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

1.

ĐK: \(-1\le x\le4\)

Đặt \(\sqrt{x+1}+\sqrt{4-x}=t\left(t\ge0\right)\)

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

\(PT\Leftrightarrow t+\frac{t^2-5}{2}=5\Rightarrow t^2+2t-15=0\) \(\Rightarrow\left[{}\begin{matrix}t=3\\t=-5\left(l\right)\end{matrix}\right.\)

\(t=3\Rightarrow\sqrt{-x^2+3x+4}=2\) \(\Leftrightarrow-x^2+3x+4=4\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\) (tm)

2.

ĐK:\(x\ge4\)

Đặt \(\sqrt{x+4}+\sqrt{x-4}=t\left(t\ge0\right)\)

\(\Rightarrow2\sqrt{x^2-16}=t^2-2x\)

\(PT\Leftrightarrow t=2x-12+t^2-2x\)

\(\Leftrightarrow t^2-t-12=0\Rightarrow\left[{}\begin{matrix}t=4\\t=-3\left(l\right)\end{matrix}\right.\) Giải tiếp như trên.

a, ĐK: ​\(\left(x+1\right)\left(x^2+2x-1\right)\ge0\)​

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

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

TH1: \(x\ge-1\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x-1=x+1\\x^2+2x-1=9x+9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2-7x-10=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

TH2: \(x< -1\)

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

\(\Leftrightarrow...\)

Bài này dài nên ... cho nhanh nha, đoạn sau dễ rồi

6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)

Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)

Phương trình sẽ trở thành là: a^2+a-42=0

=>(a+7)(a-6)=0

=>a=-7(loại) hoặc a=6(nhận)

=>2x^2+3x+9=36

=>2x^2+3x-27=0

=>2x^2+9x-6x-27=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)

1) \(\sqrt{\text{x^2− 20x + 100 }}=10\)

<=> \(\sqrt{\left(x-10\right)^2}=10\)

<=> \(\left|x-10\right|=10\)

=> \(\left[{}\begin{matrix}x-10=10\\x-10=-10\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=10+10\\x=\left(-10\right)+10\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=20\\x=0\end{matrix}\right.\)

Vậy S = \(\left\{20;0\right\}\)

2) \(\sqrt{x +2\sqrt{x}+1}=6\)

<=> \(\sqrt{\left(\sqrt{x^2}+2.\sqrt{x}.1+1^2\right)}=6\)

<=> \(\sqrt{\left(\sqrt{x}+1\right)^2}=6\)

<=> \(\left|\sqrt{x}+1\right|=6\)

=> \(\left[{}\begin{matrix}\sqrt{x}+1=6\\\sqrt{x}+1=-6\end{matrix}\right.\)=>\(\left[{}\begin{matrix}\sqrt{x}=6-1=5\\\sqrt{x}=\left(-6\right)-1=-7\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=25\\x=-49\left(loai\right)\end{matrix}\right.\)

Vậy S = \(\left\{25\right\}\)

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

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

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

<=> \(\left|x-3\right|=\sqrt{3}+1\)

=> \(\left[{}\begin{matrix}x-3=\sqrt{3}+1\\x-3=-\left(\sqrt{3}+1\right)\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=\sqrt{3}+4\\x=-\sqrt{3}+2\end{matrix}\right.\)

Vậy S = \(\left\{\sqrt{3}+4;-\sqrt{3}+2\right\}\)

4) \(\sqrt{3x+2\sqrt{3x}+1}=5\)

<=> \(\sqrt{\sqrt{3x}^2+2.\sqrt{3x}.1+1^2}=5\)

<=> \(\sqrt{\left(\sqrt{3x}+1\right)^2}=5\)

<=> \(\left|\sqrt{3x}+1\right|=5\)

=> \(\left[{}\begin{matrix}\sqrt{3x}+1=5\\\sqrt{3x}+1=-5\end{matrix}\right.\)=> \(\left[{}\begin{matrix}\sqrt{3x}=5-1=4\\\sqrt{3x}=\left(-5\right)-1=-6\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}3x=16\\3x=-6\left(loai\right)\end{matrix}\right.\)=> x = \(\dfrac{16}{3}\) Vậy S = \(\left\{\dfrac{16}{3}\right\}\)

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

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

<=> \(\left|x-\sqrt{3}\right|=\sqrt{3}-1\)

<=> \(\left[{}\begin{matrix}x-\sqrt{3}=\sqrt{3}-1\\x-\sqrt{3}=-\left(\sqrt{3}-1\right)\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=-1\\x=-2\sqrt{3}+1\end{matrix}\right.\)

Vậy S = \(\left\{-1;-2\sqrt{3}+1\right\}\)

6) \(\sqrt{6x+4\sqrt{6x}+4}=7\)

<=> \(\sqrt{\sqrt{6x}^2+2.\sqrt{6x}.2+2^2}=7\)

<=> \(\sqrt{\left(\sqrt{6}+2\right)^2}=7\)

<=> \(\left|\sqrt{6x}+2\right|=7\)

=> \(\left[{}\begin{matrix}\sqrt{6x}+2=7\\\sqrt{6x}+2=-7\end{matrix}\right.\)=>\(\left[{}\begin{matrix}\sqrt{6x}=7-2=5\\\sqrt{6x}=\left(-7\right)-2=-9\left(loai\right)\end{matrix}\right.\)

=> \(\sqrt{6x}=5=>6x=25=>x=\dfrac{25}{6}\)

7) \(\sqrt{2x^2-2x\sqrt{6}+3}-\sqrt{5-\sqrt{24}}=0\)

<=> \(\sqrt{2x^2-2x\sqrt{6}+3}=\sqrt{5-\sqrt{24}}\)

<=> \(\sqrt{\left(x\sqrt{2}\right)^2-2x\sqrt{2}.\sqrt{3}+\sqrt{3}^2}=\sqrt{5-\sqrt{4}.\sqrt{6}}\)

<=> \(\sqrt{\left(x\sqrt{2}-\sqrt{3}\right)^2}=\sqrt{\sqrt{3}^2-2\sqrt{3}.\sqrt{2}+\sqrt{2}^2}\)

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

<=> \(\left|x\sqrt{2}-\sqrt{3}\right|=\sqrt{3}-\sqrt{2}\)

=> \(\left[{}\begin{matrix}x\sqrt{2}-\sqrt{3}=\sqrt{3}-\sqrt{2}\\x\sqrt{2}-\sqrt{3}=-\left(\sqrt{3}-\sqrt{2}\right)\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x\sqrt{2}=2\sqrt{3}-\sqrt{2}\\x\sqrt{2}=\sqrt{2}\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{2\sqrt{3}-\sqrt{2}}{\sqrt{2}}\\1\end{matrix}\right.\)

Vậy S = \(\left\{\dfrac{2\sqrt{3}-\sqrt{2}}{\sqrt{2}};1\right\}\)

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

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

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

<=> \(\sqrt{2}-1=\left|x-\sqrt{2}\right|\)

=> \(\left[{}\begin{matrix}x-\sqrt{2}=\sqrt{2}-1\\x-\sqrt{2}=-\left(\sqrt{2}-1\right)\end{matrix}\right.=>\left[{}\begin{matrix}x+1=2\sqrt{2}\\x-\sqrt{2}=-\sqrt{2}+1\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=2\sqrt{2}-1\\x=1\end{matrix}\right.\) Vậy S = \(\left\{2\sqrt{2}-1;1\right\}\)

6) ĐKXĐ: \(x\le-6\)

\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)

\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)

Vậy \(x\le-6\)

7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)

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

\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)

Vậy \(x\ge\dfrac{2}{3}\)

8) ĐKXĐ: \(x\ge5\)

\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)

\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)

9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)