Bài tập theo tuần Toán 9 - Tuần 4

23−1−23+1=23+1−23−13−13+1=23+2−23+23−1=42=2

2xyy2x=4x2.yy2.2x=2xy=2xy=2xyy

32+233+2=6.3+23+2=6

A=4.13−8+17−8−16−7−16−5−12−5=4.3+89−8+7+87−8−6+76−7−6+56−5−2+54−5=4.3+8−7−8+6+7−6−5+2+5=4.5=20

2x+3=1+2x≥−32⇒2x+3=3+22⇔x=2(tm)

Ta có: sinC=ABBC;sinB=ACBC

⇒sinCsin​B=ABBC:ACBC=ABBC.BCAC=ABAC(dfcm)

tanα.cotα=ABAC.ACAB=1

sinα=45⇒cos2α=1−sin2α=1−452=925⇒cosα=925=35tanα=sinαcosα=45:35=43cotα=1tanα=34

sin210°+sin220°+.....+sin270°+sin2800=sin210°+sin220°+.....+cos220°+cos2100=sin210°+cos2100+sin220°+cos220°+....=1+1+1+1=4

Vẽ AH⊥BC,H∈BC

Xét ΔHAB có H^=900⇒sinB=AHAB

Xét ΔHAC có H^=900⇒sinC=AHAC

Do đó sinBsinC=ACAB=bc⇒bsinB=csinC

Chứng minh tương tự ⇒asinA=bsinB

Vậy asinA=bsinB=csinC

cos78°=sin12°,cos68°=sin22°,cos50°=sin40°

Vì sin54°>sin40°>sin27°>sin22°>sin19°>sin12°

⇒sin54°>cos50°>sin27°>cos68°>sin19°>cos78°

ΔABC vuông tại A ⇒tanB=ACAB hay tan470=63x⇒x=63tan470≈58,75

ΔDEF vuông tại D ⇒cosF^=DEEF⇒EF.cosF=DF

Hay x.cos380=16⇒x≈16cos380≈20,304(cm)

51225+32−51225−32=525−32−525+321225+3225−32=105−152−105−1521220−18=−30224=−524

5+55−5+5−55+5=5.5+15.5−1+5.5−15.5+1=5+15−1+5−15+1=5+12+5−125−15+1=6+25+6−255−1=124=3

33+1−1−33+1+1=3.3+1+1−33+1−13+12−1=3.3+1+1−3+1+13+1−1=3+1+1−3+1+1=2

1a4+1a=a3+1a4=a3+1a2

16x381y=1681.x3y=4xx9y=4xxy9y

16x381y=1681.x3y=4xx9y=4xxy9y

157−2=157+27−4=157+23=57+2=57+10

33+323=3.3+323=3+32

23−62−1=6.2−12−1=6

B=11+2+12+3+13+4+.....+12004+2005=2−12−1+3−23−2+4−34−3+......+2006−20052006−2005=2−1+3−2+4−3+...+2006−2005=2006−1

C=11−2−12−3+13−4−.......−148−49=1+21−2−2+32−3+3−43−4−.....−48+4948−49=−1−2+2+3−3−4−.....+48+49=−1+49=−1+7=6

Ta có:

1n+1n+nn+1=1nn+1n+1+n=n+1−nnn+1.n+1+n=n+1−nnn+1=1n−1n+1

Do đó 

D=11−12+12−13+.....+199−1100=11−1100=910

xx−yyx−y=x3−y3x−y=x−yx+xy+yx−y=x+xy+y

x−3x+3xx+33=x−3x+3x+3x−3x+3=1x+3=x−3x−3

10+3x=2+6x≥0⇒10+3x=10+463x=46⇒3x=96⇔x=32(tm)

3x−2=2−3x≥23⇒3x−2=7−43⇔3x=9−43⇒x=3−433tm

 x+1=5−3x≥−1⇒x+1=14−65⇒x=13−65(tm)

tanα=ABAC=ABBCACBC=sinαcosα

sin2α+cos2α=ABBC2+ACBC2=AB2+AC2BC2=BC2BC2=1