\(\left(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\right):\frac{\sqrt{x}}{\sqrt{x}+x}\)

\(=\left(\frac{\sqrt{x}+\sqrt{x}.\sqrt{x}}{\sqrt{x}.\left(\sqrt{x}+1\right)}\right).\frac{\sqrt{x}+x}{\sqrt{x}}\)

\(=\frac{\sqrt{x}+x}{\sqrt{x}.\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}+x}{\sqrt{x}}\)

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

\(\left(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\right):\frac{\sqrt{x}}{x+\sqrt{x}}=\frac{\sqrt{x}+1+x}{x+\sqrt{x}}\cdot\frac{x+\sqrt{x}}{\sqrt{x}}=\frac{\sqrt{x}+1+x}{\sqrt{x}}\)

ĐKXĐ:\(x\ge0\)

lam toan sai 

tự hỏi tự trả lời

1)

a) trong tam giac ABC vuong tai A co 

+)BC²=AB²+AC²

suy ra AC=12cm

+)AH.BC=AB.AC

suy ra AH=7,2cm

b) Trong tu giac AMHN co HMA=HNA=BAC=90 do suy ra AMHN la hcn suy ra AH=MN=7,2cm

suy ra MN=7,2cm

c) goi O la giao diem cu MN va AH 

Vi AMHN la hcn (cmt) nen OA=OH=7,2/2=3,6cm

suy ra S_BMCN=1/2[OH*(MN+BC)]=39,96cm²
d) Vi AMHN la hcn nen goc AMN=goc HAB 

Trong tam giac ABC vuong tai A co AK la dg trung tuyen ung voi canh huyen BC nen AK=BK=KC

suy ra tam giac AKB can tai K

suy ra goc B= goc BAK

Ta co goc B+ goc BAH=90 do 
tuong duong BAK+AMN=90 do suy ra AK vuong goc voi MN (dmcm)

bai 2 sai de ban oi sinx hay cosx chu ko phai sin hay cos

\(\frac{A}{7}\cdot3=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{43\cdot46}\)

\(\frac{A}{7}\cdot3=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{46}\)

\(\frac{A}{7}\cdot3=\frac{45}{46}\)

\(\frac{A}{7}=\frac{15}{46}\)

\(A=\frac{105}{46}\)

Học tốt~