cho x là số nguyên âm, y là số nguyên dương. thỏa mãn | x - 2 | = 12 và | y + 1 | = 2025. Tính A = 202x + y - 4
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Ta có:
\(1+2+3+...+n\)
Số lượng số hạng là: `(n-1):1+1=n` (số hạng)
Tổng của dãy số là: `(n+1)*n/2`
Áp dụng ta có:
\(\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+....+\dfrac{1}{1+2+3+...+100}\\ =\dfrac{1}{\dfrac{3\cdot\left(3+1\right)}{2}}+\dfrac{1}{\dfrac{4\cdot\left(4+1\right)}{2}}+...+\dfrac{1}{\dfrac{100\cdot\left(100+1\right)}{2}}\\ =\dfrac{2}{3\cdot4}+\dfrac{2}{4\cdot5}+...+\dfrac{2}{100\cdot101}\\ =2\left(\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{100\cdot101}\right)\\ =2\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)\\ =2\left(\dfrac{1}{3}-\dfrac{1}{101}\right)\\ =2\cdot\dfrac{98}{303}\\ =\dfrac{196}{303}\)
\(\dfrac{-5}{7}\cdot\dfrac{2}{11}+\dfrac{-5}{7}\cdot\dfrac{9}{14}+1\dfrac{5}{7}\\ =\dfrac{-5}{7}\cdot\dfrac{2}{11}+\dfrac{-5}{7}\cdot\dfrac{9}{14}+\dfrac{5}{7}+1\\ =\dfrac{5}{7}\cdot\left(\dfrac{-2}{11}+\dfrac{-9}{14}+1\right)+1\\ =\dfrac{5}{7}\cdot\dfrac{27}{154}+1\\ =\dfrac{135}{1078}+1\\ =\dfrac{1213}{1078}\)
\(x^3+ax+b\\ =\left(x^3+4x^2+3x\right)+\left(-4x^2-16x-12\right)+\left(a+13\right)x+\left(b+12\right)\\ =x\left(x^2+4x+3\right)-4\left(x^2+4x+3\right)+\left(a+13\right)x+\left(b+12\right)\\ =\left(x-4\right)\left(x^2+4x+3\right)+\left(a+13\right)x+\left(b+12\right)\)
Để `x^3+ax+b` chia hết cho `x^2+4x+3` thì:
\(\left\{{}\begin{matrix}a+13=0\\b+12=0\end{matrix}\right.=>\left\{{}\begin{matrix}a=-13\\b=-12\end{matrix}\right.\)
Gọi cạnh của hình lập phương đó là `x (cm)`
Điều kiện: `x > 0`
Diện tích toán phần của hình lập phương là:
`x . x . 6 = 6x^2`
Thể tích hình lập phương là:
`x . x . x = x^3`
Mà diện tích toàn phần của hình lập phương bằng thể tích của nó
`=> x^3 = 6x^2`
`=> x^3 - 6x^2 = 0`
`=> x^2 (x - 6) = 0`
`=> x = 0` hoặc `x = 6`
Mà `x > 0` nên `x = 6`
Vậy cạnh của hình lập phương là `6cm`
Thể tích hình lập phương là:
`6^3 = 216 (cm^3)`
Vậy ....
Bài 3: Các cặp góc so le trong là: \(\widehat{tBO};\widehat{BOC}\); \(\widehat{OBC};\widehat{yOB}\); \(\widehat{BCO};\widehat{x'OC}\); \(\widehat{t'CO};\widehat{BOC}\)
Các cặp góc đồng vị là:
\(\widehat{xBt};\widehat{xOy}\); \(\widehat{tBO};\widehat{x'Oy}\); \(\widehat{y'Ct'};\widehat{x'Oy'}\); \(\widehat{t'CO};\widehat{x'Oy}\)

Bài 2:
Các cặp góc so le trong là \(\widehat{FEC};\widehat{ACB}\)
Các cặp góc đồng vị là \(\widehat{ADE};\widehat{ABC}\); \(\widehat{AED};\widehat{ACB}\)
Các cặp góc trong cùng phía là: \(\widehat{BDE};\widehat{B}\); \(\widehat{DEC};\widehat{ECB}\)
15 I have black hair. Alice's hair is also black.
-> Alice's hair is the same color as mine.
16 The station was nearer than I thought.
-> The station was not as far as I thought.
17 Our neighbours have lived here for quite a long time, but we have lived here longer.
-> Our neighbours haven't lived here as long as we have.
18 My mother's favourite food is noodles. My favourite food is bread.
-> My mother likes noodles, but I like bread.
19 The number of students in this class is 40. That class also has 40 students.
-> This class has the same number of students as that class.
20 My favourite subject is English. My sister is also interested in Math.
-> I like English, and my sister likes Math.
15.i have black hair.alice is houris also black (same)
->......My hair's color is the same as alice's...............
16.the station was nearer than I thought (as)
->....The station was not as far as I thought..............
17.our neighbours have lived here for quite a long time,but we have lived here longer(as).
-->......Out neighbours haven't lived here as long as us.............
18.my mother is favourite food is noodles.my favourite food is bread.(like).
-->.........My mother's favorite food is not like mine......................
19.the number of student in this class is 40.thas class also has 40 students(same_
-->.......The number of students in this class is the same as that class................................
20.my favourite subject is english.My sister is also interested in Math (like)
-->......I don't like the same subject as my sister.................................
`-1/3<=x/3<=-1/6`
`=>-2/6<=2x/3<=-1/6`
`=>-2<=2x<=-1`
`=>-2/2<=x<=-1/2`
`=>-1<=x<=-1/2`
\(\dfrac{-1}{2}< \dfrac{x}{3}< \dfrac{-1}{6}\)
`=>` \(\dfrac{-3}{6}< \dfrac{2x}{6}< \dfrac{-1}{6}\)
`=> -3 < 2x < -1`
Mà `2x` là số nguyên
`=> 2x = -2`
`=> x = -1`
Vậy `x = -1`
`4^3<=2^x<=2^10`
`=>(2^2)^3<=2^x<=2^10`
`=>2^(2*3)<=2^x<=2^10`
`=>2^6<=2^x<=2^10`
`=>6<=x<=10`

\(\left|x-2\right|=12\\ =>\left[{}\begin{matrix}x-2=12\\x-2=-12\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=12+2=14\left(ktm\right)\\x=-12+2=-10\left(tm\right)\end{matrix}\right.\\ \left|y+1\right|=2025\\ =>\left[{}\begin{matrix}y+1=2025\\y+1=-2025\end{matrix}\right.\\ =>\left[{}\begin{matrix}y=2025-1=2024\left(tm\right)\\y=-2025-1=-2026\left(ktm\right)\end{matrix}\right.\)
\(A=202x+y-4=202\cdot-10+2024-4=-2020+2024-4=0\)