x-5⋮x-2

=> (x-2)-3⋮x-2

Vi x-2⋮x-2 => -3⋮x-2 => x-2∈U(-3)=(1,3,-1,-3)

=> x=3,5,1,-1

x-5=x-2

=> x-5-x+2=0

=> 0x-3=0

=> 0x=3

=> pt vo nghiem

a) Xet △ABE va △HBE co:

\(\left\{{}\begin{matrix}AB=BH\\B_1=B_2\\chugBE\end{matrix}\right.\)

=> △ABE=△HBE(c.g.c)

=> goc BAE= goc BHE = 90 do

=> EH⊥BC

b) CM duoc △ABD=△BHD(c.g.c)

=> goc BDA= goc BDH (GTU) va goc BDA+goc BDH=180

=> goc BDA=goc BDH=90

=> BD⊥AH (1)

Vi △ABD=△BHD => AD=DH va D∈AH => D là trung điểm của AH (2)

Tu (1)(2) => dpcm

c) Xet △AEK va △HEC co:

\(\left\{{}\begin{matrix}E_1=E_2\left(d^2\right)\\AE=EH\\gocKAE=gocCHE=90\end{matrix}\right.\)

=> △AEK=△HEC(g.c.g)

=> EK=EC(CTU)

d) Dung duong trung binh

e) Ta co : \(\left\{{}\begin{matrix}E_2=E_1\\BEA=MEC\left(d^2\right)\\BEH=KEM\left(d^2\right)\end{matrix}\right.\)

=> E₂+BEA+BEH=E₁+MEC+KEM (1)

Vi E∈AC nen E₂+BEA+BEH=180 (2)

Tu(1)(2) => E₁+MEC+KEM=180 => B,E,M thang hang

\(\dfrac{2.\left(\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{9}\right)}{2.\left(1+\dfrac{3}{5}-\dfrac{1}{3}\right)}\)

=\(\dfrac{\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{9}}{1+\dfrac{3}{5}-\dfrac{1}{3}}=\dfrac{\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{9}}{3.\dfrac{1}{3}+3.\dfrac{1}{5}-3.\dfrac{1}{9}}=\dfrac{\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{9}}{3.\left(\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{9}\right)}\)

=\(\dfrac{1}{3}\)

Goi so trung ma nguoi do mang ra cho ban la: a (qua) (a>60)

Ta co: \(\dfrac{a}{4}+a.\dfrac{1}{4}.1,5+60=a\)

=> \(\dfrac{a}{4}+\dfrac{1,5.a}{4}+\dfrac{240}{4}=a\)

=> \(\dfrac{a+1,5.a+240}{4}=a\)

=> 2,5.a+240=4.a

=> 240=4.a-2,5.a=1,5.a

=> a=240:1,5=160

Vay nguoi do da mang tat ca 160 qua trung ra cho ban

∆A'B'C' ∽ ∆A"B"C" theo tỉ số đồng dạng k₁ =

∆A"B"C" ∽∆ ABC theo tỉ số đồng dạng k₂ =

Theo tính chất 3 thì ∆A'B'C' ∽ ∆ABC.

Theo tỉ số K= = = .

vậy k= k₁.k₂

S△=\(\dfrac{156.65}{60}=169\left(mm^2\right)\)

a) 25-5x+3-4=0

=> 24-5x=0

=> 5x=24

=> x=\(\dfrac{24}{5}\)

b) 259-51/4x+3/=100.3+10=310

=> -51/4x+3/=310-259=51

=> /4x+3/=-51 (1)

Vi /4x+3/≥0 (2)

Tu (1)(2) => pt vo nghiem

c) 7x-21-15+5x=6

=> 12x-36-6=0

=> 12x-42=0

=> 12x=42

=> x=\(\dfrac{42}{12}=\dfrac{7}{2}\)

\(\left(x^2+\dfrac{1}{x^2}\right)+\left(y^2+\dfrac{1}{y^2}\right)+\left(z^2+\dfrac{1}{z^2}\right)=6\)

=> VT≥\(2.\sqrt{x^2.\dfrac{1}{x^2}}+2.\sqrt{y^2.\dfrac{1}{y^2}}+2.\sqrt{z^2.\dfrac{1}{z^2}}\)

= 2+2+2=6

Dau bang xay ra khi: \(\left\{{}\begin{matrix}x^2=\dfrac{1}{x^2}\\y^2=\dfrac{1}{y^2}\\z^2=\dfrac{1}{z^2}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=\pm1\\y=\pm1\\z=\pm1\end{matrix}\right.\)

101;106;.....996

có : ( 996 - 101 ) : 5 + 1 = 180 số