Câu 2,3,4 nx thôi ạ. Câu 1 có bạn giúp r ạ 

1)\(\sqrt{4x^2+12x+9}=2-x\)

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

\(\Leftrightarrow\left|2x+3\right|=2-x\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=2-x\\2x+3=x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)

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2)\(\sqrt{x^4+2x^2+1}=x^2+5x+4\)       ĐK:\(x\ge-1\)

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

\(\Leftrightarrow\left|x^2+1\right|=x^2+5x+4\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=x^2+5x+4\\x^2+1=-x^2-5x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-3\\2x^2+5x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\2\left(x+\dfrac{5}{4}\right)^2+\dfrac{15}{8}=0\left(voli\right)\end{matrix}\right.\)

a.

ĐKXĐ: \(x\ge0\)

\(\sqrt{2x^2+13x+5}-5\sqrt{x}+\sqrt{2x^2-3x+5}-3\sqrt{x}=0\)

\(\Leftrightarrow\dfrac{2x^2-12x+5}{\sqrt{2x^2+13x+5}+5\sqrt{x}}+\dfrac{2x^2-12x+5}{\sqrt{2x^2-3x+5}+3\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-12x+5\right)\left(\dfrac{1}{\sqrt{2x^2+13x+5}+5\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-3x+5}+3\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-12x+5=0\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(x^2\ge\dfrac{4}{3}\)

\(\sqrt{x^2-\dfrac{4}{3}}+\sqrt{4x^2-4}-x=0\)

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

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

\(\Leftrightarrow3x^2-4=0\)

\(\Leftrightarrow...\)

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

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

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

1/ ĐKXĐ: \(x\ge1\)

\(\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-2.3\sqrt{x-1}+9}=1\)

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

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

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

Mà \(\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}-2+3-\sqrt{x-1}\right|=1\)

Dấu "=" xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}\sqrt{x-1}\ge2\\\sqrt{x-1}\le3\end{matrix}\right.\) \(\Rightarrow5\le x\le10\)

Vậy phương trình nghiệm đúng với mọi \(x\in\left[5;10\right]\)

2/ ĐKXĐ: \(x\ge\dfrac{5}{2}\)

Nhân 2 vế với \(\sqrt{2}\) ta được:

\(\sqrt{2x+4+6\sqrt{2x-5}}+\sqrt{2x-4-2\sqrt{2x-5}}=4\)

\(\Leftrightarrow\sqrt{2x-5+2.3\sqrt{2x-5}+9}+\sqrt{2x-5-2\sqrt{2x-5}+1}=4\)

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

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

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

TH1: \(\sqrt{2x-5}\ge1\Rightarrow\sqrt{2x-5}+\sqrt{2x-5}-1=1\)

\(\Leftrightarrow\sqrt{2x-5}=1\Rightarrow2x=6\Rightarrow x=3\)

TH2: \(\sqrt{2x-5}< 1\Rightarrow\sqrt{2x-5}+1-\sqrt{2x-5}=1\Leftrightarrow1=1\) (đúng với mọi \(\dfrac{5}{2}\le x< 3\))

Vậy nghiệm của phương trình là \(\dfrac{5}{2}\le x\le3\)

@Nguyễn Việt Lâm

@Võ Hồng Phúc

\(b,\sqrt{8+\sqrt{x}}+\sqrt{5-\sqrt{x}}=5\) \(Đkxđ:0\le\sqrt{x}\le5\)

Phương trình trên tương đương với:

\(\sqrt{8+t}+\sqrt{5-t}=5\left(\sqrt{x}=t\right)\)

\(\Leftrightarrow13+2\sqrt{\left(8+t\right)\left(5-t\right)}=25\)

\(\Leftrightarrow\sqrt{40-3t-t^2}=6\)

\(\Leftrightarrow t^2+3t-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t_1=1\\t_2=-4\left(ktm\right)\end{matrix}\right.\)

\(\Leftrightarrow x=1\)

Vậy ............

Tham khảo:

1) Giải phương trình : \(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\) - Hoc24

\(\sqrt{x+3}+2\sqrt{x}=2+\sqrt{x\left(x+3\right)}\left(đk:x\ge0\right)\)

\(\Leftrightarrow x+3+4x+4\sqrt{x\left(x+3\right)}=4+x\left(x+3\right)+4\sqrt{x\left(x+3\right)}\)

\(\Leftrightarrow5x+3=4+x^2+3x\)

\(\Leftrightarrow x^2-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\left(tm\right)\)