Tìm x εIN biết
a) 390 - (x-8) = 168:13
b) (x-140) : 7 = 27 - 24
c) x- 6 :2 - ( 48 - 24 ) :2 :6 - 3 = 0
d) x+5.2-(32+16.3:6-15)=0

b) \(\left(x-140\right):7=27-24\)

\(\left(x-140\right):7=3\)

\(x-140=21\)

\(x=161\)

      vay   \(x=161\)

c) \(x-6:2-\left(48-24\right):2:6-3=0\)

\(x-3-24:2:6-3=0\)

\(x-3-2-3=0\)

\(x-8=0\)

\(x=8\)

vay \(x=8\)

d) \(x+5.2-\left(32+16.3:6-15\right)=0\)

\(x+10-\left(32+8-15\right)=0\)

\(x+10-25=0\)

\(x-15=0\)

\(x=15\)

vay \(x=15\)

a) \(390-\left(x-8\right)=168:13\)

\(390-x+8=\frac{168}{13}\)

\(x+8=390-\frac{168}{13}\)

\(x+8=\frac{5070}{13}-\frac{168}{13}\)

\(x+8=\frac{4902}{13}\)

\(x=\frac{4902}{13}-8\)

\(x=\frac{4798}{13}\)

vay \(x=\frac{4798}{13}\)

Bài 1:

\(A=7+7^3+7^5+...+7^{1999}\)

\(\Rightarrow A=\left(7+7^3\right)+\left(7^5+7^7\right)+...+\left(7^{1997}+7^{1999}\right)\)

\(\Rightarrow A=\left(7+343\right)+7^4\left(7+7^3\right)+...+7^{1996}\left(7+7^3\right)\)

\(\Rightarrow A=350+7^4.350+...+7^{1996}.350\)

\(\Rightarrow A=\left(1+7^4+...+7^{1996}\right).350⋮35\)

\(\Rightarrow A⋮35\left(đpcm\right)\)

b2:

a) \(S=1+3+3^2+...+3^{49}\)

\(\Rightarrow S=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{48}+3^{49}\right)\)

\(\Rightarrow S=\left(1+3\right)+3^2\left(1+3\right)+...+3^{48}\left(1+3\right)\)

\(\Rightarrow S=4+3^2.4+...+3^{48}.4\)

\(\Rightarrow S=\left(1+3^2+...+3^{48}\right).4⋮4\)

\(\Rightarrow S⋮4\left(đpcm\right)\)

c) \(S=1+3+3^2+...+3^{49}\)

\(\Rightarrow3S=3+3^2+3^3+...+3^{50}\)

\(\Rightarrow3S-S=\left(3+3^2+3^3+...+3^{50}\right)-\left(1+3+3^2+...+3^{49}\right)\)

\(\Rightarrow2S=3^{50}-1\)

\(\Rightarrow S=\frac{3^{50}-1}{2}\left(đpcm\right)\)

Giúp mình câu b bài 2 luôn được không?

Câu3: Ký hiệu [a,b] và (a,b) là gì ? Bạn.

Câu 1:

\(B=\frac{1}{199}+1+\frac{2}{198}+1+\frac{3}{197}+1+...+\frac{198}{2}+1+\frac{199}{1}+1-199\)

\(=\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}+1=\frac{200}{200}+\frac{200}{199}+\frac{200}{198}+...+\frac{200}{2}\)

\(=200\cdot\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)=200\cdot A\)

Vậy, \(\frac{A}{B}=\frac{1}{200}\).

1. Tính nhanh:

a) -37 + 54 + (-70) + (-163) + 246

= (-70) + {[(-163) + (-37)] + (246 + 54)}

= (-70) + [(-200) + 300]

= (-70) + 100

= 30

b) 24 - (-136) - (-70) + 15 - (-115)

= 24 + 136 + 70 + 15 + 115

= [70 + (15 + 115)] + (24 + 136)

= 70 + 130 + 160

= 200 + 160

= 360

2. Tính giá trị của các biểu thức sau bằng cách hợp lí nhất:

a) 136 . (- 47) + 36 . 47

= -136 . 47 + 36 . 47

= 47(-136 + 36)

= 47 . (-100)

= -4700

b) (- 48) . 72 + 36 . (- 304 )

= (- 48) . 72 + 72 . (-152)

= 72(-48 - 152)

= 72 . (-200)

= -14400

14. Tính tổng các số nguyên x biết:

a) - 2017 x 2018

x ∈ {-2017; -2016; ....; 2017; 2018}

Tổng các số nguyên x là :

(-2017) + (-2016) + .... + 2017 + 2018

= 2018 + [(-2017) + 2017] + [(-2016) + 2016] + ....

= 2018 + 0 + 0 + ....

= 2018

b) a + 3 x a + 2018 (a N)

x ∈ {a + 3; a + 4; ...; a + 2018}

Tổng các số nguyên x là :

a + 3 + a + 4 + .... + a + 2018

(2018 - 3) : 1 + 1 = 2016

= 2016a + (3 + 4 + .... + 2018)

(2018 + 3) . 2016 : 2 = 2037168

= 2016a + 2037168

3. Tìm x Z biết:

a) (x + 1) + (x + 3) + (x + 5) + …+ (x + 99) = 0

= (x + x + ... + x) + (1 + 3 + 5 + ... + 99) = 0

(99 - 1) : 2 + 1 = 50

(99 + 1) . 50 : 2 = 2500

x . 50 + 2500 = 0

x . 50 = -2500

x = -50

b) (x – 3) + (x - 2) + (x – 1 ) + … + 10 + 11 = 11

(x – 3) + (x - 2) + (x – 1 ) + … + 10 = 0

[(x – 3) + (x - 2) + (x – 1 )] + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) = 0

[(x – 3) + (x - 2) + (x – 1 )] + 55 = 0

(x – 3) + (x - 2) + (x – 1 ) = -55

x + x + x + (-3 - 2 - 1) = -55

3x + (-6) = -55

3x = -49

x = -49/3

c) x + (x + 1) + (x + 2) + ..... + 2018 + 2019 = 2019

x + (x + 1) + (x + 2) + ..... + 2018 = 0

(2018 - x) : 1 + 1 = 2019 - x

(2018 + x) : 2

⇒ (2019 - x) . [(2018 + x) : 2] = 0

✽ 2019 - 2019 = 0

⇒ x = 2019 (loại vì x = 2019 thì số số hạng sẽ là 0)

✽ 2018 + (-2018) = 0

⇒ x = -2018 (nhận)

x = -2018

Ukm, ko có j coi như mk học tập thêm

bài 2) a) \(2\left(x+1\right)=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\) vậy \(x=-1\)

b) \(x\left(x-2\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\) vậy \(x=0;x=2\)

c) \(\left(x-1\right)\left(x+7\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\) vậy \(x=1;x=-7\)

d) \(\left(x+2\right)\left(x^2-9\right)=0\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\x^2-9=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x^2=9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\end{matrix}\right.\) vậy \(x=-2;x=3;x=-3\)

e) \(x^2\left(x-5\right)+2\left(x-5\right)=0\Leftrightarrow\left(x^2+2\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+2=0\\x-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\in\varnothing\\x=5\end{matrix}\right.\) vậy \(x=5\)

bài 1) \(A=48+\left(-48-174\right)+\left|-74\right|=48-48-174+74=-100\)

\(B=\left(-123\right)+77+\left(-257\right)-23-43=-123+77-257-23-43=-369\)

\(C=\left(-57\right)+\left(-159\right)+47+169=-57-159+47+169=0\)

quá hợp lí

a) \(\left(x-1\right):3=2^3\) \(\Leftrightarrow\) \(\left(x-1\right):3=8\) \(x+1=24\) \(\Leftrightarrow\) \(x=23\) vậy \(x=23\)

b) \(12-2\left(x+5\right)=-10\) \(\Leftrightarrow\) \(12-2x-10=-10\)

\(\Leftrightarrow\) \(-2x=-12\) \(\Leftrightarrow\) \(x=6\) vậy \(x=6\)

c) \(x-12\left(x+5\right)=-10\) \(\Leftrightarrow\) \(x-12x-60=-10\)

\(\Leftrightarrow\) \(-11x=50\) \(\Leftrightarrow\) \(x=\dfrac{50}{-11}\) vậy \(x=\dfrac{50}{-11}\)

e) \(13-x:2=10\Leftrightarrow-x:2=-3\Leftrightarrow x=\dfrac{3}{2}\)

f) \(\left|12-x\right|-7=5\)

th1 : \(x\le12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(12-x-7=5\) \(\Leftrightarrow\) \(-x=0\Leftrightarrow x=0\)

th2 : \(x>12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(x-12-7=5\) \(\Leftrightarrow\) \(x=24\) vậy \(x=0;x=24\)

i) \(x^2-7=2\Leftrightarrow x^2=9\Leftrightarrow x=3\) vậy \(x=3\)

k) \(x^3-4=-12\) \(\Leftrightarrow\) \(x^3=-8\) \(\Leftrightarrow x=-2\) vậy \(x=-2\)

a)\(\left(x-1\right):3=2^3\Rightarrow x-1=2^3.3=24\Rightarrow x=25\)

b)\(12-2\left(x+5\right)=-10\Leftrightarrow12-2x-10=-10\Rightarrow2-2x=-10\Rightarrow2x=12\Rightarrow x=6\)c)\(x-12\left(x+5\right)=-10\Rightarrow x-12x-60=-10\Rightarrow-11x-60=-10\Rightarrow-11x=-70\Rightarrow x=\dfrac{70}{-11}\)d)\(6-\left|x\right|=5\Rightarrow\left|x\right|=1\Rightarrow x=\left\{\pm1\right\}\)

Làm nốt nha

\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)

=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)

=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)

Vậy \(x\in\left\{\frac{9}{20}\right\}\)

\(b,x+\frac{1}{4}=\frac{4}{3}\)

=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)

Vậy \(x\in\left\{\frac{13}{12}\right\}\)

\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)

=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)

=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)

Vậy \(x\in\left\{\frac{25}{42}\right\}\)

\(d,\left|x+5\right|-6=9\)

=> \(\left|x+5\right|=9+6=15\)

=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)

Vậy \(x\in\left\{10;-20\right\}\)

\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)

=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)

\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{6}\)

=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)

\(g,x^2=16\)

=> \(\left|x\right|=\sqrt{16}=4\)

=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

vậy \(x\in\left\{4;-4\right\}\)

\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)

=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)

=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)

Vậy \(x\in\left\{\frac{5}{6}\right\}\)

\(i,3^3.x=3^6\)

\(x=3^6:3^3=3^3=27\)

Vậy \(x\in\left\{27\right\}\)

\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)

=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)

Vậy \(x\in\left\{\frac{5}{27}\right\}\)

\(k,1\frac{2}{3}:x=6:0,3\)

=> \(\frac{5}{3}:x=20\)

=> \(x=\frac{5}{3}:20=\frac{1}{12}\)

Vậy \(x\in\left\{\frac{1}{12}\right\}\)