\(pt\Leftrightarrow\sqrt{3x+1}-4+1-\sqrt{6-x}+3x^2-14x-5=0\)

\(\Leftrightarrow\frac{3x+1-16}{\sqrt{3x+1}+4}+\frac{1-\left(6-x\right)}{1+\sqrt{6-x}}+\left(x-5\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left[\frac{3}{\sqrt{3x+1}+4}+\frac{1}{1+\sqrt{6-x}}+3x+1\right]=0\)

\(\Leftrightarrow x=5.\)

ĐKXĐ: \(-\dfrac{1}{3}\le x\le6\)

\(\left(\sqrt{3x+1}-4\right)+\left(1-\sqrt{6-x}\right)+\left(3x^2-14x-5\right)=0\)

\(\Leftrightarrow\dfrac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\dfrac{x-5}{1+\sqrt{6-x}}+\left(x-5\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{1+\sqrt{6-x}}+3x+1\right)=0\)

\(\Leftrightarrow x-5=0\) (do \(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{1+\sqrt{6-x}}+3x+1>0;\forall x\))

\(\Rightarrow x=5\)

ĐKXĐ: \(\left\{{}\begin{matrix}3x+1>=0\\6-x>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{3}\\x< =6\end{matrix}\right.\)

\(\sqrt{3x+1}-\sqrt{6-x}+3x^2-14x-8=0\)

=>\(\sqrt{3x+1}-4+1-\sqrt{6-x}+3x^2-14x-5=0\)

=>\(\dfrac{3x+1-16}{\sqrt{3x+1}+4}+\dfrac{1-6+x}{1+\sqrt{6-x}}+3x^2-15x+x-5=0\)

=>\(\dfrac{3\cdot\left(x-5\right)}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{6-x}+1}+\left(x-5\right)\left(3x+1\right)=0\)

=>\(\left(x-5\right)\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{\sqrt{6-x}+1}+3x+1\right)=0\)

=>x-5=0

=>x=5(nhận)

\(-\dfrac{1}{3}\le x\le6\)

\(\sqrt{3x+1}-4-\left(\sqrt{6-x}-1\right)+3x^2-14x-5=0\)

\(\Leftrightarrow\dfrac{3x-15}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{6-x}+1}+\left(x-5\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(\dfrac{3}{\sqrt{3x+1}}+\dfrac{1}{\sqrt{6-x}+1}+3x-1\right)=0\)

do \(x\ge\dfrac{-1}{3}\Rightarrow3x+1\ge0\)

\(\dfrac{3}{\sqrt{3x+1}}+\dfrac{1}{\sqrt{6-x}+1}+3x-1>0\)

\(\Rightarrow x=5\)

tại sao lại như thế hả bạn ?

b)đk:\(x\ge\dfrac{1}{2}\)

Có: \(\sqrt{2x^2-1}\le\dfrac{2x^2-1+1}{2}=x^2\)

\(x\sqrt{2x-1}=\sqrt{\left(2x^2-x\right)x}\le\dfrac{2x^2-x+x}{2}=x^2\)

=>\(\sqrt{2x^2-1}+x\sqrt{2x-1}\le2x^2\) 

Dấu = xảy ra\(\Leftrightarrow x=1\)

Vậy....

c) đk: \(x\ge0\)

\(\Leftrightarrow\sqrt{x}=\sqrt{x+9}-\dfrac{2\sqrt{2}}{\sqrt{x+1}}\)
\(\Rightarrow x=x+9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)

\(\Leftrightarrow0=9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)

Đặt \(a=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\left(a>0\right)\)

\(\Leftrightarrow\dfrac{a^2-2}{2}=\dfrac{8}{x+1}\)

pttt \(9+\dfrac{a^2-2}{2}-4a=0\) \(\Leftrightarrow a=4\) (TM)

\(\Rightarrow4=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\) \(\Leftrightarrow16=\dfrac{2\left(x+9\right)}{x+1}\) \(\Leftrightarrow x=\dfrac{1}{7}\) (TM)
Vậy ...

 

a)ĐKXĐ: x≥-1/3; x≤6

<=>\(\dfrac{3x-15}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{x-6}+1}+\left(x-5\right)\cdot\left(3x+1\right)=0\Leftrightarrow\left(x-5\right)\cdot\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{\sqrt{x-6}+1}+3x+1\right)=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)(nhận)

(vì x≥-1/3 nên3x+1≥0 )

\(DK:-\frac{1}{3}\le x\le6\)

\(\Leftrightarrow\left(\sqrt{3x+1}-4\right)-\left(\sqrt{6-x}-1\text{ }\right)+\left(3x^2-15x\right)+\left(x-5\right)=0\)

\(\Leftrightarrow\frac{3x+1-16}{\sqrt{3x+1}+4}-\frac{6-x-1}{\sqrt{6-x}+1}+3x\left(x-5\right)+\left(x-5\right)=0\)

\(\Leftrightarrow\frac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\frac{x-5}{\sqrt{6-x}+1}+3x\left(x-5\right)+\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(\frac{3}{\sqrt{3x+1}+4}+\frac{1}{\sqrt{6-x}+1}+3x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\left(n\right)\\\frac{3}{\sqrt{3x+1}+4}+\frac{1}{\sqrt{6-x}+1}+3x+1=0\left(l\right)\end{cases}}\)

Vay nghiem cua PT la \(x=5\)

Thx MaiLink

\(Pt\Leftrightarrow\sqrt{3x+1}-4+1-\sqrt{6-x}+3x^2-14x-5=0\)(ĐKXĐ: \(-\frac{1}{3}\le x\le6\))

\(\Leftrightarrow\frac{3x-15}{\sqrt{3x+1}+4}+\frac{x-5}{1+\sqrt{6-x}}+\left(x-5\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(\frac{3}{\sqrt{3x+1}+4}+\frac{1}{1+\sqrt{6-x}}+3x+1\right)=0\)

\(\Rightarrow x=5\)(tmđk)

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Câu hỏi của Lê Đức Anh - Toán lớp 9 - Học toán với OnlineMath