a, Câu này dễ quá bỏ qua nha :)

b, Ta có : \(\Delta^,=b^{,2}-ac=\left(-2\right)^2-\left(m+1\right)=4-m-1=3-m\)

- Để phương trình có 2 nghiệm phân biết thì \(\Delta^,>0\)

=> \(m< 3\)

- Theo vi ét : \(\left\{{}\begin{matrix}x_1+x_2=-\frac{b}{a}=4\\x_1x_2=\frac{c}{a}=m+1\end{matrix}\right.\)

- Để \(x^2_1+x^2_2=3\left(x_1+x_2\right)\)

<=> \(\left(x_1+x_2\right)^2-2x_1x_2=3\left(x_1+x_2\right)\)

<=> \(4^2-2\left(m+1\right)=3.4=12\)

<=> \(-2\left(m+1\right)=-4\)

<=> \(m+1=2\)

<=> \(m=1\left(TM\right)\)

Vậy ....

ĐKXĐ: \(x\ge1\)

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

\(\Leftrightarrow2x-1=x-1\left(do.x\ge1\right)\)

\(\Leftrightarrow x=0\left(ktm\right)\)

Vậy \(S=\varnothing\)

ĐK \(x\ge1\)

\(\Leftrightarrow\left|2x-1\right|=x-1\)

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

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

Gọi đường thẳng cần tìm là d1

d1 có dạng y=ax+b

Vì d1 vuông với d

=> a.(-2)=-1=> a=1/2

Mà d1 qua A(-1, 3)

=> 3=\(\frac{1}{2}.\left(-1\right)+b\)=> b=7/2

=> d1: y=\(\frac{1}{2}x+\frac{7}{2}\)

\(\left\{{}\begin{matrix}2\left(x+y\right)+\sqrt{x+1}=4\\\left(x+y\right)-3\sqrt{x+1}=-5\end{matrix}\right.\left(x\ge-1\right)\)

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

\(\Rightarrow\left\{{}\begin{matrix}2a+b=4\\a-3b=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2a+b=4\left(1\right)\\2a-6b=-10\left(2\right)\end{matrix}\right.\)

Lấy \(\left(1\right)-\left(2\right)\Rightarrow7b=14\Rightarrow b=2\Rightarrow2a=4-2=2\Rightarrow a=1\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=1\\\sqrt{x+1}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=-2\\x=3\end{matrix}\right.\)

a/ Bạn tự giải

b/ \(\Delta'=\left(m-1\right)^2-2\left(m-1\right)=m^2-4m+3\)

Biểu thức này có thể nhận giá trị âm với \(1< m< 3\) nên đề bài sai

a) \(\sqrt{1-4x+4x^2}=5\) 

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

\(\Leftrightarrow\left|1-2x\right|=5\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

b) \(\sqrt{x^2+6x+9}=3x-1\)

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

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

\(\Leftrightarrow x+3=3x-1\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)

\(a,\sqrt{1-4x+4x^2}=5\\ \Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\\ \Leftrightarrow\left|1-2x\right|=5\)

\(TH_1:x\le\dfrac{1}{2}\)

\(1-2x=5\\ \Leftrightarrow x=-2\left(tm\right)\)

\(TH_2:x\ge\dfrac{1}{2}\)

\(-1+2x=5\\ \Leftrightarrow x=3\left(tm\right)\)

Vậy \(S=\left\{-2;3\right\}\)

\(b,\sqrt{x^2+6x+9}=3x-1\\ \Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-1\\ \Leftrightarrow\left|x+3\right|=3x-1\)

\(TH_1:x\ge-3\\ x+3=3x-1\\ \Leftrightarrow-2x=-4\Leftrightarrow x=2\left(tm\right)\)

\(TH_2:x< 3\\ -x-3=3x-1\\ \Leftrightarrow-4x=2\\ \Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)

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

Sai đề rồi bạn ơi!!!

Lời giải:

ĐKXĐ: \(x\geq \frac{5}{3}\)

Ta có: \(4\sqrt{3x-5}-8=0\Leftrightarrow 4\sqrt{3x-5}=8\Leftrightarrow \sqrt{3x-5}=2\)

\(\Rightarrow 3x-5=4\Rightarrow x=\frac{4+5}{3}=3\) (thỏa mãn)

Vậy $x=3$

Đk: \(x\ge-\frac{1}{4}\)

pt <=> \(4x^2+4x+2=2\sqrt{4x-1}\)

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

Đặt \(\sqrt{2\left(2x+1\right)-1}=a\left(a\ge0\right)\)

Ta có hệ \(\left\{{}\begin{matrix}\left(2x+1\right)^2+1=2a\left(1\right)\\a^2+1=2\left(2x+1\right)\left(2\right)\end{matrix}\right.\)

Từ (1),(2)=> \(\left(2x+1\right)^2-a^2=2a-2\left(2x+1\right)\)

<=> \(\left(2x+1-a\right)\left(2x+1+a\right)=-2\left(2x+1-a\right)\)

<=> \(\left(2x+1-a\right)\left(2x+1+a\right)+2\left(2x+1-a\right)=0\)

<=> \(\left(2x+1-a\right)\left(2x+a+3\right)=0\)( *)

vì \(x\ge-\frac{1}{4}\) và \(a\ge0\)=> \(2x+a+3\ge2.\frac{-1}{4}+0+3=\frac{5}{2}>0\)

(*) => \(2x+1-a=0\)

<=> \(2x+1=a\)

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

=> \(4x^2+4x+1=2\left(2x+1\right)-1\)

<=> \(4x^2+4x+1-4x-1=0\)

<=> \(4x^2=0\)

<=> x=0 (t/m)