x . $(6 - x )^{2019}$ = $(6 - x )^{2019}$
Cho \(A=x^6-2019.x^5+2019.x^4-2019.x^3+2019.x^2-2019.x+2019\) tại x = 2018
Vì \(x=2018\Rightarrow x+1=2019\)
Thay x+1=2019 vào biểu thức A ta được :
\(A=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-...-\left(x+1\right)x+x+1\)
\(=x^6-x^6-x^5+x^5+x^4-...-x^2-x+x+1\)
\(=1\)
\(A=x^6-2019x^5+2018x^4-2019x^3+2019x^2-2019x+2019\)
\(=x^6-2018x^5-x^5+2018x^4+x^4-2018x^3-x^3+2018x^2+x^2\)
\(-2018x-x+2019\)
\(=x^5\left(x-2018\right)-x^4\left(x-2018\right)-x^3\left(x-2018\right)+x^2\left(x-2018\right)\)
\(+x\left(x-2018\right)-\left(x-2018\right)+1\)
= 1
Vì \(x=2018\Rightarrow x+1=2019\)
Thay \(x+1=2019\) vào biểu thức \(A\) ta được :
\(A=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+\left(x-1\right)\)
\(=x^6-x^6-x^5+x^5-x^4+x^4-x^3+x^3-x^2+x^2-x+x+1\)
\(=1\)
x-2017/2019 + x-2019/2017 = x+6/2021
Nếu 1/3 + 1/6 +1/10 + ...... + 1/x.(x+1) : 2 = 2019/2021
A.x = 2019/2020 B. x = 2019 C. x = 2020 D. x = 2021
x-2017/2019 + x-2019/2017 = x+6/2021
\(\dfrac{x-2017}{2019}+\dfrac{x-2019}{2017}=\dfrac{x+6}{2021}\)
\(\Rightarrow\dfrac{x-2017}{2019}-1+\dfrac{x-2019}{2017}-1=\dfrac{x+6}{2021}-2\)
\(\Rightarrow\dfrac{x-2017}{2019}-\dfrac{2019}{2019}+\dfrac{x-2019}{2017}-\dfrac{2017}{2017}=\dfrac{x+6}{2021}-\dfrac{4042}{2021}\)
\(\Rightarrow\dfrac{x-2017-2019}{2019}+\dfrac{x-2019-2017}{2017}=\dfrac{x+6-4042}{2021}\)
\(\Rightarrow\dfrac{x-4036}{2019}+\dfrac{x-4036}{2017}=\dfrac{x-4036}{2021}\)
\(\Rightarrow\dfrac{x-4036}{2021}-\dfrac{x-4036}{2019}-\dfrac{x-4036}{2017}=0\)
\(\Rightarrow\left(x-4036\right)\left(\dfrac{1}{2021}-\dfrac{1}{2019}-\dfrac{1}{2017}\right)=0\)
=> x - 4036 = 0
=> x = 4036
x − 2017/2019 + x−2019/2017 = x+6/2021
=> x − 2017/2019 + x−2019/2017 = x+6/2021
=> x − 2017/2019 − 1 + x − 2019/2017 − 1 = x + 6/2021 − 2
=> x − 2017/2019 − 1 + x − 2019/2017 − 1 = x + 6/2021 − 2
=> x − 2017/2019 − 2019/2019 + x − 2019/2017 − 2017/2017
= x + 6/2021 − 4042/2021
=> x − 2017/2019 − 2019/2019 + x − 2019/2017 − 2017/2017
= x + 6/2021 − 4042/2021
=> x − 2017 − 2019/ 2019 + x − 2019 − 2017/2017
= x + 6 − 4042/2021
=> x − 2017 − 2019/2019 + x − 2019 − 2017/2017 = x + 6 − 4042/2021
=> x − 4036/2019 + x − 4036/2017 = x − 4036/2021
=> x − 4036/2019 + x − 4036/2017 = x − 4036/2021
=> x − 4036/2021 − x − 4036/2019 − x − 4036/2017 = 0
=> x − 4036/2021 − x − 4036/2019 − x − 4036/2017 = 0
=>(x − 4036)(12021 − 12019 − 12017) = 0
=> x - 4036 = 0
=> x = 4036
cho x+y+z=6 và (x-1)^3+(y-2)^3+(z-3)^3=0 tính (x-1)^2019+(y-2)^2019+(z-3)^2019
Cho x,y thỏa mãn x^2 + y^2 = 6( x - y - 3 ) Tính M = x^2019 + y^2019 + ( x + y )^2020
\(x^2+y^2=6\left(x-y-3\right)\)\(\Rightarrow x^2+y^2-6\left(x-y-3\right)=0\)
\(\Leftrightarrow x^2+y^2-6x+6y+18=0\)\(\Leftrightarrow\left(x^2-6x+9\right)+\left(y^2+6x+9\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y+3\right)^2=0\)(1)
Vì \(\hept{\begin{cases}\left(x-3\right)^2\ge0\forall x\\\left(y+3\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x-3\right)^2+\left(y+3\right)^2\ge0\forall x,y\)(2)
Từ (1) và (2) \(\Rightarrow\left(x-3\right)^2+\left(y+3\right)^2=0\Leftrightarrow\hept{\begin{cases}x-3=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-3\end{cases}}\)
\(\Rightarrow M=3^{2019}+\left(-3\right)^{2019}+\left(3-3\right)^{2020}=0\)
\(Ta \) \(có : \) \(x ^2 + y^2 = 6. ( x - y - 3 )\)
\(\Leftrightarrow\)\(x^2 + y^2 - 6. ( x - y - 3 ) = 0\)
\(\Leftrightarrow\)\(x^2 + y^2 - 6x + 6y + 18 = 0\)
\(\Leftrightarrow\)\(( x^2 - 6x + 9 ) + ( y^2 + 6y + 9 ) = 0\)
\(\Leftrightarrow\)\(( x - 3 )^2 + ( y + 3 )^2 = 0\)
\(\Leftrightarrow\)\(( x - 3 )^2 = 0 \) \(và \) \(( y - 3 )^2 = 0\)
\(\Leftrightarrow\)\(x - 3 = 0 \) \(và \) \(y + 3 = 0\)
\(\Leftrightarrow\)\(x = 3 \) \(và \) \(y = - 3\)
\(Thay\) \(x = 3 ; y = - 3 \) \(vào \) \(M \)\(ta \) \(được :\)
\(M = 3\)\(2019\) \(+ (- 3 )\)\(2019\) \(+ [ 3 + ( - 3 ) ]\)\(2020\)
\(M = 0 \)
Cho x,y thỏa mãn x^2 + y^2 = 6( x - y - 3 ) Tính M = x^2019 + y^2019 + ( x + y )^2020
cho biết x+y+z=10 và (x+6)3+(y-7)3+(z-9)3 = 0
Tính giá trị biểu thức M= (x+6)2019+(y-7)2019+(z-9)2019
Đặt \(x+6=a;y-7=b;z-9=c\)
\(\Rightarrow\hept{\begin{cases}a+b+c=0\\a^3+b^3+c^3=0\end{cases}}\)
Bạn hiểu chưa :))
Đặt x+6=a, y-7=b, z-9=c
Vì x+y+z=10 nên a+b+c=0
Xét \(a^3+b^3+c^3=0\Leftrightarrow a^3+b^3+c^3-3abc=-3abc\)(1)
Ta có đẳng thức (bạn nên học đẳng thức này nhé vì nó cực kì thông dụng trong toán nâng cao):
\(a^3+b^3+c^3-3abc=\frac{\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{2}\)(2)
Vì a+b+c=0 nên từ (1), (2) suy ra \(\hept{\begin{cases}-3abc=0\\a+b+c=0\end{cases}\Rightarrow a=b=c=0}\)
Vậy M = a2019+b2019+c2019=0
Giải hệ phương trình:
\(\left\{{}\begin{matrix}\frac{x+2019}{x+2018}+\frac{2}{y-2019}=6\\\frac{5}{x+2016}-\frac{1}{y-2019}=3\end{matrix}\right.\)
ĐKXĐ:....
Lấy pt trên cộng 2 lần pt dưới ta được:
\(\frac{x+2019}{x+2018}+\frac{10}{x+2016}=12\)
Số to quá, đặt \(x+2016=a\Rightarrow\frac{a+3}{a+2}+\frac{10}{a}=12\)
\(\Leftrightarrow12a\left(a+2\right)=a\left(a+3\right)+10\left(a+2\right)\)
\(\Leftrightarrow12a^2+24a-a^2-3a-10a-20=0\)
\(\Leftrightarrow11a^2+11a-20=0\)
Nghiệm rất xấu, bạn tự giải tiếp