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

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

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

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

X=7,3267

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 : \(x-1\ge0\)

=> \(x\ge1\)

Ta có : \(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}=5\)

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

<=> \(\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}+\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}=5\)

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

<=> \(|\sqrt{x-1}-1|+|\sqrt{x-1}+1|=5\)

<=> \(|\sqrt{x-1}-1|+\sqrt{x-1}+1=5\) ( 1 )

+, TH 1 : \(\sqrt{x-1}-1\ge0\) <=> \(x\ge2\) . Khi đó phương trình (1) được :

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

<=> \(2\sqrt{x-1}=5\)

<=> \(\sqrt{x-1}=2,5\)

<=> \(x-1=6,25\)

<=> \(x=7,25\) ( TM )

TH 2 : \(\sqrt{x-1}-1\le0\) <=> \(x\le2\) . Khi đó phương trình (1) được :

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

<=> \(2=5\) ( Vô lý )

Vậy phương trình trên có nghiệm duy nhất là x = 7,25 .

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 \(x=0\)

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

(\(\sqrt{1+x}+\sqrt{1-x}\))\(\left(2+2\sqrt{1-x^2}\right)=8\)(1)(đk: \(-1\le x\le1\))
đặt \(\sqrt{1+x}+\sqrt{1-x}\) =a (\(a\ge0\)

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

khi đó

(1)\(\Leftrightarrow a^3=8\Leftrightarrow a=\sqrt{8}=2\) (tm)

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

\(\Leftrightarrow2+2\sqrt{1-x^2}=4\)

\(\Leftrightarrow2\sqrt{1-x^2}=2\)

\(\Leftrightarrow\sqrt{1-x^2}=1\Leftrightarrow1-x^2=1\)

\(\Leftrightarrow x^2=0\Leftrightarrow x=0\)(tm)

vậy x=0 là nghiệm của phương trình

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