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 \(S=\left\{15\right\}\)

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

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

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 .

ĐKXĐ : \(\left\{{}\begin{matrix}2x+9\ge0\\4-x\ge0\\3x+1\ge0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}2x\ge-9\\-x\ge-4\\3x\ge-1\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x\ge-\frac{9}{2}\\x\le4\\x\ge-\frac{1}{3}\end{matrix}\right.\)

<=> \(4\ge x\ge-\frac{1}{3}\)

Ta có : \(\sqrt{2x+9}=\sqrt{4-x}+\sqrt{3x+1}\)

<=> \(\left(\sqrt{2x+9}\right)^2=\left(\sqrt{4-x}+\sqrt{3x+1}\right)^2\)

<=> \(2x+9=\left(4-x\right)+2\sqrt{\left(4-x\right)\left(3x+1\right)}+\left(3x+1\right)\)

<=> \(2x+9=4-x+2\sqrt{12x-3x^2+4-x}+3x+1\)

<=> \(2x+9-4+x-3x-1=2\sqrt{12x-3x^2+4-x}\)

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

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

<=> \(16=4\left(12x-3x^2+4-x\right)\)

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

<=> \(0=12x-3x^2-x\)

<=> \(0=11x-3x^2\)

<=> \(0=x\left(11-3x\right)\)

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

ĐKXĐ : x> -2

\(\sqrt{2x+\sqrt{6x^2+1}}\) = x + 1

=> (\(\sqrt{2x+\sqrt{6x^2+1}}\))² = (x+1)²

=> 2x+\(\sqrt{6x^2+1}\) = x²+2x+1

=> \(\sqrt{6x^2+1}\) = x²+1

=> 6x² +1 = (x²+1)(x²+1)

=> 6x² +1 = x⁴+2x²+1

=> -x⁴+4x² = 0

=> x²(4-x²) = 0

=>x²(2-x)(2+x) = 0

=> x² =0, 2-x=0 , 2+x =0

=> x=0(TMĐKXĐ)

x=2(TMĐKXĐ)

x= -2 (KTMĐKXĐ)

Vậy ........

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

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

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

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

→ \(2x+1=x^2-6x+9\)

→ \(2x+1-x^2+6x-9=0\)

→ \(-x^2+8x-8=0\rightarrow x^2-8x+8=0\)

→ \(x_1=4+2\sqrt{2}\)

\(x_2=4-2\sqrt{2}\)

ĐK: \(2x+1\ge0\Leftrightarrow x\ge-\frac{1}{2}\)

\(pt\Leftrightarrow2x+1=\left(x-3\right)^2\\ \Leftrightarrow2x+1=x^2-6x+9\\ \Leftrightarrow x^2-8x+8=0\\ \Leftrightarrow x^2-2.x.4+4^2-4^2+8=0\\ \Leftrightarrow\left(x-4\right)^2-8=0\\ \Leftrightarrow\left(x-4-2\sqrt{2}\right)\left(x-4+2\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-4-2\sqrt{2}=0\\x-4+2\sqrt{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4+2\sqrt{2}\\x=4-2\sqrt{2}\end{matrix}\right.\)

Vậy...............................

\(\sqrt{2x+1}=x-3\) (ĐKXĐ:\(x\ge\frac{1}{2}\))

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

\(\Leftrightarrow2x+1=x^2-6x+9\)

\(\Leftrightarrow-x^2+8x-8=0\)

\(\Leftrightarrow-\left(x^2-8x+16-8\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-4-2\sqrt{2}=0\\x-4+2\sqrt{2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4+2\sqrt{2}\\x=4-2\sqrt{2}\end{matrix}\right.\)

Ta có \(x=4+2\sqrt{2}\) thỏa mãn ĐKXĐ

\(x=4-2\sqrt{2}\) không thỏa mãn ĐKXĐ

nên nghiệm của pt trên là \(x=4+2\sqrt{2}\)

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 \(x=0\)