Đề sai à :)

ĐKXĐ: \(x\ge1\)

Bình phương 2 vế ta có:

                         \(x-1=4\left(2x+5\right)\Leftrightarrow-7x=21\Leftrightarrow x=-3\)

Kết quả x=-3 không thỏa mãn ĐKXĐ .

Vậy PT vô nghiệm

Đề câu c ptrinh = 4 là phải riêng ra chứ

\(a,\frac{3x+2}{\sqrt{x+2}}=2\sqrt{x+2}\)

\(\Rightarrow3x+2=2\sqrt{x+2}.\sqrt{x+2}\)

\(\Rightarrow3x+2=2\left(x+2\right)\)

\(\Rightarrow3x+2=2x+4\)

\(\Rightarrow3x-2x=4-2\)

\(\Rightarrow x=2\)

\(b,\sqrt{4x^2-1}-2\sqrt{2x+1}=0\)

\(\Rightarrow\sqrt{\left(2x+1\right)\left(2x-1\right)}-2\sqrt{2x+1}=0\)

\(\Rightarrow\sqrt{2x+1}\left(\sqrt{2x-1}-2\right)=0\)

\(\Rightarrow\hept{\begin{cases}\sqrt{2x+1}=0\\\sqrt{2x-1}-2=0\end{cases}\Rightarrow\orbr{\begin{cases}2x+1=0\\\sqrt{2x-1}=2\end{cases}\Rightarrow}\orbr{\begin{cases}2x=-1\\2x-1=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\2x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{2}\end{cases}}}\)

\(c,\sqrt{x-2}+\sqrt{4x-8}-\frac{2}{5}\sqrt{\frac{25x-50}{4}}=4\)

\(\Rightarrow\sqrt{x-2}+\sqrt{4\left(x-2\right)}-\frac{2}{5}\sqrt{\frac{25\left(x-2\right)}{4}}=4\)

\(\Rightarrow\sqrt{x-2}+2\sqrt{x-2}-\frac{2}{5}.\frac{5\sqrt{x-2}}{2}=4\)

\(\Rightarrow\sqrt{x-2}+2\sqrt{x-2}-\sqrt{x-2}=4\)

\(\Rightarrow2\sqrt{x-2}=4\)

\(\Rightarrow\sqrt{x-2}=2\)

\(\Rightarrow x-2=4\)

\(\Rightarrow x=6\)

\(d,\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\)

\(\Rightarrow\sqrt{x+4}=\sqrt{1-2x}+\sqrt{1-x}\)

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

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

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

\(\Rightarrow4x+2=2\sqrt{1-3x+2x^2}\)

\(\Rightarrow2x+1=\sqrt{1-3x+2x^2}\)

\(\Rightarrow4x^2+4x+1=1-3x+2x^2\)

\(\Rightarrow4x^2-2x^2+4x+3x+1-1=0\)

\(\Rightarrow2x^2+7x=0\)

\(\Rightarrow x\left(2x+7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\2x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-7}{2}\end{cases}}}\)

\(e,\frac{2x}{\sqrt{5}-\sqrt{3}}-\frac{2x}{\sqrt{3}+1}=\sqrt{5}+1\)

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

\(\Rightarrow x\left(\sqrt{5}+\sqrt{3}\right)-x\left(\sqrt{3}-1\right)=\sqrt{5}+1\)

\(\Rightarrow\sqrt{5}x+\sqrt{3}x-\sqrt{3x}+x=\sqrt{5}+1\)

\(\Rightarrow\sqrt{5}x+x=\sqrt{5}+1\)

\(\Rightarrow x\left(\sqrt{5}+1\right)=\sqrt{5}+1\)

\(\Rightarrow x=1\)

b) \(\sqrt{2x+1}.\sqrt{2x-1}-2\sqrt{2x+1}=0\)đkxđ: x>= 1/2

<=> \(\sqrt{2x+1}.\left(\sqrt{2x-1}-2\right)=0\)

<=> \(\sqrt{2x-1}-2=0\)

<=> \(\sqrt{2x-1}=2\)

<=> \(2x-1=4\)
<=> x=5/2 ( tm đkxđ)
Vậy x=5/2

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

\(\Delta=b^2-4ac=4-4\left(-\sqrt{3}+1\right)=4\sqrt{3}>0\)

\(\rightarrow\)Phương trình có 2 nghiệm phân biệt

Theo vi-ét ta có :

\(\left\{{}\begin{matrix}S=x_1+x_2=-\dfrac{b}{a}=2\\P=x_1x_2=\dfrac{c}{a}=-\sqrt{3}+1\end{matrix}\right.\)

\(M=x_1^2x_2^2-2x_1x_2-x_1-x_2\)

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

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

\(=0\)

a) \(\sqrt{4+2x-x^2}=x-2\)

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

\(\Leftrightarrow4+2x-x^2=x^2-4x+4\)

\(\Leftrightarrow-x^2+6x=0\)

\(\Leftrightarrow x\left(6-x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\6-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}\)

hình như bài này sai đó! em mới học lớp 8 thôi !

lê thị thu huyền:

sai rồi đó em, nhưng mà nhờ em chị mới biết chị sai chỗ nào. Không hiểu đầu óc kiểu gì mà lại thấy 2x+4x=8x mới chết chứ !!!

đk: \(\left\{{}\begin{matrix}x+1\ge0\\x^2-x+1\ge0\\x+3\ge0\\x\ne-3\end{matrix}\right.\)

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

phương trình trở thành:

\(\dfrac{ab}{c}+a=b+c\Leftrightarrow ab+ac=bc+c^2\Leftrightarrow\left(a-c\right)\left(b+c\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=c\\b+c=0\left(@@\right)\end{matrix}\right.\)

Vì b >/ 0 và c > 0 => b+c >0 nên phương trình (@@) vô nghiệm

\(\Rightarrow\sqrt{x+1}=\sqrt{x+3}\Leftrightarrow x+1=x+3\left(vn\right)\)

kết luận: ptvn

(không lẽ vô nghiệm thật hả trời T_T!!)

\(ĐK:x\ge-1,x\ne-3\)

\(PT\Leftrightarrow\sqrt{\dfrac{x^3+1}{x+3}}-\sqrt{x+3}=\sqrt{x^2-x+1}-\sqrt{x+1}\\ \Rightarrow\left(\sqrt{\dfrac{x^3+1}{x+3}}-\sqrt{x+3}\right)^2=\left(\sqrt{x^2-x+1}-\sqrt{x+1}\right)^2\\ \Leftrightarrow\dfrac{x^3+1}{x+3}+x+3-2\sqrt{x^3+1}=x^2-x+1+x+1-2\sqrt{x^3+1}\\ \Leftrightarrow\dfrac{x^3+1}{x+3}-x^2+x+1=0\Leftrightarrow x^3+1-\left(x+3\right)\left(x^2-x-1\right)=0\Leftrightarrow x^3+1-x^3+x^2+x-3x^2+3x+3=0\\ \Leftrightarrow x^2-2x-2=0\Leftrightarrow\left(x-1\right)^2=3\Leftrightarrow x=\pm\sqrt{3}+1\)Thử lại thấy không đúng=>phương trình vô nghiệm

\(1,\sqrt{x+2+4\sqrt{x-2}}=5\left(x\ge2\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-2}+4\right)^2}=5\\ \Leftrightarrow\sqrt{x-2}+4=5\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\Leftrightarrow x=3\\ 2,\sqrt{x+3+4\sqrt{x-1}}=2\left(x\ge1\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}+4\right)^2}=2\\ \Leftrightarrow\sqrt{x-1}+4=2\\ \Leftrightarrow\sqrt{x-1}=-2\\ \Leftrightarrow x\in\varnothing\left(\sqrt{x-1}\ge0\right)\)

\(3,\sqrt{x+\sqrt{2x-1}}=\sqrt{2}\left(x\ge\dfrac{1}{2};x\ne1\right)\\ \Leftrightarrow x+\sqrt{2x-1}=2\\ \Leftrightarrow x-2=-\sqrt{2x-1}\\ \Leftrightarrow x^2-4x+4=2x-1\\ \Leftrightarrow x^2-6x+5=0\\ \Leftrightarrow\left(x-5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(loại\right)\end{matrix}\right.\)

\(4,\sqrt{x-2+\sqrt{2x-5}}=3\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}=6\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}=6\\ \Leftrightarrow\sqrt{2x-5}+1=6\\ \Leftrightarrow\sqrt{2x-5}=5\\ \Leftrightarrow2x-5=25\Leftrightarrow x=15\left(TM\right)\)