x.B= $x^{2018}-2017.x^{2017}+2017.x^{2016}-...+2017.x^{2}+x$

⇒ x.B+B= $x^{2018}-2017.x^{2017}+2017.x^{2016}-...+2017.x^{2}+x+x^{2017}-2017.x^{2016}+...+2017x+1$

⇔ B.( x+1)= $x^{2018}-2016.x^{2017}+2018x+1$

⇔ B= $\frac{x^{2018}-2016.x^{2017}+2018x+1}{x+1}$

Bài 2:

a, $(3x-5)^{2006} + (y² -1)^{2008} + (x-z)^{2020} = 0$

Vì $(3x-5)^{2006}≥ 0$

$(y² -1)^{2008}≥ 0$

$(x-z)^{2020}≥0$

⇒ $(3x-5)^{2006} + (y² -1)^{2008}+(x-z)^{2020}≥ 0$

⇒ Để phương trình thỏa mãn thì 

$(3x-5)^{2006}= 0$⇔ $3x-5= 0$⇔ $x= \frac{5}{3}$

$(y² -1)^{2008}= 0$⇔$ y²-1= 0$⇔ $y²=1$⇔ $y=±1$

$(x-z)^{2020}=0$⇔ $x-z= 0$⇔ $z=x=\frac{5}{3}$

b, | x-1|+| y+3|+| x²+xz|= 0

Vì | x-1|≥ 0

| y+3|≥ 0

| x²+xz|≥ 0

⇒| x-1|+| y+3|+| x²+xz| ≥ 0

Do đó để phương trình thõa mãn thì: 

| x-1|= 0⇔ x-1= 0⇔ x= 1

| y+3|= 0⇔ y+3= 0⇔ y= -3

| x²+xz|= 0⇔ x²+xz= 0⇔ 1+z= 0⇔ z= -1

Vậy x= 1; y= -3; z= -1

c, $(2x-1)^{2020} + (y-5)^{2020} + |x+y-z| = 0$

Giải thích như trên

⇒ $ 2x-1= 0$ ⇔ $x=0,5$

$y-5=0$⇔ $y=5$

$x+y-z=0$⇔ $ 0,5+5-z= 0$ ⇔$ z= 5,5$

Bài 3:

a, ||x+3|- 8 | = 20 - |x+3|

Th1: |x+3|- 8= 20 - |x+3|

⇔ 2.|x+3|= 28

⇔ |x+3|= 14

⇔ \(\left[ \begin{array}{l}x+3=14\\x+3=-14\end{array} \right.\) 
⇔ \(\left[ \begin{array}{l}x=11\\x=-17\end{array} \right.\) 

Th2: |x+3|- 8= -20 + |x+3|

⇔ 0= -12 ( vô lí)

b, $\frac{x-1}{2009}$+$\frac{x-2}{2008}$= $\frac{x-3}{2007}$+$\frac{x-4}{2006}$

⇔ $\frac{x-1}{2009}$-1+$\frac{x-2}{2008}$-1= $\frac{x-3}{2007}$-1+$\frac{x-4}{2006}$-1

⇔ $\frac{x-2010}{2009}$+$\frac{x-2010}{2008}$= $\frac{x-2010}{2007}$+$\frac{x-2010}{2006}$

⇔ $(x-2010)$.( $\frac{1}{2009}$+$\frac{1}{2008}$-$\frac{1}{2007}$-$\frac{1}{2006}$)= 0

⇔ x-2010= 0

⇔ x= 2010

 

Đáp án:

B= x2018−2017.x2017+2017.x2016−...+2017.x2+xx2018−2017.x2017+2017.x2016−...+2017.x2+x

⇒ x.B+B=x2018−2017.x2017+2017.x2016−...+2017.x2+x+x2017−2017.x2016+...+2017x+1x2018−2017.x2017+2017.x2016−...+2017.x2+x+x2017−2017.x2016+...+2017x+1

⇔ B.( x+1)= x2018−2016.x2017+2018x+1x2018−2016.x2017+2018x+1

⇔ B= x2018−2016.x2017+2018x+1x+1x2018−2016.x2017+2018x+1x+1

Bài 2:

a, $(3x-5)^{2006} + (y² -1)^{2008} + (x-z)^{2020} = 0$

Vì (3x−5)2006≥0(3x−5)2006≥0

(y²−1)2008≥0(y²−1)2008≥0

(x−z)2020≥0(x−z)2020≥0

⇒ (3x−5)2006+(y²−1)2008+(x−z)2020≥0(3x−5)2006+(y²−1)2008+(x−z)2020≥0

⇒ Để phương trình thỏa mãn thì 

(3x−5)2006=0(3x−5)2006=0⇔ 3x−5=03x−5=0⇔ x=53x=53

(y²−1)2008=0(y²−1)2008=0⇔y²−1=0y²−1=0⇔ y²=1y²=1⇔ y=±1y=±1

(x−z)2020=0(x−z)2020=0⇔ x−z=0x−z=0⇔ z=x=53z=x=53

b, | x-1|+| y+3|+| x²+xz|= 0

Vì | x-1|≥ 0

| y+3|≥ 0

| x²+xz|≥ 0

⇒| x-1|+| y+3|+| x²+xz| ≥ 0

Do đó để phương trình thõa mãn thì: 

| x-1|= 0⇔ x-1= 0⇔ x= 1

| y+3|= 0⇔ y+3= 0⇔ y= -3

| x²+xz|= 0⇔ x²+xz= 0⇔ 1+z= 0⇔ z= -1

Vậy x= 1; y= -3; z= -1

c, (2x−1)2020+(y−5)2020+|x+y−z|=0(2x−1)2020+(y−5)2020+|x+y−z|=0

Giải thích như trên

⇒ 2x−1=02x−1=0 ⇔ x=0,5x=0,5

y−5=0y−5=0⇔ y=5y=5

x+y−z=0x+y−z=0⇔ 0,5+5−z=00,5+5−z=0 ⇔z=5,5z=5,5

Bài 3:

a, ||x+3|- 8 | = 20 - |x+3|

Th1: |x+3|- 8= 20 - |x+3|

⇔ 2.|x+3|= 28

⇔ |x+3|= 14

⇔ [x+3=14x+3=−14[x+3=14x+3=−14 
⇔ [x=11x=−17[x=11x=−17 

Th2: |x+3|- 8= -20 + |x+3|

⇔ 0= -12 ( vô lí)

b, x−12009x−12009+x−22008x−22008= x−32007x−32007+x−42006x−42006

⇔ x−12009x−12009-1+x−22008x−22008-1= x−32007x−32007-1+x−42006x−42006-1

⇔ x−20102009x−20102009+x−20102008x−20102008= x−20102007x−20102007+x−20102006x−20102006

⇔ (x−2010)(x−2010).( 1200912009+1200812008-1200712007-1200612006)= 0

⇔ x-2010= 0

⇔ x= 2010

 

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