a)\(1+\sqrt{3x+1}=3x\)\(\Leftrightarrow\sqrt{3x+1}=3x-1\Leftrightarrow3x+1=\left(3x-1\right)^2\)

\(\Leftrightarrow3x-1=9x^2-6x+1\Leftrightarrow9x^2-6x+1-3x+1=0\)

\(\Leftrightarrow9x^2-9x+2=0\Leftrightarrow9x^2-6x-3x+2=0\)

\(\Leftrightarrow3x\cdot\left(3x-2\right)-\left(3x-2\right)=0\Leftrightarrow\left(3x-1\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}3x-1=0\\3x-2=0\end{cases}\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{3}\\x=\frac{2}{3\left[\right]}\end{array}\right.}\)

b. \(\frac{\sqrt{5x+7}}{x+3}=4\)

ĐKXĐ: \(x\ge-\frac{7}{5}\)

\(\Leftrightarrow\sqrt{5x+7}=4\left(x+3\right)\\ \Leftrightarrow\left(\sqrt{5x+7}\right)^2=\left[4\left(x+3\right)\right]^2\\ \Leftrightarrow5x+7=16\left(x^2+6x+9\right)\\ \Leftrightarrow5x+7=16x^2+96x+144\\ \Leftrightarrow16x^2+96x-5x+144-7=0\\ \Leftrightarrow16x^2+91x+137=0\\ \Leftrightarrow\left(4x\right)^2+2.4x.\frac{91}{8}+\frac{8281}{64}+\frac{487}{64}=0\\ \Leftrightarrow\left(4x+\frac{91}{8}\right)^2+\frac{487}{64}=0\left(1\right)\)

Mà \(\left(4x+\frac{91}{8}\right)^2\ge0\forall x\Rightarrow\left(4x+\frac{91}{8}\right)^2+\frac{487}{64}\ge\frac{487}{64}>0\forall x\)

\(\Rightarrow\) phương trình (1) không xảy ra.

Vậy không cógiá trị nào của x thỏa mãn phương trình.

ĐK:x>=5/3

PT <=> \(x^2-3x=4\left(\sqrt{3x-5}-2\right)\)

\(\Leftrightarrow x\left(x-3\right)-\frac{12\left(x-3\right)}{\sqrt{3x-5}+2}=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-\frac{12}{\sqrt{3x-5}+2}\right)=0\)

<=> x = 3 (giải cả hai cái ngoặc nó đều ra x = 3)

P/s: Sai thì thôi nha!

ĐK:....

\(x^2-3x+8=4\sqrt{3x-5}\)

\(\Leftrightarrow x^2-3x+8-4\sqrt{3x-5}=0\)

\(\Leftrightarrow3x-5-4\sqrt{3x-5}+4+x^2-6x+9=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{3x-5}-2=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-5=4\\x=3\end{matrix}\right.\)

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

\(\Leftrightarrow x=3\)

Vậy \(x=3\)

Bạn tham khảo thêm ở link sau:

https://hoc24.vn/cau-hoi/giai-phuong-trinhsqrt3x2-5x1-sqrtx2-2sqrt3leftx2-x-1right-sqrtx2-3x4.167769342831

ĐKXĐ: \(-\frac{16}{3}\le x\le4\)

\(\Leftrightarrow3x^2-12x+36=12\sqrt{4-x}+3\sqrt{3x+16}\)

\(\Leftrightarrow3x^2-9x+4\left(6-x-3\sqrt{4-x}\right)+\left(x+12-3\sqrt{3x+16}\right)=0\)

\(\Leftrightarrow3\left(x^2-3x\right)+\frac{4\left(x^2-3x\right)}{6-x+3\sqrt{4-x}}+\frac{x^2-3x}{x+12+3\sqrt{3x+16}}=0\)

\(\Leftrightarrow\left(x^2-3x\right)\left(3+\frac{4}{6-x+3\sqrt{4-x}}+\frac{1}{x+12+3\sqrt{3x+16}}\right)=0\)

\(\Leftrightarrow x^2-3x=0\)

đk: \(-x^4+3x-1\ge0\)

Có \(-\left(x^4+1\right)\le-2x^2\)

 \(\Rightarrow\sqrt{-x^4+3x-1}+\sqrt{2x^2-3x+2}\le\sqrt{3x-2x^2}+\sqrt{2x^2-3x+2}\) 

Áp dụng bunhia có: \(\sqrt{3x-2x^2}+\sqrt{2x^2-3x+2}\le\sqrt{\left(1+1\right)\left(3x-2x^{^2}+2x^2-3x+2\right)}=2\)

\(\Rightarrow\sqrt{-x^4+3x-1}+\sqrt{2x^2-3x+2}\le2\)  (*)

Có: \(x^4-x^2-2x+4=\left(x^4+1\right)-x^2-2x+3\ge2x^2-x^2-2x+3=\left(x-1\right)^2+2\ge2\) (2*)

Từ (*) (2*) dấu = xảy ra khi x=1 (TM)

Vậy x=1

Tớ đã trả lời ở câu hỏi mới nhất r nên xin phép được xóa câu hỏi này nhé