right)=0\)

\(\Leftrightarrow\)\(\left(3x+1\right)\left(5x+4\right)=0\)

đến đây tự lm nha

b)   \(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)    (1)

ĐKXĐ:    \(x\ne\pm\frac{1}{3}\)

\(\left(1\right)\)\(\Leftrightarrow\)\(\frac{12}{\left(1-3x\right)\left(1+3x\right)}=\frac{\left(1-3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}-\frac{\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}\)

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

\(\Leftrightarrow\)\(\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)=12\)

\(\Leftrightarrow\)\(-12x=12\)

\(\Leftrightarrow\)\(x=-1\)   (t/m ĐKXĐ)

Vậy....

a) \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left[2\left(3x+1\right)-\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(3x+1\right)\left(6x+2-x+2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(5x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\5x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=-\frac{4}{5}\end{cases}}}\)

b) ĐKXĐ: \(x\ne\pm\frac{1}{3}\)

\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)

\(\Leftrightarrow\frac{12}{\left(1-3x\right)\left(1+3x\right)}=\frac{\left(1-3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}-\frac{\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}\)

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

\(\Leftrightarrow\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)=12\)

\(\Leftrightarrow-12x=12\)

\(\Leftrightarrow x=-1\) (thỏa mãn)

Vậy x = -1

a) Chuyển vế và rút gọn được 4x = 16, tìm được x = 4.

b) Đưa PT về dạng 5x = 15, tìm được x = 3

c) Quy đồng, khử mẫu thu được 6x - 9 + 24 = 2 - 2x.

Từ đó tìm được x = - 13 8  

d) Quy đồng, khử mẫu thu được 30x + 9 = 36 + 24x +32

Từ đó tìm được x = 59 6

\(VT=\sqrt{3\left(x^2+2x+1\right)+9}+\sqrt{5\left(x^2-2x+1\right)+25}\ge\sqrt{9}+\sqrt{25}=8\)

Do dấu "=" ko đồng thời xảy ra ở hai bđt nên pt vô nghiệm 

\(\sqrt{3\left(x+1\right)^2+9}-3+\sqrt{5\left(x^2-1\right)^2+25}-5=0\)

\(\Leftrightarrow\frac{3\left(x+1\right)^2}{\sqrt{3\left(x+2\right)^2+9}+3}+\frac{5\left(x+1\right)^2\left(x-1\right)^2}{\sqrt{5\left(x^2-1\right)^2+25}+5}=0\)

\(\Leftrightarrow\left(x+1\right)^2\left(\frac{3}{\sqrt{3\left(x+2\right)^2+9}+3}+\frac{5\left(x-1\right)^2}{\sqrt{5\left(x^2-1\right)^2+25}+5}\right)=0\)

\(\left(\frac{3}{\sqrt{3\left(x+2\right)^2+9}+3}+\frac{5\left(x-1\right)^2}{\sqrt{5\left(x^2-1\right)^2+25}+5}\right)>0\left(\forall x\right)\)

\(\Rightarrow x=-1\)

Bạn kia làm sai rùi ạ chắc nhìn nhầm đề 

\(\Leftrightarrow\sqrt{2\left(x^2-1\right)^2+9}+\sqrt{3\left(x-1\right)^2+25}=-3\left(x-1\right)^2+8\)

Ta có:

\(\left\{{}\begin{matrix}\sqrt{2\left(x^2-1\right)^2+9}+\sqrt{3\left(x-1\right)^2+25}\ge\sqrt{9}+\sqrt{25}=8\\-3\left(x-1\right)^2+8\le8\end{matrix}\right.\)

\(\Rightarrow\sqrt{2\left(x^2-1\right)^2+9}+\sqrt{3\left(x-1\right)^2+25}\ge-3\left(x-1\right)^2+8\)

Đẳng thức xảy ra khi và chỉ khi \(x=1\)

1/ x²-3x+2=0

⇒ (x²-2x)-(x-2)=0

⇒ x(x-2)-(x-2)=0

⇒ (x-1)(x-2)=0

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

2) x²-6x+5=0

⇒x²-6x+9-4=0

⇒(x²-6x+9)-2²=0

⇒(x-3)²-2²=0

⇒(x-3-2)(x-3+2)=0

⇒(x-5)(x-1)=0

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

3) 2x²+5x+3=0

⇒ (2x²+2x)+(3x+3)=0

⇒ 2x(x+1)+3(x+1)=0

⇒ (x+1)(2x+3)=0

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

4) x²-8x+15=0

⇒ (x²-8x+16)-1=0

⇒ (x-4)²-1²=0

⇒ (x-4-1)(x-4+1)=0

⇒ (x-5)(x-3)=0

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=5\end{matrix}\right.\)

5) x²-x-12=0

⇒ (x²-4x)+(3x-12)=0

⇒ x(x-4)+3(x-4)=0

⇒ (x-4)(x+3)=0

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)

1: Ta có: \(x^2-3x+2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

2: Ta có: \(x^2-6x+5=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-5\right)=0\)

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

3: Ta có: \(2x^2+5x+3=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{2}\end{matrix}\right.\)

4: Ta có: \(x^2-8x+15=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=5\end{matrix}\right.\)

5: Ta có: \(x^2-x-12=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)