Tự nhiên trả lời làm cái gì

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Chứ không phải trả lời nha o0o I am a studious person CTV 

chuẩn không cần phải chỉnh nha bn!!!!

2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)

\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)

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

Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)

\(\Rightarrow x=3\)

c,\(pt\Leftrightarrow3\left(x-1\right)+\frac{x-1}{4x}+\left(2-\sqrt{3x+1}\right)=0\)

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

\(\Rightarrow x=1\)

\(3+\frac{1}{4x}+\frac{1}{2+\sqrt{3x+1}}=0\)

bạn làm nốt pần này nhá

giải bài nào hộ mk cx được ko cần lm hết đâu :) :) :)

1/ Đặt \(\sqrt{x^2+2}=t>0\Rightarrow x^2=t^2-2\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}t-3=0\\t+1-x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}t=3\\t=x-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+2}=3\left(1\right)\\\sqrt{x^2+2}=x-1\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x^2=7\Rightarrow x=\pm\sqrt{7}\)

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2+2=\left(x-1\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2+2=x^2-2x+1\end{matrix}\right.\) \(\Rightarrow x=\dfrac{-1}{2}\left(l\right)\)

Vậy nghiệm pt là \(x=\pm\sqrt{7}\)

2/

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

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

Đặt \(\sqrt{x^2+3}-3x=t\)

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

TH1: \(\sqrt{x^2+3}-3x=-1\Rightarrow\sqrt{x^2+3}=3x-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-1\ge0\\x^2+3=\left(3x-1\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\8x^2-6x-2=0\end{matrix}\right.\) \(\Rightarrow x=1\)

TH2: \(\sqrt{x^2+3}-3x=2\Leftrightarrow\sqrt{x^2+3}=3x+2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{-2}{3}\\x^2+3=\left(3x+2\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{-2}{3}\\8x^2+12x+1=0\end{matrix}\right.\) \(\Rightarrow x=\dfrac{-3+\sqrt{7}}{4}\)

3/ ĐKXĐ: \(\dfrac{3}{2}\le x\le\dfrac{5}{2}\)

\(1.\sqrt{2x-3}+1.\sqrt{5-2x}\le\sqrt{\left(1^2+1^2\right)\left(2x-3+5-2x\right)}=2\)

\(\Rightarrow VT\le2\)

\(VP=3\left(x^2-4x+4\right)+2=3\left(x-2\right)^2+2\ge2\)

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

Vậy pt có nghiệm duy nhất \(x=2\)

4/

ĐKXĐ: \(x\ge\dfrac{-5}{4}\)

\(x^2-2x+1+4x+5-6\sqrt{4x+5}+9=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\\sqrt{4x+5}-3=0\end{matrix}\right.\) \(\Rightarrow x=1\)

Vậy pt có nghiệm duy nhất \(x=1\)

a)\(\sqrt{\left(x-3\right)^2}=|x-3|\)(*)

TH1: x-3 \(\ge0\Leftrightarrow x\ge3\)

(*)=> |x - 3|=x-3

TH2 \(x-3< 0\Leftrightarrow x< 3\)

(*)=>|x-3|=-(x-3)=3-x

Vậy khi x\(\ge\)3 thì (*)=x-3

Khi x<3 thì (*)=3-x

b) ĐK: x<\(\dfrac{1}{3}\)

\(\sqrt{\left(3x+1\right)^2}+2x\\ =\left|3x+1\right|+2x\left(@\right)\)

TH1:3x+1\(\ge\)0\(\Leftrightarrow3x\ge-1\Leftrightarrow x\ge-\dfrac{1}{3}\)=>\(-\dfrac{1}{3}\le x\le\dfrac{1}{3}\)

(@)=>|3x+1|+2x

=3x+1+2x

=5x+1

TH2 \(3x+1< 0\Leftrightarrow3x< -1\Leftrightarrow x< -\dfrac{1}{3}\)

(@)=>|3x+1|+2x

= -3x-1+2x

= -x-1

c) tương tự như vậy

cảm ơn bn