a) \(21x+7=15x+35\)

\(\Leftrightarrow21x-15x=35-7\)

\(\Leftrightarrow6x=28\)

\(\Leftrightarrow x=\frac{28}{6}\)

b)\(|5x+3|-2x=x-17\)

\(\Leftrightarrow|5x+3|=3x-17\)

\(\Leftrightarrow\orbr{\begin{cases}5x+3=3x-17\\5x+3=17-3x\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-20\\8x=14\end{cases}}}\Leftrightarrow\orbr{\begin{cases}x=-10\\x=\frac{4}{7}\end{cases}}\)

c) \(\left(5x+2\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5x+2=0\\3x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=-2\\3x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{2}{5}\\x=\frac{4}{3}\end{cases}}}\)

a)  \(21x+7=15x+35\)

\(\Leftrightarrow\)\(6x=28\)

\(\Leftrightarrow\)\(x=\frac{14}{3}\)

Vậy...

b)  \(\left|5x+3\right|-2x=x-17\)

\(\Leftrightarrow\)\(\left|5x+3\right|=3x-17\)

Nếu  \(x\ge-\frac{3}{5}\)thì pt trở thành:

             \(5x+3=3x-17\)

      \(\Leftrightarrow\)\(2x=-20\)

      \(\Leftrightarrow\)\(x=-10\)(loại)

Nếu  \(x< -\frac{3}{5}\)thì pt trở thành:

       \(-5x-3=3x-17\)

      \(\Leftrightarrow\)\(8x=14\)

      \(\Leftrightarrow\) \(x=\frac{7}{4}\) (loại)

Vậy pt vô nghiệm

c)  \(\left(5x+2\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}5x+2=0\\3x-4=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{2}{5}\\x=\frac{4}{3}\end{cases}}\)

Vậy...

a) \(\Leftrightarrow\)21x-15x=35-7 \(\Leftrightarrow\)6x=28 \(\Leftrightarrow\)x=28:6 \(\Leftrightarrow\)x=28/6

c)\(\Leftrightarrow\)\(\orbr{\begin{cases}5\text{x}+2=0\\3\text{x}-4=0\end{cases}}\Leftrightarrow\) \(\orbr{\begin{cases}5\text{x}=-2\\3\text{x}=4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{5}\\x=\frac{4}{3}\end{cases}}\)

b) ko biết làm

a) \(\text{Đ}K\text{X}\text{Đ}:\frac{3}{2}\le x\le\frac{5}{2}\)

Áp dụng BĐT Bunhiacopxki ta có:

\(VT=\sqrt{2x-3}+\sqrt{5-2x}\le\sqrt{2\left(2x-3+5-2x\right)}=2\)

Dấu '=' xảy ra khi \(\sqrt{2x-3}=\sqrt{5-2x}\Leftrightarrow x=2\)

Lại có: \(VP=3x^2-12x+14=3\left(x-2\right)^2+2\ge2\)

Dấu '=' xảy ra khi x=2

Do đó VT=VP khi x=2

b) ĐK: \(x\ge0\). Ta thấy x=0 k pk là nghiệm của pt, chia 2 vế cho x ta có:

\(x^2-2x-x\sqrt{x}-2\sqrt{x}+4=0\Leftrightarrow x-2-\sqrt{x}-\frac{2}{\sqrt{x}}+\frac{4}{x}=0\)

\(\Leftrightarrow\left(x+\frac{4}{x}\right)-\left(\sqrt{x}+\frac{2}{\sqrt{x}}\right)-2=0\)

Đặt \(\sqrt{x}+\frac{2}{\sqrt{x}}=t>0\Leftrightarrow t^2=x+4+\frac{4}{x}\Leftrightarrow x+\frac{4}{x}=t^2-4\), thay vào ta có:

\(\left(t^2-4\right)-t-2=0\Leftrightarrow t^2-t-6=0\Leftrightarrow\left(t-3\right)\left(t+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t=3\\t=-2\end{cases}}\)

Đối chiếu ĐK  của t

\(\Rightarrow t=3\Leftrightarrow\sqrt{x}+\frac{2}{\sqrt{x}}=3\Leftrightarrow x-3\sqrt{x}+2=0\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=1\end{cases}}\)

c) Đặt \(y=\sqrt{x^2+7x+7};y\ge0\)

Pt có dạng: \(3y^2+2y-5=0\Leftrightarrow\orbr{\begin{cases}y=\frac{-5}{3}\\y=1\end{cases}\Leftrightarrow y=1}\)

Với y=1\(\Leftrightarrow\sqrt{x^2+7x+7}=1\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-6\end{cases}}\)

giải pt 

a.\(2\sqrt{12x}-3\sqrt{3x}+4\sqrt{48x}=17\)

=>\(4\sqrt{3x}-3\sqrt{3x}+16\sqrt{3x}=17\)

=>\(17\sqrt{3x}=17\)

=>\(\sqrt{3x}=1\)

=>\(x=\dfrac{1}{3}\)

b.Ta có:\(\sqrt{x^2-6x+9}=1\)

=>\(\sqrt{\left(x-3\right)^2}=1\)

=>\(\left|x-3\right|=1\)

Vậy có hai trường hợp:

TH1:\(x-3=1\)

=>\(x=4\)

TH2:\(x-3=-1\)

=>\(x=2\)

a) ĐKXĐ: \(x\ge0\)

Ta có: \(2\sqrt{12x}-3\sqrt{3x}+4\sqrt{48x}=17\)

\(\Leftrightarrow2\cdot2\cdot\sqrt{3x}-3\cdot\sqrt{3x}+4\cdot4\cdot\sqrt{3x}=17\)

\(\Leftrightarrow4\sqrt{3x}-3\sqrt{3x}+16\sqrt{3x}=17\)

\(\Leftrightarrow17\sqrt{3x}=17\)

\(\Leftrightarrow\sqrt{3x}=1\)

\(\Leftrightarrow3x=1\)

hay \(x=\dfrac{1}{3}\)(nhận)

Vậy: \(S=\left\{\dfrac{1}{3}\right\}\)

b) ĐKXĐ: \(x\in R\)

Ta có: \(\sqrt{x^2-6x+9}=1\)

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

\(\Leftrightarrow\left|x-3\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

Vậy: S={2;4}