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

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

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

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

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

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

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

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

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

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

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

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

còn lại bn tự làm nha

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 )

\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=5\)

\(\Leftrightarrow\sqrt{x-1-2.\sqrt{x-1}.2+4}+\sqrt{x-1-2.\sqrt{x-1}.3+9}=5\)

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

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

bạn giải tiếp nhé

Ta có:\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}\)(ĐK: \(x\ge1\))

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

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

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

Thay vào phương trình ta được:

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

\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\sqrt{x-1}+3=5\)(vì \(\sqrt{x-1}\ge0\Rightarrow\sqrt{x-1}+3>0\))

-TH: \(\sqrt{x-1}-2\ge0\Leftrightarrow\sqrt{x-1}\ge2\Leftrightarrow x-1\ge4\Leftrightarrow x\ge3\)thì ta có:

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

\(\Leftrightarrow2\sqrt{x-1}=4\)

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

\(\Leftrightarrow x-1=4\)

\(\Leftrightarrow x=5\)

-TH:\(\sqrt{x-1}-2< 0\Leftrightarrow x< 3\) thì ta có:

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

\(\Leftrightarrow5=5\)(luôn đúng \(\forall1\le x< 3\))

Vậy nghiệm của phương trình là \(1\le x< 3\) và \(x=5\)