( anh mod truongtiennhat mô rồi vanhsura đây )
Tính :
B = x^2017 – 2017.x^2016 + 2017.x^2015 – 2017.x^2014 + … + 2017.x + 1
Tìm x,y,z biết :
a) (3x-5)^2006 + (y² -1)^2008 + (x-z)^2020 = 0
b) |x-1| + |y+3| + |x² + xz| = 0
c) (2x-1)^2020 + (y-5)^2020 + |x+y-z| = 0
Tìm x :
a) ||x+3|- 8 | = 20 – |x+3|
b) (x-1/2009) + (x-2/2008) = (x-3/2007) + (x-4/2006)
( anh mod truongtiennhat mô rồi vanhsura đây ) Tính : B = x^2017 – 2017.x^2016 + 2017.x^2015 – 2017.x^2014 + … + 2017.x + 1 Tìm x,y,z biết :
By Madelyn
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
Giải thích các bước giải:
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