ĐKXĐ:...

a. Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+4x+16}=a>0\\\sqrt{x+70}=b\ge0\end{matrix}\right.\)

\(\Rightarrow6x^2+10x-92=3a^2-2b^2\)

Pt trở thành:

\(3a^2-2b^2+ab=0\)

\(\Leftrightarrow\left(a+b\right)\left(3a-2b\right)=0\)

\(\Leftrightarrow3a=2b\)

\(\Leftrightarrow9\left(2x^2+4x+16\right)=4\left(x+70\right)\)

\(\Leftrightarrow...\)

b. ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{1-x}=b\ge0\end{matrix}\right.\)

Phương trình trở thành:

\(a^2+2+ab=3a+b\)

\(\Leftrightarrow a^2-3a+2+ab-b=0\)

\(\Leftrightarrow\left(a-1\right)\left(a-2\right)+b\left(a-1\right)=0\)

\(\Leftrightarrow\left(a-1\right)\left(a+b-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a+b=2\end{matrix}\right.\)

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

\(\Leftrightarrow...\)

Điều kiện tự xử nhé!

\(6x^2+10x-92+\sqrt{\left(x+70\right)\left(2x^2+4x+16\right)}=0\)(*)

Đặt \(a=\sqrt{x+70};\sqrt{2x^2+4x+16}=b\), (*) trở thành:

\(6x^2+10x-92+ab=0\)

\(\Leftrightarrow6x^2+12x+48-2x-140+ab=0\)

\(\Leftrightarrow3b^2-2a^2+ab=0\)

\(\Leftrightarrow3b^2+3ab-2ab-2a^2=0\)

\(\Leftrightarrow3b\left(a+b\right)-2a\left(a+b\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(3b-2a\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\3b=2a\end{matrix}\right.\)

Tới đây dễ rồi UwU

Chú ý rằng \(3\left(2x^2+4x+16\right)-2\left(x+70\right)=6x^2+10x-92\)

ĐKXĐ: \(x\ge-70\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+70}=a\ge0\\\sqrt{2x^2+4x+16}=b>0\end{matrix}\right.\) \(\Rightarrow a+b>0\)

\(\Rightarrow6x^2+10x-92=3a^2-2b^2\)

Pt trở thành: \(3a^2+ab-2b^2=0\Leftrightarrow\left(3a-2b\right)\left(a+b\right)=0\)

\(\Leftrightarrow3a-2b=0\Rightarrow3a=2b\)

\(\Rightarrow3\sqrt{x+70}=2\sqrt{2x^2+4x+16}\Leftrightarrow9\left(x+70\right)=4\left(2x^2+4x+16\right)\)

\(\Leftrightarrow8x^2+7x-566=0\Rightarrow\left[{}\begin{matrix}x=...\\x=...\end{matrix}\right.\)

a, Ta có: \(\Delta'=1-m+3=4-m\)

Phương trình có 2 nghiệm phân biệt \(\Leftrightarrow\Delta'>0\Leftrightarrow4-m>0\Leftrightarrow m< 4\)

b, ĐXXĐ: \(x\le\frac{9}{4}\)

\(pt\Leftrightarrow\sqrt{\left(9-4x\right)\left(x-3\right)^2}=\left|-2x+5\right|\sqrt{9-4x}\)

\(\Leftrightarrow\sqrt{9-4x}\left(\left|x-3\right|-\left|-2x+5\right|\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}9-4x=0\\\left|x-3\right|=\left|-2x+5\right|\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}9-4x=0\\x-3=-2x+5\\x-3=2x-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{9}{4}\\x=\frac{8}{3}\left(l\right)\\x=2\end{matrix}\right.\)

Vậy pt đã cho có 2 nghiệm \(x=2;x=\frac{9}{4}\)

Đặt \(\left\{{}\begin{matrix}2x+3=a\\\sqrt{x^2-x+1}=b>0\end{matrix}\right.\)

Pt trở thành:

\(a^2+2b^2-3ab=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=2b\end{matrix}\right.\)

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

\(\Leftrightarrow...\)

Lời giải:

1. 

PT $\Leftrightarrow (x^2+5x)^2+2(x^2+5x)-24=0$

$\Leftrightarrow t^2+2t-24=0$ (đặt $x^2+5x=t$)

$\Leftrightarrow (t-4)(t+6)=0$

$\Rightarrow t-4=0$ hoặc $t+6=0$

Nếu $t-4=0\Leftrightarrow x^2+5x-4=0$

$\Leftrightarrow x=\frac{-5\pm \sqrt{41}}{2}$

Nếu $t+6=0$

$\Leftrightarrow x^2+5x+6=0$

$\Leftrightarrow (x+2)(x+3)=0\Rightarrow x=-2$ hoặc $x=-3$

2.

PT $\Leftrightarrow (x^2-4x+1)^2+2(x^2-4x+1)-3=0$

$\Leftrightarrow t^2+2t-3=0$ (đặt $x^2-4x+1=t$)

$\Leftrightarrow (t-1)(t+3)=0$

$\Rightarrow t-1=0$ hoặc $t+3=0$

Nếu $t-1=0\Leftrightarrow x^2-4x=0\Leftrightarrow x(x-4)=0$

$\Rightarrow x=0$ hoặc $x=4$

Nếu $t+3=0\Leftrightarrow x^2-4x+4=0$

$\Leftrightarrow (x-2)^2=0\Leftrightarrow x=2$

1:

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

=>x-3=0 hoặc \(\sqrt{x+3}=2\)

=>x=3 hoặc x+3=4

=>x=1(loại) hoặc x=3(nhận)

2:

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

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

=>\(\sqrt{4\left(4x+1\right)\left(3x-4\right)}=7x-6\)

=>4(12x^2-16x+3x-4)=(7x-6)^2

=>49x^2-84x+36=48x^2-52x-16

=>-84x+36=-52x-16

=>-32x=-52

=>x=13/8

3: =>\(\sqrt{\left(x-5\right)^2}=5-x\)

=>|x-5|=5-x

=>x-5<=0

=>x<=5

4: \(\Leftrightarrow\left|x-4\right|=x+2\)

=>\(\left\{{}\begin{matrix}x>=-2\\\left(x-4\right)^2=\left(x+2\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\x^2-8x+16=x^2+4x+4\end{matrix}\right.\)

=>x>=-2 và -8x+16=4x+4

=>x=1