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

Đặt \(a=2x-3;b=\sqrt{\left(x-1\right)\left(2x-3\right)-\left(x-3\right)}\)

Ta có hpt \(\hept{\begin{cases}a^2+x-3=\left(x-1\right)b\\b^2+x-3=\left(x-1\right)a\end{cases}}\)

Trừ 2 pt trên ta được: \(a^2-b^2=\left(x-1\right)\left(b-a\right)\Rightarrow\left(a-b\right)\left(a+b+x-1\right)=0\)

+) Nếu \(a=b\Leftrightarrow2x-3=\sqrt{2x^2-6x+6}\Leftrightarrow\hept{\begin{cases}x\ge\frac{3}{2}\\2x^2-6x+3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3-\sqrt{3}}{2}\left(loại\right)\\x=\frac{3+\sqrt{3}}{2}\left(tm\right)\end{cases}}}\)

+) Nếu \(2x-3+\sqrt{2x^2-6x+6}+x-3=0\Leftrightarrow\sqrt{2x^2-6x}=6-3x\)\(\Leftrightarrow\hept{\begin{cases}x\le2\\7x^2-30x+36=0\end{cases}\left(VN\right)}\)

Vậy pt có nghiệm duy nhất: \(x=\frac{3+\sqrt{3}}{2}\)

\(4x^2-11x+6=\left(x-1\right)\sqrt{2x^2-6x+6}\)

\(\Leftrightarrow\left(4x^2-12x+9\right)+x-3=\left(x-1\right)\sqrt{2x^2-5x+3-x+3}\)

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

đặt \(\hept{\begin{cases}t=2x-3\\y=\sqrt{\left(x-1\right)\left(2x-3\right)-\left(x-3\right)}\left(y\ge0\right)\end{cases}}\)

ta có hệ : \(\hept{\begin{cases}t^2+x-3=\left(x-1\right)y\\y^2-\left(x-1\right)t+x-3=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}t^2-\left(x-1\right)y+\left(x-3\right)=0\\y^2-\left(x-1\right)t+\left(x-3\right)=0\end{cases}}\)

\(\Rightarrow t^2-y^2-\left(x-1\right)y+\left(x-1\right)t=0\)

\(\Leftrightarrow\left(t-y\right)\left(t+y\right)+\left(x-1\right)\left(t-y\right)=0\)

\(\Leftrightarrow\left(t-y\right)\left(t+y+x-1\right)=0\)

th1 : \(t-y=0\Leftrightarrow t=y\Leftrightarrow2x-3=\sqrt{\left(x-1\right)\left(2x-3\right)-\left(x-3\right)}\) ĐK : \(x\ge\frac{3}{2}\)

\(\Leftrightarrow4x^2-12x+9=2x^2-6x+6\)

\(\Leftrightarrow2x^2-6x+3=0\)

\(\Delta=b^2-4ac=\left(-6\right)^2-4\cdot2\cdot3=12\)

\(\Rightarrow\orbr{\begin{cases}x=3+\sqrt{3}\left(tm\right)\\x=3-\sqrt{3}\left(loai\right)\end{cases}}\)

th2 : \(x+y+t-1=0\Leftrightarrow y=1-x-t\)

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

\(\Leftrightarrow\sqrt{2x^2-6x+6}=4-3x\left(đk:x\le\frac{4}{3}\right)\)

\(\Leftrightarrow2x^2-6x+6=16-24x+9x^2\)

\(\Leftrightarrow7x^2-18x+10=0\)

\(\Delta=b^2-4ac=\left(-18\right)^2-4\cdot7\cdot10=44\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{18+\sqrt{44}}{2}=9+\sqrt{11}\left(loai\right)\\x=\frac{18-\sqrt{44}}{2}=9-\sqrt{11}\left(loai\right)\end{cases}}\)

https://olm.vn/hoi-dap/detail/1610090115961.html

Đặt \(x=a^2;y=b^2;z=c^2\)

bđt \(\Leftrightarrow\sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}}\le\frac{3}{\sqrt{2}}\)

Áp dụng bđt Bunhiacopxki ta có:

\(\left(\sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}}\right)^2\)\(=\left(\sqrt{\frac{x\left(x+z\right)}{\left(x+y\right)\left(x+z\right)}}+\sqrt{\frac{y\left(y+x\right)}{\left(y+z\right)\left(y+x\right)}}+\sqrt{\frac{z\left(z+y\right)}{\left(z+x\right)\left(z+y\right)}}\right)^2\)

\(\le2\left(x+y+z\right)\left(\frac{x}{\left(x+y\right)\left(x+z\right)}+\frac{y}{\left(y+z\right)\left(y+x\right)}+\frac{z}{\left(z+x\right)\left(z+y\right)}\right)\)

\(=\frac{4\left(x+y+z\right)\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)      (*)

Ta cần CM:  (*) \(\le\frac{9}{2}\)

Hay \(8\left(x+y+z\right)\left(xy+yz+zx\right)\le9\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

hay \(8xyz\le\left(x+y\right)\left(y+z\right)\left(z+x\right)\)  (luôn đúng)

=> đpcm 

Đẳng thức xảy ra <=> a=b=c

AD bđt AM-GM ta có

\(\frac{b+c}{a^2}+\frac{4}{b+c}\ge2\sqrt{\frac{b+c}{a^2}\frac{4}{b+c}}=\frac{4}{a}\)   (1)

Tương tự có: \(\frac{c+a}{b^2}+\frac{4}{c+a}\ge\frac{4}{b}\)  (2)

\(\frac{a+b}{c^2}+\frac{4}{a+b}\ge\frac{4}{c}\)  (3)

Cộng theo vế các bđt (1), (2), (3) ta được:

\(\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}+\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}\ge\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\)     (4)

Lại có:  \(\frac{1}{a}+\frac{1}{b}\ge2\sqrt{\frac{1}{a}\frac{1}{b}}=\frac{4}{2\sqrt{ab}}\ge\frac{4}{a+b}\)   (5)

Tương tự: \(\frac{1}{b}+\frac{1}{c}\ge\frac{4}{b+c}\)  (6)

\(\frac{1}{c}+\frac{1}{a}\ge\frac{4}{c+a}\)   (7)

Cộng theo vế các bđt (4),(5),(6),(7) ta được:

\(\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}+\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}+\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\)

\(\ge\frac{4}{a}+\frac{4}{b}+\frac{4}{c}+\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}\)

=> đpcm

đk: \(x\ge-3\)

\(pt\Leftrightarrow x^2+2x\sqrt{x+3}+x+3+x+\sqrt{x+3}-12=0\)

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

\(\Leftrightarrow\left(x+\sqrt{x+3}+4\right)\left(x+\sqrt{x+3}-3\right)=0\Leftrightarrow\orbr{\begin{cases}\sqrt{x+3}=-4-x\left(1\right)\\\sqrt{x+3}=3-x\left(2\right)\end{cases}}\)

giải (1): Ta có: \(x\ge-3\Leftrightarrow VP=-4-x\le-1,VT\ge0\)

=> (1) vô nghiệm

giải (2): \(\left(2\right)\Leftrightarrow\hept{\begin{cases}x\le3\\x+3=x^2-6x+9\end{cases}\Leftrightarrow x^2-7x+6=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=6\end{cases}}}\)

    Theo điều kiện => pt có nghiệm: x=1

giúp voii mình cần gấpp