Đk: \(x\ge\frac{-3}{2}\)

Bất pt <=> \(2x+3+x+2+2\sqrt{2x^2+7x+6}\le1\)

<=> \(2\sqrt{2x^2+7x+6}\le-4-3x\)

<=> \(\hept{\begin{cases}-3-4x\ge0\\4\left(2x^2+7x+6\right)\le16+24x+9x^2\end{cases}}\)

<=> \(\hept{\begin{cases}x\le-\frac{3}{4}\\x^2-4x-8\ge0\end{cases}}\)

<=> \(\hept{\begin{cases}x\le-\frac{3}{4}\\\left(x-2\right)^2\ge12\end{cases}}\)

<=> \(x\le2-\sqrt{12}\)

Đối chiếu đk: \(-\frac{3}{2}\le x\le2-\sqrt{12}\)

ĐỀ BÀI SAI RỒI !!!!!!

1 SỐ CÓ DẠNG \(\sqrt{x}\ge0\)   VÀ     \(x\ge0\left(đkxđ\right)\)

NHƯNG \(\sqrt{3}< 2\)

=> \(\sqrt{3}-2< 0\)

=> \(\sqrt{\sqrt{3}-2}\)KO TỒN TẠI !!!!!.

Đặt  \(A=\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right).\sqrt{\sqrt{3}-2}\)

\(\Rightarrow A^2=\left(\sqrt{6}+\sqrt{2}\right)^2.\left(\sqrt{3}-2\right)^2.\left(\sqrt{3}-2\right)\)

\(\Leftrightarrow A^2=\left(8+4\sqrt{3}\right).\left(-1-4\sqrt{3}\right)\left(\sqrt{3}-2\right)\)

\(\Leftrightarrow A^2=\left(8+4\sqrt{3}\right).\left(-\sqrt{3}+2-12+8\sqrt{3}\right)\)

\(\Leftrightarrow A^2=\left(8+4\sqrt{3}\right)\left(7\sqrt{3}-10\right)\)

\(\Leftrightarrow A^2=56\sqrt{3}-80+84-40\sqrt{3}\)

\(\Leftrightarrow A^2=16\sqrt{3}+4\)

\(\Rightarrow A=\pm\sqrt{16\sqrt{3}+4}\)

\(P=x+\frac{9}{x-2}+2018\)

\(=\left(x-2\right)+\frac{9}{x-2}+2020\)

\(\ge2\sqrt{\frac{\left(x-2\right)9}{x-2}}+2020\)

\(=2\sqrt{9}+2020=2026\)

Dấu = xảy ra khi và chỉ khi \(x=5\)

Vậy \(Min_P=2026\)khi \(x=5\)

\(P=\left(x-2\right)+\frac{9}{x-2}+2020\)

\(P\ge2.\sqrt{\frac{\left(x-2\right).9}{x-2}}+2020\)

=> \(P\ge6+2020=2026\)

"=" xảy ra <=> \(x-2=\frac{9}{x-2}\)

<=> \(\left(x-2\right)^2=9\)

<=> \(\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}}\)

<=> \(\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

Do \(x>2\)    => \(x=5\)

VẬY P MIN = 2026 <=> x = 5.

Chú ý 2 điều: \(\cos45^o=\sin45^o=\frac{\sqrt{2}}{2}\) và \(\cos^2a+\sin^2a=1\)

Do đó: 

a) \(A=\cos^252^o.\frac{\sqrt{2}}{2}+\sin^252^o.\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}\left(\cos^252^o+\sin^252^o\right)=\frac{\sqrt{2}}{2}.1=\frac{\sqrt{2}}{2}\)

b) \(B=\frac{\sqrt{2}}{2}.\cos^247^o+\frac{\sqrt{2}}{2}.\sin^247^o=\frac{\sqrt{2}}{2}\left(\cos^247^o+\sin^247^o\right)=\frac{\sqrt{2}}{2}.1=\frac{\sqrt{2}}{2}\)

\(F=2x^2+2xy+5y^2-8x-22y\)

<=> \(2F=4x^2+4xy+10y^2-16x-44\)

\(=\left(4x^2+4xy+y^2-16x+16-8y\right)+9y^2-36y-16\)

\(=\left(2x+y-4\right)^2+\left(3y-6\right)^2-52\ge-52\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x+y-4=0\\3y-6=0\end{cases}}\Leftrightarrow y=2;x=1\)

=> min 2F = -52

=> min F = -26. 

pt => \(2x+5=1-x\)

<=> \(3x=4\)

<=> \(x=\frac{4}{3}\)

VẬY    \(x=\frac{4}{3}\)là no duy nhất

Hermit :) làm nhầm rồi em.

\(\sqrt{2x+5}=\sqrt{1-x}\Leftrightarrow\hept{\begin{cases}1-x\ge0\\2x+5=1-x\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le1\\3x=-4\end{cases}}\Leftrightarrow x=-\frac{4}{3}\)

Vậy x = -4/3

lớp 9 thì mình dùng cách lớp 9 

\(\sqrt{x+2\sqrt{x}-1}=2\left(đk:x\ge1\right)\)

\(< =>x+2\sqrt{x}-1=4\)(bình phương 2 vế)

Đặt \(\sqrt{x}=t\left(t\ge0\right)\)(*)

\(< =>t^2+2t-5=0\)

\(\Delta=2^2-4.\left(-5\right)=4+20=24\)

\(\orbr{\begin{cases}t_1=\frac{-2+2\sqrt{6}}{2}=-1+\sqrt{6}\left(tm\right)\\t_2=\frac{-2-2\sqrt{6}}{2}=-1-\sqrt{6}\left(ktm\right)\end{cases}}\)

Khi đó thế vào * ta được :

 \(\sqrt{x}=\sqrt{6}-1< =>x=7-2\sqrt{6}\left(tmđk\right)\)

Vậy nghiệm của phương trình trên là \(7-2\sqrt{6}\)

ĐK: \(x\ge1\)

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

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

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

<=> \(\sqrt{x-1}+1=2\)

<=> \(\sqrt{x-1}=1\)

<=> x - 1 = 1 

<=> x = 2 thỏa mãn

Ap dung \(\frac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

Ta co \(P< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2007}}-\frac{1}{\sqrt{2008}}\right)\)  

=> \(P< 2\left(1-\frac{1}{\sqrt{2008}}\right)< 2.1=2\)

Suy ra P khong phai so nguyen to

Có: \(A=\sqrt{\frac{1}{1^2}+\frac{1}{a^2}+\frac{1}{\left(-a-1\right)^2}}\)

Có: \(1+a+\left(-a-1\right)=1+a-1-a=0\)

=> \(\sqrt{\frac{1}{1^2}+\frac{1}{a^2}+\frac{1}{\left(-a-1\right)^2}}=\sqrt{\left(\frac{1}{1}+\frac{1}{a}+\frac{1}{-a-1}\right)^2}=\frac{1}{1}+\frac{1}{a}+\frac{1}{-a-1}\)

=>    \(A=1+\frac{1}{a}-\frac{1}{a+1}=1+\frac{1}{a\left(a+1\right)}\)

VẬY     \(A=1+\frac{1}{a\left(a+1\right)}\)

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

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

\(=\sqrt{\left[\frac{1}{a\left(a+1\right)}+1\right]^2}=\left|\frac{1}{a}-\frac{1}{a+1}+1\right|\)