Làm cho mk đi @Ender Dragon Boy Vcl

(x−y+z)2+(z−y)2+2(x−y+z)(y−z)(x−y+z)2+(z−y)2+2(x−y+z)(y−z)

=(x−y+z)2+(z−y)2+(x−y+z)(y−z)+(x−y+z)(y−z)=(x−y+z)2+(z−y)2+(x−y+z)(y−z)+(x−y+z)(y−z)

=(x−y+z)2+(x−y+z)(y−z)+(z−y)2+(x−y+z)(y−z)=(x−y+z)2+(x−y+z)(y−z)+(z−y)2+(x−y+z)(y−z)

=(x−y+z)2+(x−y+z)(y−z)+(z−y)2−(x−y+z)(z−y)=(x−y+z)2+(x−y+z)(y−z)+(z−y)2−(x−y+z)(z−y)

=(x−y+z)(x−y+y+z−z)+(z−y)[z−y−(x−y+z)]=(x−y+z)(x−y+y+z−z)+(z−y)[z−y−(x−y+z)]

=(x−y+z)x+(z−y)(z−y−x+y−z)=(x−y+z)x+(z−y)(z−y−x+y−z)

=x2−xy+xz+(z−y)(−x)=x2−xy+xz+(z−y)(−x)

=x2−xy+xz−xz+xy=x2−xy+xz−xz+xy

=x2

Ban tu ve hinh nha

Goi O la tam duong tron ngoai tiep tam giac ABC , ke OD,OE,OF vuong goc voi AB,BC,AC

Do ABC la tam giac can nenA,O,E thang hang ( duong phan giac dong thoi la duong cao va trung tuyen )

=> AD=DB=15 cm , BE=EC=18 cm

Xet tam giac ABE vuong o E co \(AE=\sqrt{30^2-18^2}=24\)  cm Dinh ly PYTAGO

Xet tam giac ADO vuong o D va tam giac AEB vuong o E co goc DAO= goc EAB

Suy ra tam giac ADO dong dang voi tam giac AEB (g-g) 

=>\(\frac{AD}{AB}=\frac{OD}{BE}\) <=> \(\frac{15}{24}=\frac{OD}{18}=>OD=11,25cm\) =OF do ta giac abc can tai a

Xet tam giac ODB vuong tai D co \(OB=\sqrt{\left(11,25\right)^2+15^2}=18,75cm\) dinh ly pytago

Xet tam giac OBE vuong tai E co \(OE=\sqrt{\left(18,75\right)^2-18^2}=5.25cm\) Dinh ly PYTAGO 

Vay khoang cach tu tam dong tron ngoai tiep tam giac ABC de 3 canh AB,AC,BC lan luot la 11,25 cm ,  11,25 cm    , 5,25 cm

STUDY WELL !!!

\(\frac{\left(\sqrt{x^2+15}-4\right).\left(\sqrt{x^2+15}+4\right)}{\sqrt{x^2+15}+4}=3x-3+\frac{\left(\sqrt{x^2+8}-3\right)\left(\sqrt{x^2+8}+3\right)}{\sqrt{x^2+8}+3}\)

\(\Leftrightarrow\frac{x^2-1}{\sqrt{x^2+15}+4}=3\left(x-1\right)+\frac{x^2-1}{\sqrt{x^2+8}+3}\)

\(\Leftrightarrow\left(x-1\right)\left(3+\frac{x+1}{\sqrt{x^2+8}+3}-\frac{x+1}{\sqrt{x^2+15}+4}\right)=0\)

\(\Leftrightarrow3+\frac{x+1}{\sqrt{x^2+8}+3}-\frac{x+1}{\sqrt{x^2+15}+4}=0\)hoặc x=1

Ta có: \(\sqrt{x^2+15}-\sqrt{x^2+8}=3x-2\)

Thấy: VT>0 => VP>0 => x>2/3

Xét \(3+\frac{x+1}{\sqrt{x^2+8}+3}-\frac{x+1}{\sqrt{x^2+15}+4}=0\)(1)

Ta thấy: với x>2/3 thì VT luôn dương => (1) vô lý

Vậy S={1}

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

\(\Leftrightarrow x^2-8=\left(x+3\right)\frac{\left(\sqrt{x^2+1}-3\right)\left(\sqrt{x^2+1}+3\right)}{\sqrt{x^2+1}+3}\)

\(\Leftrightarrow x^2-8=\left(x+3\right)\frac{x^2-8}{\sqrt{x^2+1}+3}\)

\(\Leftrightarrow\left(x^2-8\right)\left(1-\frac{x+3}{\sqrt{x^2+1}+3}\right)=0\)

\(\Leftrightarrow\left(x^2-8\right)\frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}+3}=0\)

Có \(\sqrt{x^2+1}-x>0\)

\(\Leftrightarrow\frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}+3}>0\)

\(\Rightarrow x=\pm2\sqrt{2}\)

Vậy...

AI GIẢI HỘ MÌNH K CHO Ạ!!!

1)  a) Căn thức có nghĩa \(\Leftrightarrow4-2x\ge0\Leftrightarrow2x\le4\Leftrightarrow x\le2\)

b) Thay x = 2 vào biểu thức A, ta được: \(A=\sqrt{4-2.2}=\sqrt{0}=0\)

Thay x = 0 vào biểu thức A, ta được: \(A=\sqrt{4-2.0}=\sqrt{4}=2\)

Thay x = 1 vào biểu thức A, ta được: \(A=\sqrt{4-2.1}=\sqrt{2}\)

Thay x = -6 vào biểu thức A, ta được: \(A=\sqrt{4-2.\left(-6\right)}=\sqrt{16}=4\)

Thay x = -10 vào biểu thức A, ta được: \(A=\sqrt{4-2.\left(-10\right)}=\sqrt{24}=2\sqrt{6}\)

c) \(A=0\Leftrightarrow\sqrt{4-2x}=0\Leftrightarrow4-2x=0\Leftrightarrow x=2\)

\(A=5\Leftrightarrow\sqrt{4-2x}=5\Leftrightarrow4-2x=25\Leftrightarrow x=\frac{-21}{2}\)

\(A=10\Leftrightarrow\sqrt{4-2x}=10\Leftrightarrow4-2x=100\Leftrightarrow x=-48\)

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

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

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

\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}+2\cdot\frac{a+b-a-b}{ab\left(a+b\right)}\)

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

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