Bạn xem lại đề câu b và c nhé !

a) \(\sqrt{x^2+2x+4}\ge x-2\) \(\left(ĐK:x\ge2\right)\)

\(\Leftrightarrow x^2+2x+4>x^2-4x+4\)

\(\Leftrightarrow6x>0\Leftrightarrow x>0\) kết hợp với ĐKXĐ

\(\Rightarrow x\ge2\) thỏa mãn đề.

d) \(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)

\(ĐKXĐ:x\ge2,y\ge3,z\ge5\)

Pt tương đương :

\(\left(x-2-2\sqrt{x-2}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-5-6\sqrt{z-5}+9\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}=1\\\sqrt{y-3}=2\\\sqrt{z-5}=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\y=7\\z=14\end{cases}}\) ( Thỏa mãn ĐKXĐ )

e) \(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\) (1)

\(ĐKXĐ:x\ge0,y\ge1,z\ge2\)

Phương trình (1) tương đương :

\(x+y+z-2\sqrt{x}-2\sqrt{y-1}-2\sqrt{z-2}=0\)

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

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x}=1\\\sqrt{y-1}=1\\\sqrt{z-2}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\)( Thỏa mãn ĐKXĐ )

\(\Leftrightarrow\frac{7x+4}{\sqrt{2\left(x-1\right)\left(x+1\right)}}+\frac{2\sqrt{2x+1}}{\sqrt{2\left(x+1\right)}}=3+\frac{3\sqrt{2x+1}}{\sqrt{x-1}}\)

\(\Leftrightarrow7x+4+2\sqrt{\left(2x+1\right)\left(x-1\right)}=3\sqrt{2\left(x-1\right)\left(x+1\right)}+3\sqrt{2\left(2x+1\right)\left(x+1\right)}\)

 \(\Leftrightarrow\left(7x+4+\sqrt{8x^2-4x-4}\right)^2=\left(\sqrt{18x^2-18}+\sqrt{36^2+54x+18}\right)^2\)

\(\Leftrightarrow\left(7x+4\right)^2+8x^2-4x-4+2\left(7x+4\right)\sqrt{8x^2-4x-4}\)\(=18x^2-18+36x^2+54x+18+2\sqrt{\left(18x^2-18\right)\left(36x^2+54x+18\right)}\)

\(\Leftrightarrow3x^2-2x+12+4\left(7x+4\right)\sqrt{\left(x-1\right)\left(2x+1\right)}=36\left(x+1\right)\sqrt{\left(x-1\right)\left(2x+1\right)}\)

\(\Leftrightarrow3x^2-2x+12=4\left(2x+5\right)\sqrt{\left(x-1\right)\left(2x+1\right)}\)

\(\Leftrightarrow\left(3x^2-2x+12\right)^2=16\left(2x+5\right)^2\left(x-1\right)\left(2x+1\right)\)

\(\Leftrightarrow119x^4+588x^3+1940x^2-672x-544=0\left(1\right)\)

Ta thấy x>1 => Vế trái (1) \(>119.1^4+588.1^3+1940.1^2-672.1-544=1431>0\)

=> pt vô nghiệm.

hello bạn

?

khó

mik ko bít phantuananh a

Đăt \(t=\sqrt{\frac{x}{2x-1}}\)    (1) ,  với \(t\ge0\) , ta được phương trình theo t như sau:

\(t+\frac{1}{t}=2\)

\(\Leftrightarrow t^2-2t+1=0\)

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

\(\Leftrightarrow t=1\)

Thay vào (1) ta tìm được x = 1.

Chú ý đặt thêm các điều kiện để phương trình có nghĩa bạn nhé.

Điều Kiện: \(x\ne0;x\ne\frac{1}{2}\)

Nhận xét \(\frac{x}{2x-1}.\frac{2x-1}{x}=1\) Đặt \(\frac{x}{2x-1}=a\Rightarrow\frac{2x-1}{x}=\frac{1}{a}\)

\(\Rightarrow\sqrt{a}+\sqrt{\frac{1}{a}}=2\Rightarrow a+1=2\sqrt{a}\)

BP 2 vế PT \(\Rightarrow a^2+2a+1=4a\Leftrightarrow a^2-2a+1=0\)

\(\Leftrightarrow\left(a-1\right)^2=0\Rightarrow a=1\Rightarrow\frac{x}{2x-1}=1\)

\(\Leftrightarrow x=2x-1\Leftrightarrow x=1\)

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

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

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

\(\Rightarrow x^3+\frac{x^2}{2}+x+\frac{1}{2}-x-\frac{1}{2}=x^3+\frac{x^2}{2}=0\Rightarrow\frac{2x^3+x^2}{2}=0\)

\(\Rightarrow2x^3+x^2=0\Rightarrow x^2\left(2x+1\right)=0\Rightarrow\hept{\begin{cases}x^2=0\Rightarrow x=0\\2x+1=0\Rightarrow2x=-1\Rightarrow x=-\frac{1}{2}\end{cases}}\)

vậy x=0 và x=-1/2