cần gấp thì mình làm cho 

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

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

\(< =>x+1=\sqrt{x+1}\)

\(< =>\frac{x+1}{\sqrt{x+1}}=1\)

\(< =>\sqrt{x+1}=1< =>x=0\left(ktm\right)\)

ĐKXĐ : \(x\ge-1\)

Bình phương 2 vế , ta có :

\(x^2+2x+1=x+1\)

\(\Leftrightarrow x^2+2x+1-x-1=0\)

\(\Leftrightarrow x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}\left(TM\right)}\)\

Vậy ...............................

a) ĐK: \(x\ge2\)

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

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

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

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(chon\right)\\x=-1\left(loai\right)\end{matrix}\right.\)

b) ĐK: \(\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)

\(pt\Leftrightarrow\sqrt{x^2-9}=\sqrt{x+3}\)

\(\Leftrightarrow x^2-9=x+3\)

\(\Leftrightarrow x^2-x-12=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(chon\right)\\x=-3\left(chon\right)\end{matrix}\right.\)

Vậy...

\(T=\sum\frac{a}{1+9b^2}=\sum\frac{a\left(1+9b^2\right)-9ab^2}{1+9b^2}=\sum\left(a-\frac{9ab^2}{1+9b^2}\right)\ge\sum\left(a-\frac{9ab^2}{6b}\right)=\sum\left(a-\frac{3}{2}ab\right)\)

\(T\ge a+b+c-\frac{3}{2}\left(ab+ac+bc\right)\ge a+b+c-\frac{1}{2}\left(a+b+c\right)^2=\frac{1}{2}\)

\(\Rightarrow T_{min}=\frac{1}{2}\) khi \(a=b=c=\frac{1}{3}\)

a) x -\(\sqrt{2x-9}=0\) ĐKXĐ: x\(\ge\frac{9}{2}\)

<=> x=\(\sqrt{2x-9}\)

<=> x²=2x-9 (vì x>0)

<=> x²-2x+1=8

<=>(x-1)²=8

<=>\(\left[{}\begin{matrix}x-1=2\sqrt{2}\\x-1=-2\sqrt{2}\end{matrix}\right.\)

<=>x=\(2\sqrt{2}+1\)(vì x>0) (thỏa mãn)

0:0

Lời giải:

a) ĐKXĐ: $x\in\mathbb{R}$

\(\sqrt{x^2-2x+4}=2x-2\Leftrightarrow \left\{\begin{matrix} 2x-2\geq 0\\ x^2-2x+4=(2x-2)^2\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ 3x^2-6x=0\end{matrix}\right.\Rightarrow x=2\)

b) ĐKXĐ: $-x^2+x+4\geq 0$

\(\sqrt{-x^2+x+4}=x-3\Leftrightarrow \left\{\begin{matrix} x-3\geq 0\\ -x^2+x+4=(x-3)^2\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 3\\ 2x^2-7x+5=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 3\\ (2x-5)(x-1)=0\end{matrix}\right.\) (không thỏa mãn)

Vậy pt vô nghiệm

c) ĐK: $x\leq 0$

PT $\Rightarrow x^2-2x=2-3x$

$\Leftrightarrow x^2+x-2=0$

$\Leftrightarrow (x+2)(x-1)=0$

Vì $x\leq 0$ nên $x=-2$ là nghiệm duy nhất của pt.

d) ĐK: $x\geq 3$

PT $\Leftrightarrow \sqrt{x-3}-2\sqrt{(x-3)(x+3)}=0$

$\Leftrightarrow \sqrt{x-3}(1-2\sqrt{x+3})=0$

$\Rightarrow \sqrt{x-3}=0$ hoặc $1-2\sqrt{x+3}=0$

Nếu $\sqrt{x-3}=0\Rightarrow x=3$ (thỏa mãn)

Nếu $1-2\sqrt{x+3}=0\Rightarrow x=\frac{-11}{4}< 3$ (không thỏa ĐKXĐ)

Vậy ...........

a)  \(\sqrt{7+4\sqrt{3}}=\sqrt{2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}\)

    \(=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)

b)   \(\sqrt{13-4\sqrt{3}}=\sqrt{\left(2\sqrt{3}\right)^2-2.2\sqrt{3}+1}\)

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

c)  \(\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}\)

     \(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}-\sqrt{2}\)

d)  \(\sqrt{3+2\sqrt{2}+\sqrt{6-4\sqrt{2}}}\)

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

\(=\sqrt{3+2\sqrt{2}+2-\sqrt{2}}\)

\(=\sqrt{5+\sqrt{2}}\)

e)  \(2+\sqrt{17-4\sqrt{9+4\sqrt{5}}}\)

\(=2+\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}\)

\(=2+\sqrt{17-4\left(\sqrt{5}+2\right)}\)

\(=2+\sqrt{9-4\sqrt{5}}\)

\(=2+\sqrt{\left(\sqrt{5}-2\right)^2}\)

\(=2+\sqrt{5}-2=\sqrt{5}\)

f)   đề sai nhé:  

\(\sqrt{3a}.\sqrt{12a}=\sqrt{36a^2}=6a\)\(\left(a\ge0\right)\)

g)  \(\sqrt{16a^2b^8}=4b^4\left|a\right|\)

h)  \(\sqrt{7a}.\sqrt{63a^3}=\sqrt{441.a^4}=21a^2\)

1: ĐKXĐ: x<>0

\(\Leftrightarrow x^2-6\left(m-1\right)x+9m^2=0\)

\(\text{Δ}=\left(6m-6\right)^2-4\cdot1\cdot9m^2\)

\(=36m^2-72m+36-36m^2=-72m+36\)

Để pt vô nghiệm thì -72m+36<0

=>-72m<-36

hay m>1/2

2:ĐKXD: x<>9/8

\(\Leftrightarrow2x^2-\left(m+1\right)x+\dfrac{1}{8}m^2+1=0\)

\(\text{Δ}=\left(m+1\right)^2-4\cdot2\cdot\left(\dfrac{1}{8}m^2+1\right)\)

\(=m^2+2m+1-m^2-8=2m-7\)

Để pt vô nghiệm thì 2m-7<0

hay m<7/2