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a) |2x + 1| - 19 = -7
=> \(\left|2x+1\right|=-7+19=12\)
=> \(\left[{}\begin{matrix}2x+1=12\\2x+1=-12\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}2x=12-1=11\\2x=-12-1=-13\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{11}{2}\\x=-\frac{13}{2}\end{matrix}\right.\)
Vậy:............
b) -28 – 7. |- 3x + 15| = -70
=> \(\text{7. |- 3x + 15| = -28 - (-70) = -28 + 70 = 42}\)
=> \(\left|-3x+15\right|=42:7=6\)
=> \(\left[{}\begin{matrix}-3x+15=6\\-3x+15=-6\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}-3x=6-15=-9\\-3x=-6-15=-6+\left(-15\right)=-21\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-9:\left(-3\right)=3\\x=x=-21:\left(-3\right)=7\end{matrix}\right.\)
Vậy:.....................
c) |18 – 2. |-x + 5|| = 12
=> \(\left[{}\begin{matrix}18-2.\left|-x+5\right|=12\\18-2.\left|-x+5\right|=-12\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}2.\left|-x+5\right|=18-12=6\\2.\left|-x+5\right|=18-\left(-12\right)=18+12=30\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left|-x+5\right|=6:2=3\\\left|-x+5\right|=30:2=15\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}-x+5=3\\-x+5=-3\end{matrix}\right.\\\left[{}\begin{matrix}-x+5=15\\-x+5=-15\end{matrix}\right.\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}-x=3-5=-2\\-x=-3-5=-8\end{matrix}\right.\\\left[{}\begin{matrix}-x=15-5=10\\-x=-15-5=-20\end{matrix}\right.\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=8\end{matrix}\right.\\\left[{}\begin{matrix}x=-10\\x=20\end{matrix}\right.\end{matrix}\right.\)
bài 1:a,
\(3^9.3:3^{10}+\left|2010^0\right|\)
=> \(3^9.3:3^{10}+\left|1\right|\)
=> \(3^9.3:3^{10}+1\)
=> \(3^{10}:3^{10}+1\)
=> 1+1
=> 2
b, \([\left(4^9:4^7\right):8-735^0]^{2011}\)
=> \([4^2:8-735^0]^{2011}\)
=> \([2^4:2^3-735^0]^{2011}\)
=> \([2-1]^{2011}\)
=> 1
c, \(8^{2x}:8=512\)
=> \(8^{2x}:8=8^3\)
=> \(8^{2x}=8^4\)
=> 2x=4
=> x=2
bài 2:
Theo đề ta có:
\(\left(7^0+7^1+7^2+7^3+......+7^{2010}+7^{2011}\right)\)
=> \((7^0+7^1)+(7^2+7^3)+......+(7^{2010}+7^{2011})\)
=> \(7^0.\left(1+7\right)+7^2\left(1+7\right)+..+7^{2010}\left(1+7\right)\)
=> \(7^0.8+7^2.8+..+7^{2010}.8\)
Mà \(7^0.8+7^2.8+..+7^{2010}.8\) \(⋮\) 8 ( vì có thừa số 8 nên chia hết cho 8)
nên \(\left(7^0+7^1+7^2+7^3+......+7^{2010}+7^{2011}\right)\)\(⋮\) 8
\(a)\dfrac{3}{4}-\dfrac{-5}{2}-\dfrac{7}{-24}\)
\(=\dfrac{13}{4}-\dfrac{7}{-24}\)
\(=\dfrac{85}{24}\)
\(b)\dfrac{4}{7}+\dfrac{-5}{8}-\dfrac{3}{28}\)
\(=\dfrac{-3}{56}-\dfrac{3}{28}\)
\(=\dfrac{-9}{56}\)
\(c)\dfrac{7}{36}-\dfrac{8}{-9}+\dfrac{-2}{3}\)
\(=\dfrac{13}{12}\)\(+\dfrac{-2}{3}\)
\(=\dfrac{5}{12}\)
\(d)\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)
\(=\dfrac{-1}{14}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)
\(=\dfrac{-23}{126}+\dfrac{-7}{18}+\dfrac{4}{7}\)
\(=\dfrac{-4}{7}+\dfrac{4}{7}\)
\(=0\)
\(e)\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)
\(=\dfrac{-5}{56}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)
\(=\dfrac{83}{56}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)
\(=\dfrac{305}{168}+\dfrac{1}{7}+\dfrac{5}{-8}\)
\(=\dfrac{47}{24}+\dfrac{5}{-8}\)
\(=\dfrac{4}{3}\)
Bài 2 : Tính
a) \(\dfrac{3}{4}-\dfrac{-5}{2}-\dfrac{7}{-24}\)
\(=\dfrac{18}{24}-\dfrac{-60}{24}-\dfrac{-4}{24}\)
\(=\dfrac{18-\left(-60\right)-\left(-7\right)}{24}\)
\(=\dfrac{85}{24}\)
b) \(\dfrac{4}{7}+\dfrac{-5}{8}-\dfrac{3}{28}\)
\(=\dfrac{32}{56}+\dfrac{-35}{56}-\dfrac{6}{56}\)
\(=\dfrac{32+\left(-35\right)-6}{56}\)
\(=\dfrac{-9}{56}\)
c) \(\dfrac{7}{36}-\dfrac{8}{9}+\dfrac{-2}{3}\)
\(=\dfrac{7}{36}-\dfrac{32}{36}+\dfrac{-24}{36}\)
\(=\dfrac{7-32+\left(-24\right)}{36}\)
\(=\dfrac{-49}{36}\)
d) \(\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)
\(=\dfrac{-9}{18}+\dfrac{3}{7}-\dfrac{2}{18}+\dfrac{-7}{18}+\dfrac{4}{7}\)
\(=\left(\dfrac{-9}{18}+\dfrac{-7}{18}-\dfrac{2}{18}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\)
\(=\left(-1\right)+1\)
\(=0\)
e) \(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)
\(=\left(\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{11}{7}\right)+\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\dfrac{1}{3}\)
\(=2+\left(-1\right)+\dfrac{1}{3}\)
\(=1+\dfrac{1}{3}\)
\(=\dfrac{4}{3}\)
1/ (a – b + c) – (a + c) = -b
a-b+c-a-c=-b
-b=-b
2/ (a + b) – (b – a) + c = 2a + c
a+b-b+a+c=2a+c
2a+c=2a+c
3/ - (a + b – c) + (a – b – c) = -2b
-a-b+c+a-b-c=-2b
-(b.2)=-2b
-2b=-2b
4/ a(b + c) – a(b + d) = a(c – d)
ab+ac-ab+ad=a(c-d)
ac-ad=a(c-d)
a(c-d)=a(c-d)
5/ a(b – c) + a(d + c) = a(b + d)
ab-ac+ad+ac=a(b+d)
ab+ad=a(b+d)
a(b+d)=a(b+d)
6/ a.(b – c) – a.(b + d) = -a.( c + d)
ab-ac-ab=ad=-a(c+d)
-ac+ad=-a(c+d)
-a(c+d)=-a(c+d)
Bài 1:
a)
\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)
b)
\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)
c)
\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)
d)
\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)
e)
\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)
f)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)
g)
\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)
h)
\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)
i)
\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)
D=(7*1+7*7)+(73*1+7*7)+...+(72009*1+72009*7)
D=7*(1+7)+73*(1+7)+...+72009*(1+7)
D=7*8+73*8+...+72009*8
D=(7+73+...+72009)*8 chia hết cho 8(vì 8chia hết cho 8)
vậy D chia hết cho 8
bạn hãy làm thử chia hết cho 57 đi
bằng cách gộp 3 số hạng đó mà.
Bài 2
a, \(\dfrac{4}{5}+\dfrac{2}{7}-\dfrac{7}{10}=\) \(\dfrac{27}{70}\)
câu a
\(A=\frac{33.10^3}{2^3.5.10^3+7000}=\frac{33.10^3}{2^3.5.10^3+7.10^3}=\frac{33.10^3}{10^3\left(2^3.5+7\right)}=\frac{33.10^3}{10^3.47}=\frac{33}{47}\)
\(B=\frac{3774}{5217}=\frac{34.111}{47.111}=\frac{34}{47}\)
\(\Rightarrow\frac{33}{47}< \frac{34}{47}\)
=> A<B
Bài 1:
a) Cho A = 1+14+...+142014
=> 14A = 14 + 142 +...+142015
=> 14A - A = 142015 - 1
13A = 142015 - 1
mà 13 A chia hết cho 13
=> đpcm
b) làm tương tự
c) 1+3+32 +...+32015 ( có 2016 số hạng)
= (1+3+32 +33) + ...+ (32012 + 32013 +32014 +32015)
= 40 + ...+ 32012.(1+3+32+33)
...
Bài 2:
N = 7+72 + 73 +...+ 7n
=> 7N = 72 + 73 +74 +...+ 7n+1
=> \(6N=7^{n+1}-7\)
Thay vào biểu thức
=> 7n+1 -7 + 7 = 22016
7n+1 = 22016
...
Để chứng minh S chia hết cho 2 và S chia hết cho 57, ta sẽ xem xét từng thành phần trong công thức của S.
Đầu tiên, ta xét dãy từ 71 đến 72025. Trong dãy này, có 72025 - 71 + 1 = 71955 số.
Ta biết rằng nếu một số chia hết cho 2, thì số đó là số chẵn. Trong dãy từ 71 đến 72025, ta có 2 số lẻ liên tiếp (71 và 72), sau đó là 2 số chẵn liên tiếp (73 và 74), và tiếp tục lặp lại quy luật này. Vì vậy, trong 71955 số này, ta có 71955/2 = 35977.5 cặp số chẵn và lẻ.
Do đó, tổng của các số chẵn trong dãy này là 35977.5 * 2 = 71955.
Tiếp theo, ta xét số 72024. Ta biết rằng 72024 chia hết cho 2.
Cuối cùng, ta xét số 72025. Ta biết rằng 72025 chia hết cho 57, vì 72025 = 57 * 1265.
Vậy tổng S chia hết cho 2 và chia hết cho 57.