Em thử nhá, ko chắc đâu

ĐK: \(x\ge\frac{3}{4}\)

PT \(\Leftrightarrow4x^2+12x-9-7x\sqrt{4x-3}=0\)

\(\Leftrightarrow4x^2-9x-9-7x\left(\sqrt{4x-3}-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(4x+3\right)-\frac{28x\left(x-3\right)}{\sqrt{4x-3}+3}=0\)

\(\Leftrightarrow\left(x-3\right)\left(4x+3-\frac{28x}{\sqrt{4x-3}+3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\4x+3=\frac{28x}{\sqrt{4x-3}+3}\left(1\right)\end{matrix}\right.\)

Giải (1): \(\Leftrightarrow\left(4x+3\right)\sqrt{4x-3}-16x+9=0\)

\(\Leftrightarrow\left(4x+3\right)\left(\sqrt{4x-3}-1\right)-12\left(x-1\right)=0\)

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

\(\Leftrightarrow\left(x-1\right)\left[\frac{4\left(4x+3\right)}{\sqrt{4x-3}+1}-12\right]=0\)

Nhận xét rằng cái ngoặc to luôn > 0 với mọi \(x\ge\frac{3}{4}\). Suy ra x = 1

Vậy tập hợp nghiệm của pt: S = {1;3}

Cách 2:

ĐK: \(x\ge\frac{3}{4}\)

\(4x^2+12x-9-7x\sqrt{4x-3}=0\)

\(\Leftrightarrow4x^2-16x+12+7\left[\left(4x-3\right)-x\sqrt{4x-3}\right]=0\)

\(\Leftrightarrow4\left(x-1\right)\left(x-3\right)-7\sqrt{4x-3}\left(x-\sqrt{4x-3}\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(4-\frac{7\sqrt{4x-3}}{x+\sqrt{4x-3}}\right)=0\)

Cái ngoặc to phía sau \(=\frac{4x-3\sqrt{4x-3}}{MS>0}=\frac{16x^2-36x+27}{\left(4x+3\sqrt{4x-3}\right).MS>0}>0\) cái ngoặc to vô nghiệm

Do đó x = 1 (Thỏa mãn) hoặc x = 3 (thỏa mãn)

Ngắn gọn hơn nhỉ:)

Ta có ; \(4x^2+12x=9+7x\sqrt{4x-3}\)(ĐKXĐ : \(x\ge\frac{3}{4}\))

\(\Leftrightarrow4x^2+5x-9=7x\left(\sqrt{4x-3}-1\right)\)

Xét vế trái : \(4x^2+5x-9=4\left(x-1\right)\left(x+\frac{9}{4}\right)=\left[\left(4x-3\right)-1\right]\left(x+\frac{9}{4}\right)=\left(\sqrt{4x-3}-1\right)\left(\sqrt{4x-3}+1\right)\left(x+\frac{9}{4}\right)\)

Suy ra phương trình : \(\left(\sqrt{4x-3}-1\right)\left(\sqrt{4x-3}+1\right)\left(x+\frac{9}{4}\right)=7x\left(\sqrt{4x-3}-1\right)\)

\(\Leftrightarrow\left(\sqrt{4x-3}-1\right)\left[\left(\sqrt{4x-3}+1\right)\left(x+\frac{9}{4}\right)-7x\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{4x-3}-1=0\\\left(\sqrt{4x-3}+1\right)\left(x+\frac{9}{4}\right)-7x=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)(TMDK)

Bài này liên hợp

ĐKXĐ: \(x\ge\frac{3}{4}\)

\(4x^2+12x-16-7x\sqrt{4x-3}+7=0\)

\(\Rightarrow\frac{\left(4x^2+12x\right)^2-16^2}{4x^2+12x+16}-\frac{\left(7x\sqrt{4x-3}\right)^2-7^2}{7x\sqrt{4x-3}+7}=0\)

\(\Rightarrow\frac{16\left(x-1\right)\left(x+4\right)\left(x^2+3x+4\right)}{4x^2+12x+16}-\frac{196x^3-147x^2-49}{7x\sqrt{4x-3}+7}=0\)

\(\Rightarrow\frac{16\left(x-1\right)\left(x+4\right)\left(x^2+3x+4\right)}{4x^2+12x+6}-\frac{\left(x-1\right)\left(4x^2+x+1\right)49}{7x\sqrt{4x-3}+7}=0\)

\(\Rightarrow\left(x-1\right)\left[\frac{16\left(x+4\right)\left(x^2+3x+4\right)}{4x^2+12x+6}-\frac{49\left(4x^2+x+1\right)}{7x\sqrt{4x-3}+7}\right]=0\)

Vì \(\frac{16\left(x+4\right)\left(x^2+3x+4\right)}{4x^2+12x+6}-\frac{49\left(4x^2+x+1\right)}{7x\sqrt{4x-3}+7}>0\)

=> x - 1 = 0 => x = 1

                                                                 Vậy x = 1

Cảm ơn 2 bạn nhìu. Nhưng mà sao kết quả # nhau thế?????

em ms học lớp 9 thôi