Dạng 1: Bài luyện tập 1 có đáp án

2x+12x−1−2x−12x+1:4x10x+5   (dkxd:x≠±12)=2x+122x−12x+1−2x−122x+12x−1.10x+54x

=2x+12−2x−122x−12x+1.10x+54x

=2x+1+2x−12x+1−2x+12x−12x+1.10x+54x

=8x2x−12x+1.10x+54x

=8x2x−12x+1.52x+14x

=10x2x−1

1x2+x−2−xx+1:1x+x−2 DK:x≠0;x≠−1

=1xx+1−2−xx+1:1+x2−2xx

=1xx+1−x2−xxx+1:x−12x

=x2−2x+1xx+1:x−12x

=x−12xx+1:x−12x

=x−12xx+1.xx−12

=1x+1

1x−1−x3−xx2+1.1x2−2x+1+11−x2DK:x≠±1

=1x−1−xx+1x−1x2+1.1x−12−1x+1x−1

=1x−1−xx+1x−1x2+1.x+1x+1x−12−x−1x+1x−12

=1x−1−xx+1x−1x2+1.x+1−x+1x+1x−12

=1x−1−xx+1x−1x2+1.2x+1x−12

=1x−1−2xx2+1x−1

=x2+1−2xx2+1x−1

=x−12x2+1x−1=x−1x2+1

P=2+xx−2+2x+2−x2+5xx2−4:1−x+1x+2DK:x≠±2

=2+x2x−2x+2+2x−2x+2x−2−x2+5xx+2x−2:x+2−x−1x+2

=x2+4x+4+2x−4−x2−5xx−2x+2:x+2−x−1x+2

=xx−2x+2:1x+2

=xx−2x+2x+2=xx−2

x2−2x=0⇔xx−2=0⇔x=0(tm)x=2(l)

Thay x=0 vào P , ta có P=00−2=0

Vậy P=0 khi  x=0(tm)

A=xx−2+3x−22x−x2:x+2x+4−xx−2DK:x≠0;x≠2

=xx−2+3x−2x2−x:x+2x+4−xx−2

=x2xx−2−3x−2xx−2:x+2x−2xx−2+4−xxxx−2

=x2−3x+2xx−2:x2−4+4x−x2xx−2

=x−2x−1xx−2:4x−1xx−2

=x−2x−1xx−2.xx−24x−1

=x−24

x2−5x+6=0⇔x−2x−3=0⇔x=2(l)x=3tm

Thay x=3(tm) vào A , ta có

A=3−24=14

VậyA=14  khi  x=3(tm)

P=x−12:x2+2x3−1+xx2+x+1+11−xDK:x≠1

P=x−12:x2+2x−1x2+x+1+xx2+x+1+11−x

P=x−12:x2+2x−1x2+x+1+xx−1x−1x2+x+1−x2+x+1x−1x2+x+1

=x−12:x2+2+x2−x−x2−x−1x−1x2+x+1

=x−12:x2−2x+1x−1x2+x+1

=x−12:x−12x−1x2+x+1

=x−12:x−1x2+x+1=x−12.x2+x+1x−1

=x2+x+12

x2+x+12=12x2+2.x.12+14+34=12x+122+38

Ta có x+122≥0,∀x∈TXD

⇒12x+122≥0⇒12x+122+38≥38>0

⇒P>0

P=x2+x+12=12x2+2.x.12+14+34=12x+122+38

Ta có x+122≥0,∀x∈TXD

⇒12x+122≥0⇒12x+122+38≥38>0 

 ⇒P≥38

 minP=38, dấu bằng xảy ra ⇔x=−12

P=x+11−x−1−x1+x−4x2x2−1:4x2−4x2−2x+1DK:x≠±1

=−x+12x−1x+1+x−12x−1x+1−4x2x−1x+1:4x+1x−1x−12

=x−12−x+12−4x2x−1x+1:4x+1x−1

=−4x−4x2x−1x+1.x−14x+1

=−4xx+1x−1x+1.x−14x+1

=−xx+1

x2+4x=5⇔x2+4x−5=0⇔x+5x−1=0⇔x=−5tmx=1  l

Thay x=−5(tm)vào P , ta cóP=5−5+1=−54

Vậy  P=−54khi  x=−5(tm)

P=−xx+1=−x−1+1x+1=−1+1x+1

P∈Z⇔1x+1∈Z⇔x+1∈U1=±1

Vậy  x∈0;−2

P=x2−3xx2−9−1:9−x2x2+x−6−x−32−x−x−2x+3DK:x≠±3;x≠2

=xx−3x+3x−3−1:9−x2x−2x+3+x−3x−2−x−2x+3

=xx+3−1:9−x2x−2x+3+x−3x+3x−2x+3−x−22x−2x+3

=x−x−3x+3:9−x2+x2−9−x2+4x−4x−2x+3

=−3x+3:−x2+4x−4x−2x+3

=−3x+3.x−2x+3−x−22

=3x−2

x3–4x=0⇔xx+2x−2=0⇔x=0tmx=2lx=−2tm

·         Thay x=0(tm) vào P , ta có

 P=30−2=−32

Vậy P=−32 khi  x=0(tm)

·         Thay x=−2(tm) vào P , ta có

 P=3−2−2=−34

Vậy P=−34 khi  x=−2(tm)

P=x−1x−3+x−3x+3+4x−2x2−9:1−x+1x+3DK:x≠±3

=x−1x+3x+3x−3+x−3x−3x+3x−3+4x−2x+3x−3:x+3−x−1x+3

=x2+2x−3+x2−6x+9+4x−2x+3x−3:2x+3

=2x2+4x+3x−3.x+32

=2x2+2x+3x−3.x+32

=x2+2x−3

2x−1−1=0⇔2x−1=1⇔2x−1=±1⇔2x−1=12x−1=−1⇔x=1tmx=0tm

·         Thay x=0(tm) vào P  , ta có

 P=02+20−3=−23

Vậy P=−23 khi  x=0(tm)

·         Thay x=1(tm) vào P , ta có

 P=12+21−3=−32

Vậy P=−32 khi  x=0(tm)

P=x2+2x−3=x+3+11x−3P∈Z⇔11x−3∈Z⇔x−3∈U11=±1;±11

Vậy  x∈2;14;4;−8

2aba2−b2+a−b2a+2b.2aa+b+bb−a=2aba+ba−b+a−b2(a+b).2aa+b+bb−a=4ab2a+ba−b+a−b22a+ba−b.2aa+b+bb−a=4ab+a2−2ab+b22a+ba−b.2aa+b+bb−a=a2+2ab+b22a+ba−b.2aa+b+bb−a=a+b22a+ba−b.2aa+b+bb−a=a+b2a−b.2aa+b+bb−a=aa−b+bb−a=a−ba−b=1

xx−y−x3−xy2x2+y2.xx−y2−yx2−y2=xx−y−xx+yx−yx2+y2.xx−y2−yx+yx−y=xx−y−xx+yx−yx2+y2.xx+yx−y2x+y−yx−yx+yx−y2=xx−y−xx+yx−yx2+y2.x2+xy−xy+y2x−y2x+y=xx−y−xx+yx−yx2+y2.x2+y2x−y2x+y=xx−y−xx−y=0

y3−y+y2+3y2y+3.y+3y2−3y−yy2−9=y3−y+y2+3y2y+3.y+3yy−3−yy+3y−3=y3−y+yy+32y+3.y+32yy−3y+3−y2yy+3y−3=y3−y+yy+32y+3.y+3+yy+3−yyy−3y+3=y3−y+yy+32y+3.32y+3yy−3y+3=y3−y+3y−3=y−33−y=−1

xx2−36−x−6x2+6x:2x−6x2+6x+x6−x=xx+6x−6−x−6xx+6:2x−3xx+6+x6−x=x2xx+6x−6−x−62xx+6x−6.xx+62x−3+x6−x=x+x−6x−x+6xx+6x−6.xx+62x−3+x6−x=12x−3xx+6x−6.xx+62x−3+x6−x=6x−6+x6−x=6−xx−6=−1