ròi kêu lmj?

Đề ??

a: ΔBHA vuông tại H

=>\(BH^2+HA^2=BA^2\)

=>\(BH^2=\left(3\sqrt5\right)^2-3^2=45-9=36=6^2\)

=>BH=6(cm)

Xét ΔBAC vuông tại B có BH là đường cao

nên \(BH^2=HA\cdot HC\)

=>\(HC=\frac{BH^2}{HA}=\frac{6^2}{3}=12\left(\operatorname{cm}\right)\)

Xét ΔBCD vuông tại C có CH là đường cao

nên \(CH^2=HB\cdot HD\)

=>\(HD=\frac{12^2}{6}=\frac{144}{6}=24\left(\operatorname{cm}\right)\)

HA:HB:HC:HD=3:6:12:24=1:2:4:8

b: \(\frac{1}{HB^2}-\frac{1}{HC^2}=\frac{1}{BC^2}+\frac{1}{BA^2}-\frac{1}{BC^2}-\frac{1}{CD^2}\)

\(=\frac{1}{BA^2}-\frac{1}{CD^2}\)

Đặt \(\dfrac{ab+ac}{4}=\dfrac{bc+ab}{6}=\dfrac{ca+cb}{8}=k\)

=>ab+ac=4k; bc+ab=6k; ac+bc=8k

=>ac-bc=-2k; ac+bc=8k; ab+ac=4k

=>ac=3k; bc=5k; ab=k

=>c/b=3; c/a=5

=>c=3b=5a

=>a/3=b/5=c/15

a) \(A=2A-A\)

\(=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)

\(=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2021}}-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)

\(=1-\dfrac{1}{2^{2022}}\)

b) \(B=\dfrac{20+15+12+17}{60}=\dfrac{4}{5}=1-\dfrac{1}{5}\)

\(A>B\left(Vì\left(\dfrac{1}{2^{2022}}< \dfrac{1}{5}\right)\right)\)

a) A = 2 A − A = 2 ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) − ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) = 1 + 1 2 + . . . + 1 2 2021 − ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) = 1 − 1 2 2022 b) B = 20 + 15 + 12 + 17 60 = 4 5 = 1 − 1 5 A > B ( V ì ( 1 2 2022 < 1 5 ) )

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

\(=ab+\dfrac{2ab}{a}-\sqrt{a^2+1}=ab+2b-\sqrt{a^2+1}\)

c: \(=2\sqrt{6b}-6\sqrt{18}+10\sqrt{12}-\sqrt{48}\)

\(=2\sqrt{6b}-18\sqrt{2}+20\sqrt{3}-4\sqrt{3}\)

\(=2\sqrt{6n}-18\sqrt{2}+16\sqrt{3}\)

d: \(=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{21}}{7}\)

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a: AB/EF=4/8

BC/DF=1/2

AC/DE=1/2

=>AB/EF=BC/DF=AC/DE

=>ΔABC đồng dạng vơi ΔEFD

b: \(\dfrac{BC}{DE}=\dfrac{3}{4}=\dfrac{AB}{EF}=\dfrac{AC}{DF}\)

=>ΔBCA đồng dạng với ΔEDF

c: EF/AB=2/3

DF/BC=2/3

ED/AC=12/18=2/3

=>EF/AB=FD/BC=ED/AC

=>ΔEFD đồng dạg với ΔABC

d: AB=3k; BC=4k; AC=5k

DE=3h; EF=4h; DF=5h

=>AB/DE=BC/EF=AC/DF=k/h

=>ΔABC đồng dạng với ΔDEF

\(e,\)

\(\left(\dfrac{1}{3}a^3b+\dfrac{1}{3}a^2b^2-\dfrac{1}{4}ab^3\right):5ab\)

\(=\dfrac{1}{15}a^2+\dfrac{1}{15}ab-\dfrac{1}{20}b^2\)

\(f,\)

\(\left(-\dfrac{2}{3}x^5y^2+\dfrac{3}{4}x^4y^3-\dfrac{4}{5}x^3y^4\right):6x^2y^2\)

\(=-\dfrac{1}{9}x^3+\dfrac{1}{8}x^2y-\dfrac{2}{15}xy^2\)

\(g,\)

\(\left(\dfrac{3}{4}a^6b^3+\dfrac{6}{5}a^3b^4-\dfrac{5}{10}ab^5\right):\left(\dfrac{3}{5}ab^3\right)\)

\(=\dfrac{5}{4}a^5+2a^2b-\dfrac{5}{6}b^2\)

cam on