Đề thiếu dữ liệu giả thiết =))

e chx đến tầm 

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

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

\(\frac{x+2-x-\sqrt{x}}{\sqrt{x}+1}.\frac{4-\sqrt{x}-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\frac{2-\sqrt{x}}{\sqrt{x}+1}.\frac{4-\sqrt{x}-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

đề lên sửa thành phép chia thì dễ hơn 

\(\frac{2-\sqrt{x}}{\sqrt{x}+1}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{4-x}\)

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

Ta có: Đk: x \(\ge\)-5/4

Ta có: \(2x^2-6x-1=\sqrt{4x+5}\)

<=> \(4x^2-12x-2-2\sqrt{4x+5}=0\)

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

<=> \(\left(2x-2\right)^2-\left(\sqrt{4x+5}+1\right)^2=0\)

<=> \(\left(2x-2-\sqrt{4x+5}-1\right)\left(2x-2+\sqrt{4x+5}+1\right)=0\)

<=> \(\orbr{\begin{cases}2x-3-\sqrt{4x+5}=0\left(1\right)\\2x-1+\sqrt{4x+5}=0\left(2\right)\end{cases}}\)

Giải pt (1) Ta có: \(2x-3=\sqrt{4x+5}\) (đk: x \(\ge\)3/2)

<=> \(4x^2-12x+9=4x+5\)

<=> \(4x^2-16x+4=0\)

<=> \(x^2-4x+1=0\)

\(\Delta'=\left(-2\right)^2-1=3>0\) => pt có 2 nghiệm pb

\(x_1=2+\sqrt{3}\)(tm) ; \(x_2=2-\sqrt{3}\)(ktm)

Giải pt (2) ta có: \(1-2x=\sqrt{4x+5}\) (đk: \(-\frac{5}{4}\le x\le\frac{1}{2}\))

<=> \(4x+5=4x^2-4x+1\)

<=> \(4x^2-8x-4=0\)

<=> \(x^2-2x-1=0\)

\(\Delta'=\left(-1\right)^2+1=2>0\) 

=> pt có 2 nghiệm pb 

\(x_1=1+\sqrt{2}\)ktm); \(x_2=1-\sqrt{2}\left(tm\right)\)

Vậy \(S=\left\{1-\sqrt{2};2+\sqrt{3}\right\}\)

ĐKXĐ:\(x\ge-5\)

\(x^2-4x-3=\sqrt{x+5}\)

\(\Leftrightarrow x^2-4x-3+x+5+\frac{1}{4}=\left(x+5\right)+\sqrt{x+5}+\frac{1}{4}\)

\(\Leftrightarrow x^2-3x+\frac{9}{4}=\left(x+5\right)+\sqrt{x+5}+\frac{1}{4}\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=\left(\sqrt{x+5}+\frac{1}{2}\right)\)

Xét :

+)\(x-\frac{3}{2}=\sqrt{x+5}+\frac{1}{2}\)

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

\(\Leftrightarrow\hept{\begin{cases}x\ge2\\x^2-5x-1=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x\ge2\\\left(x-\frac{5}{2}\right)^2=\frac{29}{4}\end{cases}}\)\(\Leftrightarrow x=\frac{5+\sqrt{29}}{2}\left(TM\right)\)

+)\(x-\frac{3}{2}=-\sqrt{x+5}-\frac{1}{2}\)

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

\(\Leftrightarrow\hept{\begin{cases}x\le1\\\left(x-4\right)\left(x+1\right)=0\end{cases}}\)\(\Leftrightarrow x=-1\left(TM\right)\)

Vậy tập nghiệm của PT là \(x\in\left\{\frac{5+\sqrt{29}}{2};-1\right\}\)

ĐK: \(x\ge-5\).

\(x^2-4x-3=\sqrt{x+5}\)

\(\Rightarrow\left(x^2-4x-3\right)^2=x+5\)

\(\Leftrightarrow x^4-8x^3+10x^2+23x+4=0\)

\(\Leftrightarrow x^4+x^3-9x^3-9x^2+19x^2+19x+4x+4=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3-9x^2+19x+4\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3-4x^2-5x^2+20x-x+4\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-4\right)\left(x^2-5x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1;x=4\\x=\frac{1}{2}\left(5\pm\sqrt{29}\right)\end{cases}}\)

Thử lại chỉ có \(x=-1\)và \(x=\frac{1}{2}\left(5+\sqrt{29}\right)\)thỏa mãn. 

\(\sqrt{16}=4\)

\(\sqrt{49}=7\)

\(\sqrt{121}=11\)

\(\sqrt{169}=13\)

\(\sqrt{196}=14\)

trả lời 

\(\sqrt{16}\);\(\sqrt{49}\);\(\sqrt{121}\);\(\sqrt{169}\);\(\sqrt{196}\)

chúc bn hok tốt 

\(\hept{\begin{cases}\left(x+1\right)\left(y+1\right)=xy+13\\\left(x-2\right)\left(y-1\right)=xy-15\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}xy+x+y+1=xy+13\\xy-x-2y+2=xy-15\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x+y=xy+13-xy-1\\-x-2y=xy-15-xy-2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x+y=12\\-x-2y=-17\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-y=-5\\x+y=12\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=5\\x=7\end{cases}}\)

Vậy hệ phương trình có nghiệm duy nhất (7;5)

Ta có: \(a+b=\sqrt{1-a^2}+\sqrt{1-b^2}\)

<=> \(\sqrt{1-a^2}-b+\sqrt{1-b^2}-a=0\)

<=> \(\frac{1-a^2-b^2}{\sqrt{1-a^2}+b}+\frac{1-b^2-a^2}{\sqrt{1-b^2}+a}=0\) (Do 0 < a,b < 1)

<=> \(\left(1-a^2-b^2\right).\left(\frac{1}{\sqrt{1-a^2}+b}+\frac{1}{\sqrt{1-b^2}+a}\right)=0\)

<=> \(1-a^2-b^2=1\) (vì 0 < a,b < 1 => \(\frac{1}{\sqrt{1-a^2}+b}+\frac{1}{\sqrt{1-b^2}+a}>0\))

<=> a² + b² = 1

\(\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{17+12\sqrt{2}}\)

\(\left|2\sqrt{2}-1\right|-\sqrt{17+6\sqrt{8}}\)

\(2\sqrt{2}-1-\sqrt{3^2+6\sqrt{8}+\sqrt{8}^2}\)

\(2\sqrt{2}-1-\left|3+\sqrt{8}\right|\)

\(2\sqrt{2}-1-3-\sqrt{8}\)
\(2\sqrt{2}-4-\sqrt{8}\)

\(=-4\)

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

\(\left|x^2+x+4\right|=4\)

\(\orbr{\begin{cases}x^2+x+4=4\\x^2+x+4=-4\end{cases}}\)

ta có \(x^2+x+4=\left(x+1\right)^2+3>0\)

\(< =>x^2+x+4=-4\left(ktm\right)\)

\(x^2+x+4=4\)

\(x^2+x=0\)

\(x\left(x+1\right)=0\)

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