\(\sqrt{\frac{x^2}{4}+\sqrt{x^2-4}}=8-x^2\)

\(\sqrt{\left(\frac{x^2-4+4\sqrt{x^2-4}+4}{4}\right)}=\sqrt{\left(\sqrt{x^2-4}+2\right)^2}=2\left(8-x^2\right)\)

Điều kiện: \(\left\{\begin{matrix}\left|x\right|\ge2\\\left|x\right|\le2\sqrt{2}\end{matrix}\right.\Rightarrow\left[\begin{matrix}-2\sqrt{2}\le x\le-2\\2\le x\le2\sqrt{2}\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{x^2-4}+2=2\left(8-x^2\right)\) đặt \(\sqrt{x^2-4}=t\)

\(\Leftrightarrow2\left(t^2-4\right)+t+2=0\Leftrightarrow2t^2+t-6=0\){delta =1+48=7^2}

\(\Rightarrow\left[\begin{matrix}t=\frac{-1-7}{4}\left(loiaj\right)\\t=\frac{-1+7}{4}=\frac{3}{2}\end{matrix}\right.\)

\(\Leftrightarrow x^2-4=\frac{9}{4}\Rightarrow x^2=\frac{25}{4}\Rightarrow\left[\begin{matrix}x=-\frac{5}{2}\\x=\frac{5}{2}\end{matrix}\right.\) nhận hết

ĐKXĐ : x\(\ge0\)

ADBĐT BCS ta được

\(\left(\frac{x^2}{3}+4\right)\left(3+1\right)\ge\left(x+2\right)^2\)

\(\Rightarrow4\sqrt{\frac{x^2}{3}+4}\ge2x+4\)(do x\(\ge0\))    (1)

Do x\(\ge0\)nên ADBĐT Cauchy ta được:

\(\sqrt{6x}\le\frac{x+6}{2}\)\(\Rightarrow1+\frac{3x}{2}+\sqrt{6x}\le1+\frac{3x}{2}+\frac{x+6}{2}=1+\frac{4x+6}{2}=2x+4\)(2)

Từ (1) và (2) \(\Rightarrow4\sqrt{\frac{x^2}{3}+4}\ge1+\frac{3x}{2}+\sqrt{6x}\)

Dấu = xảy ra \(\Leftrightarrow x=6\)(thỏa mãn ĐKXĐ)

3) ĐKXĐ \(-1\le x\le1\)

Khi đó phương trình đã cho \(\Leftrightarrow4\left(\sqrt{1+x}+\sqrt{1-x}\right)=8-x^2\)

\(\Leftrightarrow\hept{\begin{cases}16\left(2+2\sqrt{1-x^2}\right)=\left(7+1-x^2\right)\left(2\right)\\8-x^2\ge0\end{cases}}\)

Đặt \(\sqrt{1-x^2}=a\ge0\)

Khi đó phương trình (2) trở thành: 

\(\hept{\begin{cases}16\left(2+2a\right)=\left(7+a^2\right)\\x^2\le8\end{cases}}\)

\(\Leftrightarrow a^4+14a^2+49=32+32a\)

\(\Leftrightarrow a^4+14a^2-32a+17=0\)

\(\Leftrightarrow a^4-2a^2+1+16a^2-32a+16=0\)

\(\Leftrightarrow\left(a^2-1\right)^2+16\left(a-1\right)^2=0\)

\(\Leftrightarrow a=1\)

hay \(\sqrt{1-x^2}=1\)

\(\Leftrightarrow x=0\)(thỏa mãn)

b, Đặt \(\sqrt[3]{x}=t\)

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

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

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

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

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 \(\sqrt{x^2-6x+6}=a\left(a\ge0\right)\)

\(\Rightarrow a^2+3-4a=0\)

=> (a - 3).(a - 1) = 0

=> \(\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)

Bình phương lên giải tiếp nhé!

c) Tương tư câu b nhé

1)

a) \(\left\{{}\begin{matrix}2x-y=5\\x+y=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2x-y+x+y=5+4\\x+y=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3x=9\\x+y=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

Vậy (x;y)=(3;1)

b) \(16x^5-8x^3+x=0\Leftrightarrow x\left(16x^4-8x^2+1\right)=0\Leftrightarrow x\left[\left(4x^2\right)^2-2.4x^2.1+1^2\right]=0\Leftrightarrow x\left(4x^2-1\right)^2=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\4x^2-1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=\frac{\pm1}{2}\end{matrix}\right.\)

Vậy S={\(-\frac{1}{2};0;\frac{1}{2}\)}

2)

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

B=\(\frac{4}{3+\sqrt{5}}-\frac{8}{1+\sqrt{5}}+\frac{15}{\sqrt{5}}=\frac{4\left(3-\sqrt{5}\right)}{9-5}-\frac{8\left(1-\sqrt{5}\right)}{1-5}+3\sqrt{5}=\frac{4\left(3-\sqrt{5}\right)}{4}-\frac{8\left(\sqrt{5}-1\right)}{4}+3\sqrt{5}=3-\sqrt{5}-2\sqrt{5}+2+3\sqrt{5}=5\)

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

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

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

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

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

 \(\Leftrightarrow x^2-4x+2=0\)

\(\Leftrightarrow\hept{\begin{cases}2-\sqrt{2}\\2+\sqrt{2}\left(l\right)\end{cases}}\) 

mình mới học lớp 5 ko biết làm 

tks alibaba