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1.
3A= 3(3+32+ ...+ 32017)
3A= 32 + 33 + .... + 32018
Lấy 3A - A = (32 + 33 +...+ 32018) - (3+32+...+32017)
2A = 32018 - 3
2A+3 = 32018 - 3 +3 = 32018
=> 2A+3 là một lũy thừa của 3
a) ( 195 - x3 ) : 7 + 105 : 102 = 28 . 22
=> (195 - x3 ) : 7 + 1000 = 1024
=> (195 - x3 ) : 7 = 1024 - 1000 = 24
=> 195 - x3 = 24 . 7 = 168
=> x3 = 195 - 168 = 27
=> x = 3 hoặc x = - 3
Vậy : x = 3 hoặc - 3
b) 3 . ( x2 - 65 ) + 15 = 810 : 88 - 1690
=> 3 . ( x2 - 65 ) + 15 = 64 - 1 = 63
=> 3 . ( x2 - 65 ) = 63 - 15 = 48
=> x2 - 65 = 48 : 3 = 16
=> x2 = 16 + 65 = 81
=> x = 9 hoặc x = - 9
Vậy : x = 9 hoặc x = - 9
1. She is very beautiful.
2. Mai’s sister is a singer.
3. Her teeth are small and white.
4. Is your father a farmer ?
5. It’s very noisy in the city.
6. Our children are in the bookstore.
7. Don’t go straight ahead.It’svery dangerous.
8. I’m watching television now.
~~~~Good luck~~~~
* Three sentences use tobe
1, She is a doctor
2, I am a student
3, He is singing a song
* From 0 to 20
+)zero , one , two , three , four , five, six , seven , eight , nine , ten , eleven , twelve , thirteen , fourteen , fifteen , sixteen , seventeen ,eighteen , nineteen , twenty .
+) first , second , third , fourth , fifth , sixth , seventh , eighth , ninth , tenth , eleventh , twelfth , thirteenth , fourteenth, fifteenth , sixteenth , seventeenth , eighteenth , nineteenth , twentieth .
2.
Theo đề bài ta có
480 \(⋮\) a
600 \(⋮\) a
a lớn nhất
\(\Rightarrow\) a \(\in\) ƯCLN ( 480 ; 600 )
480 = 25 . 3 .5
600 = 23 . 3 . 52
a \(\in\){ 480 ; 600 } = 23 . 3 . 5 = 120
6.
Gọi số người tham gia buổi đồng diễn là a ( a \(\in\) N* )
Theo đề bài ta có
350 < a < 400
a \(⋮\) 5
a\(⋮\) 6
a\(⋮\) 8
\(\Rightarrow\) a \(\in\) BC ( 5 ; 6 ; 8 ) , 350 < a < 400
5 = 5
6 = 2 . 3
8 = 23
BCNN ( 5 ; 6 ; 8 ) = 5 . 23 . 3 = 120
a \(\in\) BC ( 5; 6 ; 8 ) = B ( 120 ) = { 0 ; 120 ; 240 ; 360 ; 480 ; 600 ...}
mà 350 < a < 400
\(\Rightarrow\) a \(\in\) { 360 }
Vậy số người tham gia buổi đồng diễn là 360 người
a) \(\left|x\right|< 1\Rightarrow-1< x< 1\Rightarrow x=0\)
b) \(\left|x+3\right|=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
c) \(\left|x+2\right|=\left|12-10\right|\)
\(\Leftrightarrow\left|x+2\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=-2\\x+2=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\left(-2\right)-2\\x=2-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=0\end{matrix}\right.\)
d) \(\left|x+3\right|=2x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-2\ge0\\\left[{}\begin{matrix}x+3=2x-2\\x+3=\left(-2x\right)+2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x\ge2\\\left[{}\begin{matrix}x-2x=-2-3\\x-\left(-2x\right)=2-3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\left[{}\begin{matrix}-x=-5\\3x=-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\left[{}\begin{matrix}x=5\left(tm\right)\\x=\dfrac{-1}{3}\end{matrix}\right.\end{matrix}\right.\)
Vì \(\dfrac{-1}{3}< 1\) nên \(x=5\) thỏa mãn đề bài.
e) \(\left|x+1\right|>4\)
\(\Rightarrow\left[{}\begin{matrix}x+1>4\\x+1< 4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>3\\x< 3\end{matrix}\right.\)
f) \(\left|x-3\right|=\left|2x-1\right|\)
(cho thời gian suy nghĩ, mình chưa làm dạng này bao giờ)
g) \(\left|2x-1\right|-1+2x=0\)
\(\Rightarrow\left|2x-1\right|=-2x+1\)
Mà \(\left|2x-1\right|=\left|-2x+1\right|\)
\(\Rightarrow\left|-2x+1\right|=-2x+1\)
\(\Rightarrow-2x+1\ge0\)
\(\Rightarrow-2x\ge-1\)
\(\Rightarrow x\ge\dfrac{-1}{-2}=\dfrac{1}{2}\)
h) \(\left|3-2x\right|=2x-3\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3\ge0\\\left[{}\begin{matrix}3-2x=2x-3\\3-2x=-2x+3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x\ge3\\\left[{}\begin{matrix}3+3=2x+2x\\3-3=-2x+2x\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\left[{}\begin{matrix}6=4x\\0=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\left[{}\begin{matrix}x=\dfrac{3}{2}\\0=0\end{matrix}\right.\end{matrix}\right.\)
Vì \(0=0\) luôn đúng nên ta có \(x=\dfrac{3}{2}\)
j) \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|=5x\)
(đầu hàng)
a)x2 +x =0
x(x+1)=0
* x=0
* x+1=0
x=0-1
x=-1
Vậy x=0 hoặc x=-1
2.(x-1) +3(x-2) =x-4
2x-2+3x-6=x-4
x+3x-x=-4+2+6
3x=4
x=\(\frac{4}{3}\)
a,
x2 +x =0
\(\orbr{\begin{cases}x^2=0\\x=0\end{cases}}\)
Vậy x= 0
b,2.(x-1) +3(x-2) =x-4
2x - 2 + 3x - 6 = x - 4
2x + 3x - x = -4 +6 + 2
4x = 0
Vậy x = 0
Tìm x, biết:
a. 12 + x = 122
=> x=122-12
=> x= 110
b. | x | + 4 = 8
=> /x/=8-4
=> /x/=4
=> x={4;-4}
c. 3x - 5 . 32 = 32 . 4
=> 3x=32.(4+5)
=> 3x=32.32
=> 3x=34
=> x=4
a)
\(12+x=122\)
=> \(x=110\)
Vậy \(x=110\)
b)
\(\left|x\right|+4=8\)
\(\Rightarrow\left|x\right|=4\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=4\\x=-4\end{array}\right.\)
Vậy x=4 ; x = - 4
c)
\(3^x-5.3^2=4.3^2\)
\(\Rightarrow3^x=3^2\left(4+5\right)\)
\(\Rightarrow3^x=3^4\)
=> x=4
Vậy x=4
Bạn ơi, đăng qua môn toán nhé ở đây là môn tiếng anh, lần sau chú ý hơn nha