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

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

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

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

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

\(\)

1: ĐKXĐ: x<=31

\(\sqrt{31-x}=x-1\)

=>\(\begin{cases}x-1\ge0\\ \left(x-1\right)^2=31-x\end{cases}\Rightarrow\begin{cases}x\ge1\\ x^2-2x+1-31+x=0\end{cases}\)

=>\(\begin{cases}1\le x\le31\\ x^2-x-30=0\end{cases}\Rightarrow\begin{cases}1\le x\le31\\ \left(x-6\right)\left(x+5\right)=0\end{cases}\Rightarrow x=6\)

3: ĐKXĐ: x∈R
\(\sqrt{x^2-3x+5}+x=3x+7\)

=>\(\sqrt{x^2-3x+5}=2x+7\)

=>\(\begin{cases}2x+7\ge0\\ \left(2x+7\right)^2=x^2-3x+5\end{cases}\Rightarrow\begin{cases}x\ge-\frac72\\ 4x^2+28x+49-x^2+3x-5=0\end{cases}\)

=>\(\begin{cases}x\ge-\frac72\\ 3x^2+31x+44=0\end{cases}\Rightarrow\begin{cases}x\ge-\frac72\\ x^2+\frac{31}{3}+\frac{44}{3}=0\end{cases}\)

=>\(\begin{cases}x\ge-\frac72\\ x^2+2\cdot x\cdot\frac{31}{6}+\frac{961}{36}=\frac{433}{36}\end{cases}\Rightarrow\begin{cases}x\ge-\frac72\\ \left(x+\frac{31}{6}\right)^2=\frac{433}{36}\end{cases}\)

=>\(\begin{cases}x\ge-\frac72\\ x+\frac{31}{6}\in\left\lbrace\frac{\sqrt{433}}{6}^{\prime};-\frac{\sqrt{433}}{6}\right\rbrace\end{cases}\Rightarrow\begin{cases}x\ge-\frac72\\ x\in\left\lbrace\frac{\sqrt{433}-31}{6};\frac{-\sqrt{433}-31}{6}\right\rbrace\end{cases}\)

=>\(x=\frac{\sqrt{433}-31}{6}\)

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

5: ĐKXĐ: \(\frac{x+3}{x-7}>0\)

=>x>7 hoặc x<-3

Ta có: \(\left(x-7\right)\cdot\sqrt{\frac{x+3}{x-7}}=x+4\)

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

=>\(\begin{cases}x+4\ge0\\ \left(x+3\right)\left(x-7\right)=\left(x+4\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-4\\ x^2-4x-21=x^2+8x+16\end{cases}\)

=>\(\begin{cases}x\ge-4\\ -12x=37\end{cases}\Rightarrow x=-\frac{37}{12}\) (nhận)

6: ĐKXĐ: x>=4

Ta có: \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)

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

=>\(\sqrt{2x-3}=\sqrt{x-1}\)

=>2x-3=x-1

=>2x-x=-1+3

=>x=2(loại)

7: ĐKXĐ: x>=1

Ta có: \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\frac{x+3}{2}\)

=>\(\sqrt{x-1+2\cdot\sqrt{x-1}+1}+\sqrt{x-1-2\cdot\sqrt{x-1}\cdot1+1}=\frac{x+3}{2}\)

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

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

TH1: \(\sqrt{x-1}-1\ge0\)

=>\(\sqrt{x-1}\ge1\)

=>x-1>=1

=>x>=2

(1) sẽ trở thành: \(\sqrt{x-1}+1+\sqrt{x-1}-1=\frac{x+3}{2}\)

=>\(2\sqrt{x-1}=\frac{x+3}{2}\)

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

=>\(16\left(x-1\right)=\left(x+3\right)^2\)

=>\(x^2+6x+9=16x-16\)

=>\(x^2-10x+25=0\)

=>\(\left(x-5\right)^2=0\)

=>x-5=0

=>x=5(nhận)

TH2: \(\sqrt{x-1}-1<0\)

=>\(\sqrt{x-1}<1\)

=>0<=x-1<1

=>1<=x<2

(1) sẽ trở thành: \(\sqrt{x-1}+1+1-\sqrt{x-1}=\frac{x+3}{2}\)

=>\(\frac{x+3}{2}=2\)

=>x+3=4

=>x=1(nhận)

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

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

ghê thậc, còn cái còn lại thì seo?

1: ĐKXĐ: x>=8/3

\(\sqrt{3x-8}-\sqrt{x+1}=\frac{2x-11}{5}\)

=>\(\sqrt{3x-8}-1+2-\sqrt{x+1}=\frac{2x-11}{5}+1\)

=>\(\frac{3x-8-1}{\sqrt{3x-8}+1}+\frac{4-x-1}{2+\sqrt{x+1}}=\frac{2x-11+5}{5}\)

=>\(\left(x-3\right)\left(\frac{3}{\sqrt{3x-8}+1}-\frac{1}{2+\sqrt{x+1}}-\frac25\right)=0\)

=>x-3=0

=>x=3(nhận)

3: ĐKXĐ: -5/2<=x<=5/2

Đặt \(a=\sqrt{5+2x};b=\sqrt{5-2x}\)

=>\(ab=\sqrt{\left(5+2x\right)\left(5-2x\right)}=\sqrt{25-4x^2}\)

Theo đề, ta có: a+b+5=3ab

=>3ab-a-b-5=0

=>a(3b-1)-b+1/3-16/3=0

=>\(3a\left(b-\frac13\right)-\left(b-\frac13\right)=\frac{16}{3}\)

=>\(\left(b-\frac13\right)\left(3a-1\right)=\frac{16}{3}\)

=>(3a-1)(3b-1)=16

=>(3a-1;3b-1)∈{(1;16);(16;1);(2;8);(8;2);(4;4)}

=>(3a;3b)∈{(2;17);(17;2);(3;9);(9;3);(5;5)}

=>(a;b)∈{(2/3;17/3);(17/3;2/3);(1;3);(3;1);(5/3;5/3)}

mà a<>b

nên (a;b)∈{(2/3;17/3);(17/3;2/3);(1;3);(3;1)}

TH1: a=2/3 và b=17/3

=>\(\begin{cases}5+2x=\frac49\\ 5-2x=\frac{289}{9}\end{cases}\Rightarrow\begin{cases}2x=\frac49-5=\frac49-\frac{45}{9}=-\frac{41}{9}\\ 2x=5-\frac{289}{9}=-\frac{244}{9}\end{cases}\)

=>x∈∅

TH2: a=17/3 và b=2/3

=>\(\begin{cases}5+2x=\frac{289}{9}\\ 5-2x=\frac49\end{cases}\Rightarrow\begin{cases}2x=\frac{289}{9}-5=\frac{244}{9}\\ 2x=5-\frac49=\frac{41}{9}\end{cases}\)

=>x∈∅

TH3: a=1 và b=3

=>5+2x=1 và 5-2x=9

=>2x=-4 và 2x=5-9=-4

=>x=-2(nhận)

TH4: a=3 và b=1

=>5+2x=9 và 5-2x=1

=>2x=4 và 2x=4

=>x=2(nhận)