giải xog thì chớt

Đề đúng chưa v

1. \(\sqrt{x^2-4}-x^2+4=0\)( ĐK: \(\orbr{\begin{cases}x\ge2\\x\le-2\end{cases}}\))

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

\(\Leftrightarrow\left(x^2-4\right)^2=x^2-4\)

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

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-4-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=4\\x^2=5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\pm2\left(tm\right)\\x=\pm\sqrt{5}\left(tm\right)\end{cases}}\)

Vậy pt có tập no \(S=\left\{2;-2;\sqrt{5};-\sqrt{5}\right\}\)

2. \(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\)ĐK: \(\hept{\begin{cases}x^2-4x+5\ge0\\x^2-4x+8\ge0\\x^2-4x+9\ge0\end{cases}}\)

\(\Leftrightarrow\sqrt{x^2-4x+5}-1+\sqrt{x^2-4x+8}-2+\sqrt{x^2-4x+9}-\sqrt{5}=0\)

\(\Leftrightarrow\frac{x^2-4x+4}{\sqrt{x^2-4x+5}+1}+\frac{x^2-4x+4}{\sqrt{x^2-4x+8}+2}+\frac{x^2-4x+4}{\sqrt{x^2-4x+9}+\sqrt{5}}=0\)

\(\Leftrightarrow\left(x-2\right)^2\left(\frac{1}{\sqrt{x^2-4x+5}+1}+\frac{1}{\sqrt{x^2-4x+8}+2}+\frac{1}{\sqrt{x^2}-4x+9+\sqrt{5}}\right)=0\)

Từ Đk đề bài \(\Rightarrow\frac{1}{\sqrt{x^2-4x+5}+1}+\frac{1}{\sqrt{x^2-4x+8}+2}+\frac{1}{\sqrt{x^2}-4x+9+\sqrt{5}}>0\)

\(\Rightarrow\left(x-2\right)^2=0\)

\(\Leftrightarrow x=2\left(tm\right)\)

Vậy pt có no x=2

\(a,\) Sửa đề: \(\sqrt{3x^2-12x+16}+\sqrt{y^2-4y+13}=5\)

Ta thấy \(3x^2-12x+16=3\left(x-2\right)^2+4\ge4\Leftrightarrow\sqrt{3x^2-12x+16}\ge\sqrt{4}=2\)

\(y^2-4y+13=\left(y-2\right)^2+9\ge9\Leftrightarrow\sqrt{y^2-4y+13}\ge\sqrt{9}=3\)

Cộng vế theo vế 2 BĐT trên:

\(\sqrt{3x^2-12x+16}+\sqrt{y^2-4y+13}\ge2+3=5\)

Dấu \("="\Leftrightarrow x=y=2\)

Vậy pt có nghiệm \(\left(x;y\right)=\left(2;2\right)\)

\(b,x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\\ \Leftrightarrow x+y+z+4-2\sqrt{x-2}-4\sqrt{y-3}-6\sqrt{z-5}=0\\ \Leftrightarrow\left(x-2-2\sqrt{x-2}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-5+6\sqrt{z-5}+9\right)=0\\ \Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}-1=0\\\sqrt{y-3}-2=0\\\sqrt{z-5}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\y-3=4\\z-5=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\\z=14\end{matrix}\right.\)

a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\)) 

\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

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

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

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

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\left(tm\right)\)

b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))

\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\)

\(\Leftrightarrow\sqrt{x+1}=4\)

\(\Leftrightarrow x+1=16\)

\(\Leftrightarrow x=15\left(tm\right)\)

ĐKXĐ: \(x\ge1\).

Phương trình đã cho tương đương:

\(\sqrt{x+3}+\sqrt{x-1}=\dfrac{8}{\sqrt{4x^4-12x^3+9x^2+16}-\left(2x^2-3x\right)}\)

\(\Leftrightarrow\sqrt{x+3}+\sqrt{x-1}=\dfrac{\sqrt{4x^4-12x^3+9x^2+16}+\left(2x^2-3x\right)}{2}\)

\(\Leftrightarrow\sqrt{4x^4-12x^3+9x^2+16}+\left(2x^2-3x\right)-2\sqrt{x+3}-2\sqrt{x-1}=0\)

\(\Leftrightarrow\left(\sqrt{4x^4-12x^3+9x^2+16}-2\sqrt{x+3}\right)+\left(2x^2-3x-2\sqrt{x-1}\right)=0\)

\(\Leftrightarrow\dfrac{4x^4-12x^3+9x^2-4x+4}{\sqrt{4x^4-12x^3+9x^2+16}+2\sqrt{x+3}}+\dfrac{4x^4-12x^3+9x^2-4x+4}{2x^2-3x+2\sqrt{x-1}}=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x^3-4x^2+x-2\right)\left(\dfrac{1}{\sqrt{4x^4-12x^3+9x^2+16}+2\sqrt{x+3}}+\dfrac{1}{2x^2-3x+2\sqrt{x-1}}\right)=0\).

Do \(x\ge1\) nên ta có \(\dfrac{1}{\sqrt{4x^4-12x^3+9x^2+16}+2\sqrt{x+3}}+\dfrac{1}{2x^2-3x+2\sqrt{x-1}}>0\).

Do đó \(\left[{}\begin{matrix}x-2=0\Leftrightarrow x=2\left(TMĐK\right)\\4x^3-4x^2+x-2=0\left(1\right)\end{matrix}\right.\).

Giải phương trình bậc 3 ở (1) ta được \(x=\dfrac{\sqrt[3]{36\sqrt{13}+53\sqrt{6}}}{\sqrt[6]{279936}}+\dfrac{1}{\sqrt[6]{7776}\sqrt[3]{36\sqrt{13}+53\sqrt{6}}}+\dfrac{1}{3}\approx1,157298106\left(TMĐK\right)\).

Vậy...

Vì trong bài làm của mình có một số dòng khá dài nên bạn có thể vào trang cá nhân của mình để đọc tốt hơn!

a) \(3x-2\sqrt{x-1}=4\) (ĐK: x ≥ 1)

\(\Rightarrow3x-2\sqrt{x-1}-4=0\)

\(\Rightarrow3x-6-2\sqrt{x-1}+2=0\)

\(\Rightarrow3\left(x-2\right)-2\left(\sqrt{x-1}-1\right)=0\)

\(\Rightarrow3\left(x-2\right)-2.\dfrac{x-2}{\sqrt{x-1}+1}=0\)

\(\Rightarrow\left(x-2\right)\left[3-\dfrac{2}{\sqrt{x-1}+1}\right]=0\)

*TH1: x = 2 (t/m)

*TH2: \(3-\dfrac{2}{\sqrt{x-1}+1}=0\)

\(\Rightarrow3=\dfrac{2}{\sqrt{x-1}+1}\)

\(\Rightarrow3\sqrt{x-1}+3=2\)

\(\Rightarrow3\sqrt{x-1}=-1\) (vô lí)

Vậy S = {2}

b) \(\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\) (ĐK: \(-\dfrac{1}{4}\le x\le3\) )

\(\Rightarrow\sqrt{4x+1}-3-\sqrt{x+2}+2-\sqrt{3-x}+1=0\)

\(\Rightarrow\dfrac{4x-8}{\sqrt{4x+1}+3}-\dfrac{x-2}{\sqrt{x+2}+2}+\dfrac{x-2}{\sqrt{3-x}+1}=0\)

\(\Rightarrow\left(x-2\right)\left(\dfrac{4}{\sqrt{4x+1}+3}-\dfrac{1}{\sqrt{x+2}+2}+\dfrac{1}{\sqrt{3-x}+1}\right)=0\)

=> x = 2

\(a,3x-2\sqrt{x-1}=4\left(x\ge1\right)\\ \Leftrightarrow-2\sqrt{x-1}=4-3x\\ \Leftrightarrow4\left(x-1\right)=16-24x+9x^2\\ \Leftrightarrow9x^2-28x+20=0\\ \Leftrightarrow\left(x-2\right)\left(9x-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=\dfrac{10}{9}\left(tm\right)\end{matrix}\right.\)

\(b,\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\left(-\dfrac{1}{4}\le x\le3\right)\\ \Leftrightarrow4x+1+x+2-2\sqrt{\left(4x+1\right)\left(x+2\right)}=3-x\\ \Leftrightarrow-2\sqrt{\left(4x+1\right)\left(x+2\right)}=2-6x\\ \Leftrightarrow\sqrt{4x^2+9x+2}=3x-1\\ \Leftrightarrow4x^2+9x+2=9x^2-6x+1\\ \Leftrightarrow5x^2-15x-1=0\\ \Leftrightarrow\Delta=225+20=245\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15-\sqrt{245}}{10}=\dfrac{15-7\sqrt{5}}{10}\left(ktm\right)\\x=\dfrac{15+\sqrt{245}}{10}=\dfrac{15+7\sqrt{5}}{10}\left(tm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{15+7\sqrt{5}}{10}\)