Bài tập theo tuần Toán 8 - Tuần 15

x2−y26x2y2:x+y3xy=x−yx+y6x2y2.3xyx+y=x−y2xy

27−x35x+5:2x−63x+3=3−x9+3x+x25x+1.3x+1−23−x=x2+3x+9−32.5=−3x2+3x+910

5x+yx2−5xy+5x−yx2+5xy.x2−25y2x2+y2=5x+yxx−5y+5x−yxx+5y.x2−5y2x2+y2=5x+yx+5y+x−5y5x−yxx−5yx+5y.x−5yx+5yx2+y2=5x2+26xy+5y2+5x2−26xy+5y2xx2+y2=10x2+y2xx2+y2=10x

12x−y2+24x2−y2+12x+y2.4x2+4xy+y216x=2x+y2+22x−y2x+y+2x−y22x−y2.2x+y2.2x+y216x=2x+y+2x−y22x−y2.16x=4x22x−y.16x=16x22x−y.16x=x2x−y

a) Điều kiện để A có nghĩa : x≠2,x≠−3b) A=x+2x+3−5x2+x−6+12−x=x+2x+3−5x+3x−2−1x−2=x+2x−2−5−x+3x+3x−2=x2−4−5−x−3x+3x−2=x2−x−12x+3x−2=x−4x+3x+3x−2=x−4x−2c) A=−34⇔x−4x−2=−34⇔4x−16=6−3x⇔x=227(tm)d) A=x+1x+3=x+3−2x+3=1−2x+3Để A∈ℤ⇔2x+3∈ℤ⇔x+3∈U(2)=−1;1;−2;2⇒x∈−2;−4;−1;−5(tm)Vậy x∈−2;−4;−1;−5 thì A∈ℤe) x2−9=0⇔x=3(tm)x=−3(ktm)⇒A=1−41−2=3 

a) P xác định khi x≠±1,x≠0

b) P=x2+1x+1−14x−1−2x=x2+1−x−1x+1.4x−2x+2xx−1=x2−xx+1.2x+2xx−1=xx−1.2x+1x+1.x.x−1=2

a) M có nghĩa khi x≠2,x≠0

b) M=x2x−2.x2+4x−4+3=x2x−2.x2−4x+4x+3=xx−22x−2+3=xx−2+3=x2−2x+3

AD2=16⇒AD=4⇒AB=284=7mVậy AD=4cm, AB=7cm

Ta có: BC = a, theo Pytago, ta có:

AB2+AC2=a2   hay  2AB2=a2⇒AB2=a22⇒AB=AC=a22

⇒SABC=12AB.AC=12.a22.a22=a24(dvdt)