Câu 1: Thực hiện phép tính
A= -125 x 2^3 + 71 x 53 + 53 x (- 29) - 42 x 53
Câu 2: Tính giá trị biểu thức
A= 2019/1x2 + 2019/2x3 + 2019/3x4 +...........+ 2019/2018x2019
Tính giá trị biểu thức
A= 2019/1x2 + 2019/2x3 + 2019/3x4 +.............+ 2019/2018x2019
Ai nhanh m tick nha
\(\frac{2019}{1\times2}+\frac{2019}{2\times3}+\frac{2019}{3\times4}+...+\frac{2019}{2018\times2019}\)
\(=2019\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2018\times2019}\right)\)
\(=2019\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=2019\left(1-\frac{1}{2019}\right)\)
\(=2019\left(\frac{2019}{2019}-\frac{1}{2019}\right)\)
\(=2019\times\frac{2018}{2019}\)\(=\frac{2019\times2018}{2019}=2018\)
Cứu e vs :<
Thực hiện phép tính
a) 58.57 + 58.150 – 58.125 b) 32.5 - 22.7 + 83.20190
c) 2019 + (-247) + (-53) – 2019 d) 13.70 – 50 [(19 - 32) : 2 + 23]
a: =58(57+150-125)=58x82=4756
b: \(=9\cdot5-4\cdot7+83=45-28+83=100\)
c: =(2019-2019)+(-247-53)=-300
d: \(=13\cdot70-50\cdot\left[10:2+8\right]=910-50\cdot13=910-650=260\)
\(a,=58.\left(57+150-125\right)\\ =58.82=4756\\ b,=9.5-4.7+83.1\\ =45-28+83=100\)
a, Tìm một số biết 0,125 của số đó là: 5,32
b, Tính giá trị biểu thức A= \(\dfrac{2020}{2019}\) - \(\dfrac{2019}{2018}\) + \(\dfrac{1}{2018x2019}\) =
\(x\) là dấu nhân
a: Số cần tìm là 5,32:0,125=42,56
b: \(A=1+\dfrac{1}{2019}-1-\dfrac{1}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}=0\)
cho 2 số thực x,y thỏa mãn (x+\(\sqrt{x^2+2019}\))\(\left(y+\sqrt{y^2+2019}\right)\)=2019. tính giá trị biểu thức P=x4+x3y+3x2+xy-2y2+1
Tính giá trị biểu thức sau tại x=2018
x5-2019.x4+2019.x3-2019.x2+2019.x-2020
giá trị biểu thức là 174
Ta có x = 2018
=> x + 1 = 2019
\(x^5-2019.x^4+2019.x^3-2019.x^2+2019.x-2020\)
\(=x^5-\left(x+1\right).x^4+\left(x+1\right).x^3-\left(x+1\right).x^2+\left(x+1\right).x-2020\)
\(=x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-2020\)
\(=x-2020\)
Thay x = 2018 vào biểu thức , ta được
\(2018-2020=-2\)
Vậy giá trị biểu thức là -2
cho các số thực x,y,z thỏa mãn : x3 + y3 = z( 3xy - z2) và x + y + z = 3 . Tính giá trị của biểu thức A = 673.(x2019 + y2019 + z2019) + 1
Ta có : x3 + y3 = z(3xy - z2)
=> x3 + y3 = 3xyz - z3
=> x3 + y3 + z3 - 3xyz = 0
=> (x + y)(x2 - xy + y2) + z3 - 3xyz = 0
=> (x + y)3 - 3xy(x + y) + z3 - 3xyz = 0
=> [(x + y)3 + z3] - 3xy(x + y) - 3xyz = 0
=> (x + y + z)[(x + y)2 - (x + y)z + z2] - 3xy(x + y + z) = 0
=> (x + y +z)(x2 + y 2 + 2xy - xz - yz + z2) - 3xy(x + y + z) = 0
=> (x + y + z)(x2 + y2 + z2 - xy - yz - zx) = 0
=> x2 + y2 + z2 - xy - yz - zx = 0 (Vì x + y + z = 3)
=> 2(x2 + y2 + z2 - xy - yz - zx) = 0
=> 2x2 + 2y2 + 2z2 - 2xy - 2yz - 2zx = 0
=> (x2 - 2xy + y2) + (y2 - 2yz + z2) + (x2 - 2zx + z2) = 0
=> (x - y)2 + (y - z)2 + (x - z)2 = 0
=> \(\hept{\begin{cases}x-y=0\\y-z=0\\x-z=0\end{cases}}\Rightarrow x=y=z\)
mà x + y + z = 3
=> x = y = z = 1
Khi đó A = 673(x2019 + y2019 + z2019) + 1
= 673(12019 + 12019 + 12019) + 1
= 673.3 + 1 = 2020
Vậy A = 2020
1. 2019/2020-(2019/2020-2020/2021)
2.2/9+7/9 :(42/5-7/5
3.a)3/4+x/4=5/8
4./3x+1/-1/4=-1/4
1. \(\dfrac{2019}{2020}-\left(\dfrac{2019}{2020}-\dfrac{2020}{2021}\right)\)
\(=\dfrac{2019}{2020}-\dfrac{2019}{2020}+\dfrac{2020}{2021}\)
\(=0+\dfrac{2020}{2021}=\dfrac{2020}{2021}\)
Giải:
1) \(\dfrac{2019}{2020}-\left(\dfrac{2019}{2020}-\dfrac{2020}{2021}\right)\)
\(=\dfrac{2019}{2020}-\dfrac{2019}{2020}+\dfrac{2020}{2021}\)
\(=\left(\dfrac{2019}{2020}-\dfrac{2019}{2020}\right)+\dfrac{2020}{2021}\)
\(=0+\dfrac{2020}{2021}\)
\(=\dfrac{2020}{2021}\)
2) \(\dfrac{2}{9}+\dfrac{7}{9}:\left(\dfrac{42}{5}-\dfrac{7}{5}\right)\)
\(=\dfrac{2}{9}+\dfrac{7}{9}:7\)
\(=\dfrac{2}{9}+\dfrac{1}{9}\)
\(=\dfrac{1}{3}\)
3) \(\dfrac{3}{4}+\dfrac{x}{4}=\dfrac{5}{8}\)
\(\dfrac{x}{4}=\dfrac{5}{8}-\dfrac{3}{4}\)
\(\dfrac{x}{4}=\dfrac{-1}{8}\)
\(\Rightarrow x=\dfrac{4.-1}{8}=\dfrac{-1}{2}\)
4) \(\left|3x+1\right|-\dfrac{1}{4}=\dfrac{-1}{4}\)
\(\left|3x-1\right|=\dfrac{-1}{4}+\dfrac{1}{4}\)
\(\left|3x-1\right|=0\)
\(3x-1=0\)
\(3x=0+1\)
\(3x=1\)
\(x=1:3\)
\(x=\dfrac{1}{3}\)
Chúc bạn học tốt!
4) \(\left|3x+1\right|-\dfrac{1}{4}=\dfrac{-1}{4}\)
\(\left|3x+1\right|=\dfrac{-1}{4}+\dfrac{1}{4}\)
\(\left|3x+1\right|=0\)
\(3x+1=0\)
\(3x=0-1\)
\(3x=-1\)
\(x=-1:3\)
\(x=\dfrac{-1}{3}\)
bài 1 : 1/1x2+1/2x3+1/3x4+1/5x6+1/6x7
Bài 2:1/2+1/6+1/12+1/30+1/42+1/56+1/72
Bài 3: (1-1/2)x(1-1/3)x....x(1-2019)
Bài 4:1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/512
1) 1/1.2 + 1/2.3 + ... + 1/6.7
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/6 - 1/7
= 1 - 1/7
= 6/7
2) 1/2 + 1/6 + 1/12 + .. + 1/72
= 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9
= 1 - 1/9
= 8/9
3) \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2019}\right)\)
= \(\frac{1}{2}.\frac{2}{3}...\frac{2019}{2020}\)
= \(\frac{1.2....2019}{2.3...2020}\)
= \(\frac{1}{2020}\)
4) A = \(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{512}\)
= \(\frac{1}{2^2}+\frac{2}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}\)
=> 2A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\)
Lấy 2A - A = \(\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}\right)\)
A = \(\frac{1}{2}-\frac{1}{2^9}\)
Bài 1: Thực hiện phép tính
a) (134 – 34).(-28) + 72.[(-55) – 45]
b) 4.25 – 12.5 + 170:10
c) 369 – (206 – 15) – (-206 + 369)
d) 345 – 150:[(33 – 24)2 – (-21)] +20160
e) 307 - [(180.40 - 160) : 22 + 9] : 2
f) 350.12.173 + 12.27
h, 2019 + (-247) + (-53) – 2019
g) 13.70 – 50.[(19-32) : 2 + 23]
h) 32 : 4 + [60 – (12 – 7)2]
i) 17.85 + 15.17 – 120