Điều kiện xác định : \(1\le x\le7\)

Bất phương trình chuyển thành :

\(x-1+2\sqrt{7-x}-2\sqrt{x-1}-\sqrt{\left(x-1\right)\left(7-x\right)}\le0\)

Đặt \(a=\sqrt{x-1};b=\sqrt{7-x}\) ta có :

\(a^2-2a-ab+2b\le0\)

\(\Leftrightarrow\left(a-b\right)\left(a-2\right)\le0\)

\(\Leftrightarrow\left[{}\begin{matrix}a\le b\\a\le2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1\le7-x\\x-1\le4\end{matrix}\right.\)

Sau đó tìm x

ĐKXĐ:\(\hept{\begin{cases}x-2>0\\y-1>0\\z-5>0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x>2\\y>1\\z>5\end{cases}}\)

pt\(\Leftrightarrow\frac{4}{\sqrt{x-2}}+\frac{1}{\sqrt{y-1}}+\frac{25}{\sqrt{z-5}}+\sqrt{x-2}+\sqrt{y-1}+\sqrt{z-5}=16\)

Áp dụng BĐT Cauchy:

\(\frac{4}{\sqrt{x-2}}+\sqrt{x-2}+\frac{1}{\sqrt{y-1}}+\sqrt{y-1}+\frac{25}{\sqrt{z-5}}+\sqrt{z-5}\)

\(\ge2\sqrt{\frac{4}{\sqrt{x-2}}.\sqrt{x-2}}+2\sqrt{\frac{1}{\sqrt{y-1}}.\sqrt{y-1}}+2\sqrt{\frac{25}{\sqrt{z-5}}.\sqrt{z-5}}\)

\(=2\sqrt{4}+2\sqrt{1}+2\sqrt{25}=2.2+2.1+2.5\)

\(=4+2+10=16\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x-2=4\\y-1=1\\z-5=25\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=2\\z=30\end{cases}}\)

câu a tớ giải được rồi, các bn giải câu b giùm mk

a/ \(x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=4\)

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

\(\Leftrightarrow x+\sqrt{x+\frac{1}{4}}+\frac{1}{2}=4\)

Làm nốt

b/ \(\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)

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

Làm nốt

a/ ĐKXĐ: \(x\ge-1\)

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

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

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

\(\Rightarrow x=3\)

b/ ĐKXĐ: \(x\ge1\)

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

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

Ta có \(VT\ge\left|\sqrt{x-1}-1+2-\sqrt{x-1}\right|=1\)

Nên dấu "=" xảy ra khi và chỉ khi:

\(1\le\sqrt{x-1}\le2\Rightarrow2\le x\le5\)

Vậy nghiệm của pt là \(2\le x\le5\)

c/ ĐKXĐ: \(x\ge1\)

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

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

- Với \(\sqrt{x-1}\ge1\Rightarrow x\ge2\) ta có:

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

\(\Leftrightarrow2=2\) (luôn đúng)

- Với \(1\le x< 2\) ta có:

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

\(\Leftrightarrow\sqrt{x-1}=1\Rightarrow x=2\left(l\right)\)

Vậy nghiệm của pt là \(x\ge2\)

d/ ĐKXĐ: \(-\le x\le1\)

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

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

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

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

TH1: \(\sqrt{4-4x^2}\ge1\Rightarrow-\frac{\sqrt{3}}{2}\le x\le\frac{\sqrt{3}}{2}\) ta có:

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

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

\(\Leftrightarrow4-4x^2=x^2+2x+1\)

\(\Leftrightarrow5x^2+2x-3=0\Rightarrow\left[{}\begin{matrix}x=-1\left(l\right)\\x=\frac{3}{5}\end{matrix}\right.\)

TH2: \(\left[{}\begin{matrix}-1\le x< -\frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2}< x\le1\end{matrix}\right.\) ta có:

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

\(\Leftrightarrow2x=0\Rightarrow x=0\left(l\right)\)

Vậy pt có nghiệm duy nhất \(x=\frac{3}{5}\)

ĐKXĐ:...

\(\Leftrightarrow\frac{36}{\sqrt{x-2}}+4\sqrt{x-2}+\frac{4}{\sqrt{y-1}}+\sqrt{y-1}=28\)

Ta có:

\(VT\ge2\sqrt{\frac{36.4\sqrt{x-2}}{\sqrt{x-2}}}+2\sqrt{\frac{4\sqrt{y-1}}{\sqrt{y-1}}}=28\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}\frac{9}{\sqrt{x-2}}=\sqrt{x-2}\\\frac{4}{\sqrt{y-1}}=\sqrt{y-1}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=11\\y=5\end{matrix}\right.\)

b, Đặt \(\sqrt[3]{x}=t\)

Ta có: \(\sqrt[3]{x^2}-8\sqrt[3]{x}=20\)

\(\Leftrightarrow t^2-8t=20\Leftrightarrow t^2-8t-20=0\)

\(\Leftrightarrow\left(t+2\right)\left(t-10\right)=0\)

\(\orbr{\begin{cases}t=-2\\t=10\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt[3]{x}=-2\\\sqrt[3]{x}=10\end{cases}\Leftrightarrow}}\orbr{\begin{cases}x=-8\\x=1000\end{cases}}\)