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 \(x^2+x+4=\left(x+1\right)^2+3>0\)

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

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

\(x^2+x=0\)

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

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

a) \(\left(5\sqrt{2}+2\sqrt{5}\right).\sqrt{5}-\sqrt{250}=5\sqrt{10}+10-5\sqrt{10}=10\)

b) \(\left(2\sqrt{3}-5\sqrt{2}\right).\sqrt{3}-\sqrt{36}=6-5\sqrt{6}-6=-5\sqrt{6}\)

c) \(6\sqrt{\frac{1}{3}}+\frac{9}{\sqrt{3}}-\frac{2}{\sqrt{3}-1}=6\cdot\frac{\sqrt{3}}{3}+\frac{9\sqrt{3}}{3}-\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

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

d) \(4\sqrt{\frac{1}{2}}-\frac{6}{\sqrt{2}}+\frac{2}{\sqrt{2}+1}=4\cdot\frac{\sqrt{2}}{2}-\frac{6\sqrt{2}}{2}+\frac{2\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)

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

a, \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)

\(=5\sqrt{10}+10-\sqrt{250}\)

\(=5\sqrt{10}+10-5\sqrt{10}\)

\(=10\)

b, \(\left(2\sqrt{3}-5\sqrt{2}\right)\sqrt{3}-\sqrt{36}\)

\(=6-5\sqrt{6}-6\)

\(=-5\sqrt{6}\)

c, \(6\sqrt{\frac{1}{3}}+\frac{9}{\sqrt{3}}-\frac{2}{\sqrt{3}-1}\)

\(=\sqrt{\frac{36}{3}}+\frac{9\sqrt{3}}{3}-\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=\sqrt{12}+3\sqrt{3}-\frac{2\sqrt{3}+2}{2}\)

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

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

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

\(=4\sqrt{3}-1\)

d,\(4\sqrt{\frac{1}{2}}-\frac{6}{\sqrt{2}}+\frac{2}{\sqrt{2}+1}\)

\(=\sqrt{\frac{16}{2}}-\frac{6\sqrt{2}}{2}+\frac{2\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}\)

\(=\sqrt{8}-3\sqrt{2}+2\sqrt{2}-2\)

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

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

Xét tam giác \(ABD\)vuông tại \(A\):

\(BD^2=AB^2+AD^2\)(định lí Pythagore) 

\(=4^2+10^2=116\)

\(\Rightarrow BD=\sqrt{116}=2\sqrt{29}\left(cm\right)\)

Lấy \(E\)thuộc \(CD\)sao cho \(AE\perp AC\)

Suy ra \(ABDE\)là hình bình hành. 

\(AE=BD=2\sqrt{29}\left(cm\right),DE=AB=4\left(cm\right)\).

Xét tam giác \(AEC\)vuông tại \(A\)đường cao \(AD\):

\(\frac{1}{AD^2}=\frac{1}{AE^2}+\frac{1}{AC^2}\Leftrightarrow\frac{1}{AC^2}=\frac{1}{AD^2}-\frac{1}{AE^2}=\frac{1}{100}-\frac{1}{116}=\frac{1}{715}\)

\(\Rightarrow AC=\sqrt{715}\left(cm\right)\)

\(AE^2=ED.EC\Leftrightarrow EC=\frac{AE^2}{ED}=\frac{116}{4}=29\left(cm\right)\)suy ra \(DC=25\left(cm\right)\)

Hạ \(BH\perp CD\).

\(BC^2=HC^2+BH^2=21^2+10^2=541\Rightarrow BC=\sqrt{541}\left(cm\right)\)

\(S_{ABCD}=\left(AB+CD\right)\div2\times AD=\frac{4+25}{2}\times10=145\left(cm^2\right)\)

a) Xét tam giác ADB vuông tại D có: \(cos\widehat{A}=\frac{AD}{AB}\)

Xét tam giác AEC vuông tại  C có: \(cos\widehat{A}=\frac{AE}{AC}\)

=> \(\frac{AD}{AB}=\frac{AE}{AC}\) => AE.AB = AD.AC

b) Xét tam giác ADE và tam giác ABC

có: \(\widehat{A}\) :chung

  \(\frac{AD}{AB}=\frac{AE}{AC}\) (cmt)

=> tam giác ADE ∽ tam giác ABC (c.g.c)

=> \(\widehat{AED}=\widehat{ACB}\)mà \(\widehat{AED}=\widehat{QEB}\)(đối đỉnh) => \(\widehat{QEB}=\widehat{QCD}\)

Xét tam giác QEB và tam giác QCD

có: \(\widehat{QEB}=\widehat{QCD}\)(cmt); \(\widehat{Q}\) : chung

=> tam giác QEB ∽ tam giác QCD (g.g)

=> \(\frac{QE}{QC}=\frac{QB}{QD}\) => QB. QC = QE . QD

c) CMTT: \(\widehat{BKE}=\widehat{BAC}\); \(\widehat{DKC}=\widehat{BAC}\)

Ta có: \(\widehat{BKE}+\widehat{K_2}=90^0\) (phụ nhau))

   \(\widehat{K_1}+\widehat{DKC}=90^0\) (phụ nhau) 

==> \(\widehat{K_1}=\widehat{K_2}\) => KA là phân giác của \(\widehat{DKE}\)

=> \(\frac{KE}{KD}=\frac{EP}{ED}\)(1)

Gọi Kx là tia đối của tia KD => \(\widehat{DKC}=\widehat{QKx}\) mà \(\widehat{DKC}=\widehat{EKB}\) => \(\widehat{EKQ}=\widehat{QKx}\)

=> KQ là tia phân giác của \(\widehat{EKx}\) => \(\frac{EK}{KD}=\frac{QE}{QD}\)(2)

Từ (1) và (2) => \(\frac{EP}{PD}=\frac{QE}{QD}\) => PD. QE = PE. QD