ĐK: \(x\ge\frac{1}{2}\)

Đặt \(t=\sqrt{2x-1}\Leftrightarrow x=\frac{t^2+1}{2}\)(ĐK: \(t\ge0\)) thay vao phương trình ta được:

\(\sqrt{\frac{t^2+1}{2}+4+3t}\)+\(\sqrt{\frac{t^2+1}{2}+12-5t}=7\sqrt{2}\)

\(\Leftrightarrow\sqrt{\frac{t^2+6t+9}{2}}+\sqrt{\frac{t^2-10t+25}{2}}=7\sqrt{2}\)

\(\Leftrightarrow\frac{\sqrt{\left(t+3\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(t-5\right)^2}}{\sqrt{2}}=7\sqrt{2}\)

\(\Leftrightarrow\frac{\left|t+3\right|+\left|t-5\right|}{\sqrt{2}}=7\sqrt{2}\)

\(\Leftrightarrow t+3+\left|t-5\right|=14\)(vì \(t\ge0\Rightarrow t+3>0\))

\(\Leftrightarrow t+\left|t-5\right|=11\)

Xét TH: \(t-5\ge0\Leftrightarrow t\ge5\) thì ta có:

\(t+t-5=11\)

\(\Leftrightarrow2t=16\)

\(\Leftrightarrow t=8\)(chọn)

Xét TH: \(t-5< 0\Leftrightarrow t< 5\) thì ta có:

\(t-t+5=11\)

\(\Leftrightarrow5=11\)(vô lí nên loại)

Lại có: \(t=8\)

\(\Leftrightarrow\sqrt{2x-1}=8\)

\(\Leftrightarrow2x-1=64\)

\(\Leftrightarrow2x=63\)

\(\Leftrightarrow x=\frac{63}{2}=31\frac{1}{2}\)

Vậy nghiệm của phương trình là x=31\(\frac{1}{2}\)

32,5

ĐKXĐ : \(\left\{{}\begin{matrix}x+4\ge0\\1-x\ge0\\1-2x\ge0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge-4\\x\le1\\x\le0,5\end{matrix}\right.\)

=> \(-4\le x\le0,5\)

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

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

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

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

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

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

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

ĐKXĐ : \(x+2\ge0\)

\(x\ge-2\)

=> ĐKXĐ là : \(-2\le x\le0,5\)

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

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

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

<=> \(-7x-2x^2=0\)

<=> \(x\left(7+2x\right)=0\)

<=> \(\left\{{}\begin{matrix}x=0\\7+2x=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=0\left(TM\right)\\x=-\frac{7}{2}\left(L\right)\end{matrix}\right.\)

Vậy phương trình trên có nghiệm là x = 0 .

\(ĐK:\left\{{}\begin{matrix}x+4\ge0\\1-x\ge0\\1-2x\ge0\end{matrix}\right.\Leftrightarrow-4\le x\le\frac{1}{2}\)

Phương trình đc viết dưới dạng:

\(\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\Leftrightarrow\sqrt{\left(x+4\right)\left(1-x\right)}=2+x\\ \Leftrightarrow2+x\ge0\\ \left(x+4\right)\left(1-x\right)=\left(2+x\right)^2\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\2x^2+5x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\x=0\\x=-\frac{5}{2}\end{matrix}\right.\Leftrightarrow x=0\)

Vậy phương trình có nghiệm \(x=0\)

ĐKXĐ : \(\left\{{}\begin{matrix}x+9\ge0\\2x+4\ge0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x\ge-9\\2x\ge-4\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x\ge-9\\x\ge-2\end{matrix}\right.\)

<=> \(x\ge-2\)

Ta có : \(\sqrt{x+9}=5-\sqrt{2x+4}\)

<=> \(\sqrt{x+9}+\sqrt{2x+4}=5\)

<=> \(\left(\sqrt{x+9}+\sqrt{2x+4}\right)^2=5^2\)

<=> \(\left(x+9\right)+2\sqrt{\left(x+9\right)\left(2x+4\right)}+\left(2x+4\right)=25\)​

ĐKXĐ : \(x\le4\)

=> \(-2\le x\le4\)

<=> \(x+9+2\sqrt{\left(x+9\right)\left(2x+4\right)}+2x+4=25\)

<=> \(2\sqrt{\left(x+9\right)\left(2x+4\right)}=25-x-9-2x-4\)

<=> \(2\sqrt{\left(x+9\right)\left(2x+4\right)}=12-3x\)

<=> \(\left(2\sqrt{\left(x+9\right)\left(2x+4\right)}\right)^2=\left(12-3x\right)^2\)

<=> \(4\left(x+9\right)\left(2x+4\right)=\left(12-3x\right)^2\)

<=> \(4\left(2x^2+18x+4x+36\right)=144-72x+9x^2\)

<=> \(8x^2+72x+16x+144=144-72x+9x^2\)

<=> \(8x^2+72x+16x+144-9x^2-144+72x=0\)

<=> \(-x^2+160x=0\)

<=> \(x\left(160-x\right)=0\)

<=> \(\left\{{}\begin{matrix}x=0\\160-x=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=0\left(TM\right)\\x=160\left(L\right)\end{matrix}\right.\)

Vậy phương trình trên có nghiệm là x = 0 .

\( \sqrt {2x + 1} + \sqrt {5 - x} = \sqrt {5x - 4} \left( {5 \ge x \ge \dfrac{4}{5}} \right)\\ \Leftrightarrow 2x + 1 + 2\sqrt {\left( {2x + 1} \right)\left( {5 - x} \right)} + 5 - x = 5x - 4\\ \Leftrightarrow 2\sqrt {9x - 2{x^2} + 5} = 4x - 10\\ \Leftrightarrow \sqrt {9x - 2{x^2} + 5} = 2x - 5\\ \Leftrightarrow 9x - 2{x^2} + 5 = 4{x^2} - 20x + 25\\ \Leftrightarrow 6{x^2} - 29x + 20 = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 4\left( {tm} \right)\\ x = \dfrac{5}{6}\left( {ktm} \right) \end{array} \right. \)

\(\sqrt{3x-2}+\sqrt{3+x}=\sqrt{5x+4}\)

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

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

➝ \(4x+3+2\sqrt{6x+2x^2-6-2x}=5x+4\)

→ \(2\sqrt{2x^2+4x-6}=5x+4-4x-3\)

→ \(2\sqrt{2x^2+4x-6}=x+1\)

→ \(\left(2\sqrt{2x^2+4x-6}\right)^2=\left(x+1\right)^2\)

→ \(4\left(2x^2+4x-6\right)=x^2+2x+1\)

→ \(8x^2+16x-24=x^2+2x+1\)

→ \(8x^2+16x-24-x^2-2x-1=0\)

→ \(7x^2+14x-25=0\)

→ \(x_1=\frac{-7+4\sqrt{14}}{7}\)

\(x_2=\frac{-7-4\sqrt{14}}{7}\)

\( \sqrt {3x - 2} + \sqrt {3 + x} = \sqrt {5x + 4} \left( {x \ge \dfrac{2}{{3}}} \right)\\ \Leftrightarrow 3x - 2 + 2\sqrt {\left( {3x - 2} \right)\left( {3 + x} \right)} + 3 + x = 5x + 4\\ \Leftrightarrow 2\sqrt {7x + 3{x^2} - 6} = x + 3\\ \Leftrightarrow 4\left( {7x + 3{x^2} - 6} \right) = {x^2} + 6x + 9\\ \Leftrightarrow 28x + 12{x^2} - 24 = {x^2} + 6x + 9\\ \Leftrightarrow {x^2} + 2x - 3 = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 1\left( {tm} \right)\\ x = - 3\left( {ktm} \right) \end{array} \right. \)

ĐKXĐ : \(\left\{{}\begin{matrix}3x-2\ge0\\3+x\ge0\\5x+4\ge0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge\frac{2}{3}\\x\ge-3\\x\ge-\frac{4}{5}\end{matrix}\right.\)

=> \(x\ge\frac{2}{3}\) (1)

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

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

<=> \(\left(3x-2\right)+2\sqrt{\left(3x-2\right)\left(3+x\right)}+\left(3+x\right)=5x+4\)

<=> \(3x-2+2\sqrt{\left(3x-2\right)\left(3+x\right)}+3+x=5x+4\)

<=> \(2\sqrt{\left(3x-2\right)\left(3+x\right)}=5x+4+2-3-x-3x\)

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

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

ĐKXĐ : \(\frac{x+3}{2}\ge0\)

=> \(x+3\ge0\)

=> \(x\ge-3\) (2)

Từ (1) và (2)

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

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

<=> \(\left(3x-2\right)\left(3+x\right)=\frac{\left(x+3\right)^2}{4}\)

<=> \(9x-6+3x^2-2x=\frac{x^2+6x+9}{4}\)

<=> \(\frac{4\left(9x-6+3x^2-2x\right)}{4}=\frac{x^2+6x+9}{4}\)

<=> \(4\left(9x-6+3x^2-2x\right)=x^2+6x+9\)

<=> \(36x-24+12x^2-8x=x^2+6x+9\)

<=> \(36x-24+12x^2-8x-x^2-6x-9=0\)

<=> \(22x-33+11x^2=0\)

<=> \(11x^2+33x-11x-33=0\)

<=> \(11x\left(x-1\right)+33\left(x-1\right)=0\)

<=> \(\left(11x+33\right)\left(x-1\right)=0\)

<=> \(\left\{{}\begin{matrix}11x+33=0\\x-1=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=-3\left(L\right)\\x=1\left(TM\right)\end{matrix}\right.\)

Vậy phương trình trên có nghiệm là x = 1 .

Vũ Minh TuấnLê Thị Thục Hiền@Nk>↑@

\(\sqrt{x-2+\sqrt{2x-5}}+\sqrt{x+2+3\sqrt{2x-5}}=7\sqrt{2}\) (*) (đk : \(x\ge\frac{5}{2}\))

Đặt \(\sqrt{2x-5}=a\left(a\ge0\right)\)

=> 2x-5=a²

<=> \(x=\frac{a^2+5}{2}\)

Có \(\sqrt{\frac{a^2+5}{2}-2+a}+\sqrt{\frac{a^2+5}{2}+2+3a}=7\sqrt{2}\)

<=> \(\sqrt{\frac{a^2+5-4+2a}{2}}+\sqrt{\frac{a^2+5+4+6a}{2}}=7\sqrt{2}\)

<=>\(\sqrt{\frac{a^2+2a+1}{2}}+\sqrt{\frac{a^2+6a+9}{2}}=7\sqrt{2}\)

<=> \(\frac{\sqrt{\left(a+1\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(a+3\right)^2}}{\sqrt{2}}=7\sqrt{2}\)

<=> \(\left|a+1\right|+\left|a+3\right|=7\sqrt{2}.\sqrt{2}\)

<=> \(a+1+a+3=14\)(do a\(\ge\)0)

<=> \(2a=10\) <=> a=5(t/m)

<=> \(\sqrt{2x-5}=5\)

<=> \(2x-5=25\) <=> \(x=15\)(tm pt (*))

Vậy pt (*) có tập nghiệm \(S=\left\{15\right\}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x+1\ge0\\2x+3\ge0\\x+20\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ge-\frac{3}{2}\\x\ge-20\end{matrix}\right.\)

\(\sqrt{x+1}+\sqrt{2x+3}=\sqrt{x+20}\)

\(\Leftrightarrow\left(\sqrt{x+1}+\sqrt{2x+3}\right)^2=\left(\sqrt{x+20}\right)^2\)

\(\Leftrightarrow x+1+2\sqrt{\left(x+1\right)\left(2x+3\right)}+2x+3=x+20\)

\(\Leftrightarrow3x+4+2\sqrt{\left(x+1\right)\left(2x+3\right)}=x+20\)

\(\Leftrightarrow2\sqrt{\left(x+1\right)\left(2x+3\right)}=-2x+16\)

\(\Leftrightarrow2\sqrt{2x^2+5x+3}=16-2x\)

\(\Leftrightarrow\left\{{}\begin{matrix}16-2x\ge0\\4\left(2x^2+5x+3\right)=\left(16-2x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le8\\8x^2+20x+12=256-64x+4x^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le8\\4x^2+84x-244=0\end{matrix}\right.\)

còn lại bn tự làm nha