a.

\(3\sqrt{-x^2+x+6}\ge2\left(1-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-x^2+x+6\ge0\\1-2x< 0\end{matrix}\right.\\\left\{{}\begin{matrix}1-2x\ge0\\9\left(-x^2+x+6\right)\ge4\left(1-2x\right)^2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-2\le x\le3\\x>\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\25\left(x^2-x-2\right)\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}< x\le3\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\-1\le x\le2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-1\le x\le3\)

b.

ĐKXĐ: \(x\ge0\)

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

\(\Leftrightarrow\dfrac{2x^2+8x+5-16x}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-4x+5-4x}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

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

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

\(\Leftrightarrow2x^2-8x+5=0\)

\(\Leftrightarrow x=\dfrac{4\pm\sqrt{6}}{2}\)

Câu b còn 1 cách giải nữa:

Với \(x=0\) không phải nghiệm

Với \(x>0\) , chia 2 vế cho \(\sqrt{x}\) ta được:

\(\sqrt{2x+8+\dfrac{5}{x}}+\sqrt{2x-4+\dfrac{5}{x}}=6\)

Đặt \(\sqrt{2x-4+\dfrac{5}{x}}=t>0\Leftrightarrow2x+8+\dfrac{5}{x}=t^2+12\)

Phương trình trở thành:

\(\sqrt{t^2+12}+t=6\)

\(\Leftrightarrow\sqrt{t^2+12}=6-t\)

\(\Leftrightarrow\left\{{}\begin{matrix}6-t\ge0\\t^2+12=\left(6-t\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\le6\\12t=24\end{matrix}\right.\)

\(\Rightarrow t=2\)

\(\Rightarrow\sqrt{2x-4+\dfrac{5}{x}}=2\)

\(\Leftrightarrow2x-4+\dfrac{5}{x}=4\)

\(\Rightarrow2x^2-8x+5=0\)

\(\Leftrightarrow...\)

2:

a: =>2x^2-4x-2=x^2-x-2

=>x^2-3x=0

=>x=0(loại) hoặc x=3

b: =>(x+1)(x+4)<0

=>-4<x<-1

d: =>x^2-2x-7=-x^2+6x-4

=>2x^2-8x-3=0

=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)

a, ĐKXĐ : \(\left[{}\begin{matrix}x\le-3\\x\ge0\end{matrix}\right.\)

TH1 : \(x\le-3\) ( LĐ )

TH2 : \(x\ge0\)

BPT \(\Leftrightarrow x^2+2x+x^2+3x+2\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge4x^2\)

\(\Leftrightarrow\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge x^2-\dfrac{5}{2}x\)

\(\Leftrightarrow2\sqrt{\left(x+2\right)\left(x+3\right)}\ge2x-5\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\4x^2+20x+24\ge4x^2-20x+25\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}0\le x< \dfrac{5}{2}\\x\ge\dfrac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow x\ge0\)

Vậy \(S=R/\left(-3;0\right)\)

a) \(\dfrac{a^2+3}{\sqrt{a^2+2}}=\sqrt{a^2+2}+\dfrac{1}{\sqrt{a^2+2}}\ge2\sqrt{\sqrt{a^2+2}.\dfrac{1}{\sqrt{a^2+2}}}=2\)

Dấu = xảy ra khi \(\sqrt{a^2+2}=\dfrac{1}{\sqrt{a^2+2}}\Leftrightarrow a^2=-1\left(vn\right)\)

\(\Rightarrow\) Dấu "=" không xảy ra

Vậy \(\dfrac{a^2+3}{\sqrt{a^2+2}}>2\)

b)Với x,y>0,ta cm bđt phụ sau:

\(x^3+y^3\ge xy\left(x+y\right)\) (1)

Thật vậy (1)\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\ge0\)

\(\Leftrightarrow\cdot\left(x+y\right)\left(x^2-2xy+y^2\right)\ge0\)

\(\Leftrightarrow\left(x+y\right)\left(x-y\right)^2\ge0\) (lđ)

Áp dụng (1) có:

\(\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}=\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}.\sqrt{b}}\ge\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}=\sqrt{a}+\sqrt{b}\)

Dấu "=" xra khi a=b

Vậy...

\(\sqrt{7x+7}+\sqrt{7x-6}=t\ge0\)

\(bpt\Leftrightarrow t+t^2< 182\Leftrightarrow-14< t< 13\Leftrightarrow t< 13\Leftrightarrow\sqrt{7x+7}+\sqrt{7x-6}< 13\left(đk:x\ge\dfrac{6}{7}\right)\Leftrightarrow14x+1+2\sqrt{\left(7x+7\right)\left(7x-6\right)}< 169\Leftrightarrow2\sqrt{\left(7x+7\right)\left(7x-6\right)}< 168-14x\Leftrightarrow\left\{{}\begin{matrix}\left(7x+7\right)\left(7x-6\right)\ge0\\168-14x\ge0\\4\left(7x+7\right)\left(7x-6\right)< \left(168-14x\right)^2\end{matrix}\right.\)

\(giảibpt\Rightarrowđáp\) \(số\)

2:

a: Sửa đề: \(\dfrac{a^2+3}{\sqrt{a^2+2}}>2\)

\(A=\dfrac{a^2+3}{\sqrt{a^2+2}}=\dfrac{a^2+2+1}{\sqrt{a^2+2}}=\sqrt{a^2+2}+\dfrac{1}{\sqrt{a^2+2}}\)

=>\(A>=2\cdot\sqrt{\sqrt{a^2+2}\cdot\dfrac{1}{\sqrt{a^2+2}}}=2\)

A=2 thì a^2+2=1

=>a^2=-1(loại)

=>A>2 với mọi a

b: \(\Leftrightarrow\sqrt{a}+\sqrt{b}< =\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}\)

=>\(a\sqrt{a}+b\sqrt{b}>=a\sqrt{b}+b\sqrt{a}\)

=>\(\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)-\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)>=0\)

=>(căn a+căn b)(a-2*căn ab+b)>=0

=>(căn a+căn b)(căn a-căn b)^2>=0(luôn đúng)

1

ĐK: `x>1`

PT trở thành:

\(\sqrt{\dfrac{2x-3}{x-1}}=2\\ \Leftrightarrow\dfrac{2x-3}{x-1}=2^2=4\\ \Leftrightarrow4x-4-2x+3=0\\ \Leftrightarrow2x-1=0\\ \Leftrightarrow x=\dfrac{1}{2}\left(KTM\right)\)

Vậy PT vô nghiệm.

b

ĐK: \(x\ge2\)

Đặt \(t=\sqrt{x-2}\) (\(t\ge0\))

=> \(x=t^2+2\)

PT trở thành: \(t^2+2-5t+2=0\)

\(\Leftrightarrow t^2-5t+4=0\)

nhẩm nghiệm: `a+b+c=0` (`1+(-5)+4=0`)

\(\Rightarrow\left\{{}\begin{matrix}t=1\left(nhận\right)\\t=4\left(nhận\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{x-2}=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3\left(TM\right)\\x=18\left(TM\right)\end{matrix}\right.\)

a, ĐK: \(x\le-1,x\ge3\)

\(pt\Leftrightarrow2\left(x^2-2x-3\right)+\sqrt{x^2-2x-3}-3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-2x-3}=-\dfrac{3}{2}\left(l\right)\\\sqrt{x^2-2x-3}=1\end{matrix}\right.\)

\(\Leftrightarrow x^2-2x-3=1\)

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

\(\Leftrightarrow x=1\pm\sqrt{5}\left(tm\right)\)

b, ĐK: \(-2\le x\le2\)

Đặt \(\sqrt{2+x}-2\sqrt{2-x}=t\Rightarrow t^2=10-3x-4\sqrt{4-x^2}\)

Khi đó phương trình tương đương:

\(3t-t^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2+x}-2\sqrt{2-x}=0\\\sqrt{2+x}-2\sqrt{2-x}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2+x=8-4x\\2+x=17-4x+12\sqrt{2-x}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\left(tm\right)\\5x-15=12\sqrt{2-x}\left(1\right)\end{matrix}\right.\)

Vì \(-2\le x\le2\Rightarrow5x-15< 0\Rightarrow\left(1\right)\) vô nghiệm

Vậy phương trình đã cho có nghiệm \(x=\dfrac{6}{5}\)

c, ĐK: \(0\le x\le9\)

Đặt \(\sqrt{9x-x^2}=t\left(0\le t\le\dfrac{9}{2}\right)\)

\(pt\Leftrightarrow9+2\sqrt{9x-x^2}=-x^2+9x+m\)

\(\Leftrightarrow-\left(-x^2+9x\right)+2\sqrt{9x-x^2}+9=m\)

\(\Leftrightarrow-t^2+2t+9=m\)

Khi \(m=9,pt\Leftrightarrow-t^2+2t=0\Leftrightarrow\left[{}\begin{matrix}t=0\\t=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}9x-x^2=0\\9x-x^2=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=9\left(tm\right)\\x=\dfrac{9\pm\sqrt{65}}{2}\left(tm\right)\end{matrix}\right.\)

Phương trình đã cho có nghiệm khi phương trình \(m=f\left(t\right)=-t^2+2t+9\) có nghiệm

\(\Leftrightarrow minf\left(t\right)\le m\le maxf\left(t\right)\)

\(\Leftrightarrow-\dfrac{9}{4}\le m\le10\)

a,

\(\Leftrightarrow\sqrt{1-x}=\frac{x-1}{\sqrt{6-x}+\sqrt{-5-2x}}\)

\(\Leftrightarrow-\sqrt{1-x}=\sqrt{6-x}+\sqrt{-5-2x}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{1-x}=\sqrt{6-x}-\sqrt{-5-2x}\\-\sqrt{1-x}=\sqrt{6-x}+\sqrt{-5-2x}\end{cases}}\)

b,tự nàm

c,

\(\Leftrightarrow64x^2-64x-64=64\sqrt{8x+1}\)

\(\Leftrightarrow\left(8x+1\right)^2=10\left(8x+1\right)+64\sqrt{8x+1}+55\)

đặt \(\sqrt{8x+1}=a\)

=>a⁴=10a²+64a+55

nhận thấy phương trình có dạng x⁴=ax²+bx+c

tìm số m sao cho b²-4(2m+a)(m²+c)=0

sau đó đưa về (x²+m)²=k² với k là 1 số bất kì,sau đó giải ra

b)đk \(x\ge1\)

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

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

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

\(=\frac{x^2+x+1}{x+1}+\frac{x}{x+1}=x+1\)

\(\Rightarrow\sqrt{1+2012^2+\frac{2012^2}{2013^2}}+\frac{2012}{2013}=2013\)

\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=2013\)

\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2013\)

\(\Leftrightarrow x+\left|x-2\right|=2014\)

giai 2 pt 

pt1 x+x-2=2014

x=1008

pt2 x+2-x=2014(vô lý)

ý c cái a đặt có 1 nghiệm là a=5 đó