Bài làm:

a) \(\sqrt{x^2+6x+9}=4\)

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

\(\Leftrightarrow\left|x+3\right|=4\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=4\\x+3=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-7\end{cases}}\)

b) \(\sqrt{x^2+4x+4}=x-3\)

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

\(\Leftrightarrow\left|x+2\right|=x-3\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=x-3\\x+2=3-x\end{cases}}\Leftrightarrow\orbr{\begin{cases}0x=-5\left(vl\right)\\2x=1\end{cases}}\Rightarrow x=\frac{1}{2}\)

cách khác 

a,\(\sqrt{x^2+6x+9}=4\)

\(< =>x^2+6x+9=16\)

\(< =>x^2-x+7x-7=0\)

\(< =>x\left(x-1\right)+7\left(x-1\right)=0\)

\(< =>\left(x+7\right)\left(x-1\right)=0\)

\(< =>\orbr{\begin{cases}x=-7\\x=1\end{cases}}\)

b,\(\sqrt{x^2+4x+4}=x-3\)

\(< =>x^2+4x+4=x^2-6x+9\)

\(< =>x^2+4x+4-x^2+6x-9=0\)

\(< =>10x-5=0< =>x=\frac{1}{2}\)

\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)

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

\(A=\sqrt{5}-1-\sqrt{5}-1\)

\(A=-2\)

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

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

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

\(B=4\)

Sửa đề :

\(C=\sqrt{14-6\sqrt{5}}-\sqrt{14+6\sqrt{5}}\)

\(C=\sqrt{\left(3-\sqrt{5}\right)^2}-\sqrt{\left(3+\sqrt{5}\right)^2}\)

\(C=3-\sqrt{5}-3-\sqrt{5}\)

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

mình không biết bạn ơi

a. \(11+2\sqrt{10}=\left(\sqrt{10}+1\right)^2\)

b. \(12-2\sqrt{11}=\left(\sqrt{11}-1\right)^2\)

c.\(23+2\sqrt{22}=\left(\sqrt{22}+1\right)^2\)

Bài làm:

a) \(x^2-7=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)

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

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

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

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

f) \(9x-4=\left(3\sqrt{x}-2\right)\left(3\sqrt{x}+2\right)\)

a) ĐKXĐ: thỏa mãn với mọi a thực

b) ĐKXĐ: \(\frac{1}{2a+1}>0\)

\(\Rightarrow2a+1>0\Rightarrow2a>-1\Leftrightarrow a>-\frac{1}{2}\)

c) ĐKXĐ: \(a\left(1-a\right)\ge0\)

+ Nếu: \(\hept{\begin{cases}a\ge0\\1-a\ge0\end{cases}}\Leftrightarrow1\ge a\ge0\)

+ Nếu: \(\hept{\begin{cases}a\le0\\1-a\le0\end{cases}\Rightarrow}\hept{\begin{cases}a\le0\\a\ge1\end{cases}}\)(vô lý)

Vậy \(0\le a\le1\)

d) ĐKXĐ: \(\frac{2}{\left(a-2\right)\left(a+3\right)}>0\)

\(\Rightarrow\left(a-2\right)\left(a+3\right)>0\)

+ Nếu: \(\hept{\begin{cases}a-2>0\\a+3>0\end{cases}}\Rightarrow a>2\)

+ Nếu: \(\hept{\begin{cases}a-2< 0\\a+3< 0\end{cases}}\Rightarrow a< -3\)

Vậy \(\orbr{\begin{cases}a>2\\a< -3\end{cases}}\)

Để biểu thức có nghĩa thì :

\(\sqrt{4+a^2}\left(đk:\forall a-tmđk\right)\)

\(\sqrt{\frac{1}{2a+1}}\left(đk:a\ne-\frac{1}{2};a\ge-\frac{1}{2}\Leftrightarrow a>-\frac{1}{2}\right)\)

\(\sqrt{a\left(1-a\right)}\left(đk:a\ge0\right)\)

\(\sqrt{\frac{2}{\left(a-2\right)\left(a+3\right)}}\left(đk:a\ge2;a\ne2\Leftrightarrow a>2\right)\)

\(A=x-\frac{2x-2\sqrt{x}}{\sqrt{x}-1}+\frac{x\sqrt{x}+1}{x-\sqrt{x}+1}+1\left(đk:x\ne1;x\ge0\right)\)

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

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

\(=1+\frac{x^2+3x+1-2x\sqrt{x}-2\sqrt{x}}{x-\sqrt{x}+1}\)

\(=\frac{x-\sqrt{x}+1+x^2+3x+1-2x\sqrt{x}-2\sqrt{x}}{x-\sqrt{x}+1}\)

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

bạn thử chia đa thức cho đa thức xem

Bg

Ta có: 2x + 3y = 0   (x, y thuộc Q)

=> 2x = -3y

=> x = -3y ÷ 2

=> x = \(\frac{-3}{2}\)y

Vậy với mọi y thuộc Q và x = \(\frac{-3}{2}\)y

Ta có : \(2x+3y=0\Leftrightarrow2x=-3y\)

Ta có tỉ lệ : \(\frac{x}{-3}=\frac{y}{2}\)(K) 

Từ K Suy ra : \(x=-\frac{3y}{2};y=-\frac{2x}{3}\)