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17 tháng 12 2023

m: \(\left(-7x+7\right)\left(2x+100\right)=0\)

=>\(\left[{}\begin{matrix}-7x+7=0\\2x+100=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}-7x=-7\\2x=-100\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-50\end{matrix}\right.\)

n: \(2x\left(x+2023\right)\left(-4x+8\right)=0\)

=>\(2\cdot x\left(x+2023\right)\cdot\left(-4\right)\left(x-2\right)=0\)
=>x(x-2)(x+2023)=0

=>\(\left[{}\begin{matrix}x=0\\x-2=0\\x+2023=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2023\end{matrix}\right.\)

o: \(-2024x\left(4x-4\right)\left(-2x+6\right)=0\)

=>\(x\left(4x-4\right)\left(-2x+6\right)=0\)

=>\(x\cdot4\left(x-1\right)\cdot\left(-2\right)\left(x-3\right)=0\)

=>x(x-1)(x-3)=0

=>\(\left[{}\begin{matrix}x=0\\x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=3\end{matrix}\right.\)

p: -17-2x=199

=>2x=-17-199=-216

=>x=-216/2=-108

q: -24+2x=100

=>2x=100+24=124

=>\(x=\dfrac{124}{2}=62\)

r: \(119-\left(x+5\right)=-21\)

=>\(x+5=119-\left(-21\right)=119+21=140\)

=>x=140-5=135

s: \(-24+\left(-7+x\right)=45\)

=>\(\left(x-7\right)-24=45\)

=>x-31=45

=>x=45+31=76

t: \(-146-\left(-5+x\right)=-6\)

=>\(x-5=-146-\left(-6\right)=-140\)

=>x=-140+5=-135

u: \(-29+4\left(1-5x\right)=-45\)

=>4(1-5x)=-45+29=-16

=>1-5x=-4

=>5x=1+4=5

=>\(x=\dfrac{5}{5}=1\)

v: \(24-5\left(-3+2x\right)=59\)

=>\(24-5\left(2x-3\right)=59\)

=>5(2x-3)=24-59=-35

=>2x-3=-7

=>2x=-4

=>x=-4/2=-2

17 tháng 12 2023

-17-2x=199

<=>-2x=216

<=>x=-108

 

-24+2x=100

<=>2x=124

<=>x=62

 

-24+(-7+x)=45

<=>-7+x=69

<=>x=76

 

-146-(-5+x)=-6

<=>-146+5-x=-6

<=>x=-135

 

-29+4(1-5x)=-45

<=>-29+4+20x=-45

<=>20x=-20

<=>x=-1

 

24-5(-3+2x)=59

<=>24+15-10x=59

<=>-10x=20

<=>x=-2

25 tháng 12 2023

a: \(\left(-256\right)\cdot45-256\cdot56+256\)

\(=256\left(-45-56+1\right)\)

\(=256\left(-100\right)=-25600\)

b: \(\left(-2\right)^3\cdot1975\cdot\left(-4\right)\cdot\left(-5\right)^3\cdot25\)

\(=\left(-8\right)\cdot\left(-125\right)\cdot\left(-4\right)\cdot25\cdot1975\)

\(=1000\cdot\left(-100\right)\cdot1975=-197500000\)

c: \(2076-1976\cdot65-1976\cdot35\)

\(=2076-1976\left(65+35\right)\)

\(=2076-1976\cdot100=2076-197600=-195524\)

d: \(-437-25\cdot78+25\cdot178\)

\(=-437+25\left(178-78\right)\)

\(=-437+2500=2063\)

AH
Akai Haruma
Giáo viên
3 tháng 2

c/

$C=\frac{11}{2}(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{91.93})$

$=\frac{11}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{93-91}{91.93}\right)$

$=\frac{11}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{91}-\frac{1}{93}\right)$

$=\frac{11}{2}(1-\frac{1}{93})$

$=\frac{11}{2}.\frac{92}{93}=\frac{506}{93}$

AH
Akai Haruma
Giáo viên
3 tháng 2

d/

$D=5\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{675}\right)$

$=\frac{5}{2}\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{675}\right)$

$=\frac{5}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{27-25}{25.27}\right)$

$=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{25}-\frac{1}{27}\right)$
$=\frac{5}{2}\left(1-\frac{1}{27}\right)$
$=\frac{5}{2}.\frac{26}{27}=\frac{65}{27}$

23 tháng 2 2023

11 tháng 12 2021

d: \(\Leftrightarrow x+5=5\)

hay x=0

7 tháng 12 2023

Bài 3:

a) \(159-\left(25-x\right)=43\)

\(\Rightarrow25-x=159-43\)

\(\Rightarrow25-x=116\)

\(\Rightarrow x=25-116\)

\(\Rightarrow x=-91\)

b) \(\left(79-x\right)-43=-\left(17-52\right)\)

\(\Rightarrow\left(79-x\right)-43=-\left(-35\right)\)

\(\Rightarrow79-x=35+43\)

\(\Rightarrow79-x=78\)

\(\Rightarrow x=79-78\)

\(\Rightarrow x=1\)

c) \(-\left(-x+13-142\right)+18=55\)

\(\Rightarrow-\left(-x+13-142\right)=55-18\)

\(\Rightarrow x-13+142=37\)

\(\Rightarrow x+129=37\)

\(\Rightarrow x=37-129\)

\(\Rightarrow x=-92\)

o: \(\dfrac{-35}{120}=\dfrac{-35\cdot3}{120\cdot3}=\dfrac{-105}{360}\)

\(\dfrac{19}{-45}=\dfrac{-19}{45}=\dfrac{-19\cdot8}{45\cdot8}=\dfrac{-152}{360}\)

mà -105>-152

nên \(-\dfrac{35}{120}>-\dfrac{19}{45}\)

p: \(\dfrac{-12}{48}=\dfrac{-1}{4}=\dfrac{-13}{52}\)

\(\dfrac{-7}{26}=\dfrac{-7\cdot2}{26\cdot2}=\dfrac{-14}{52}\)

mà -13>-14

nên \(-\dfrac{12}{48}>-\dfrac{7}{26}\)

q: \(\dfrac{-8}{-90}=\dfrac{8}{90}\)

\(\dfrac{14}{42}=\dfrac{1}{3}=\dfrac{30}{90}\)

mà 8<30

nên \(\dfrac{-8}{-90}< \dfrac{14}{42}\)

r: \(\dfrac{-34}{68}=\dfrac{-1}{2}=\dfrac{-28}{56}\)

\(\dfrac{28}{-51}=\dfrac{-28}{51}\)

56>51

=>\(\dfrac{28}{56}< \dfrac{28}{51}\)

=>\(-\dfrac{28}{56}>-\dfrac{28}{51}\)

=>\(\dfrac{-34}{68}>\dfrac{28}{-51}\)

t: \(\dfrac{-19}{60}< 0\)

\(0< \dfrac{42}{45}=\dfrac{-42}{-45}\)

Do đó: \(\dfrac{-19}{60}< \dfrac{-42}{-45}\)

v: \(\dfrac{1}{-2024}< 0\)

\(0< \dfrac{-2023}{-45}=\dfrac{2023}{45}\)

Do đó: \(\dfrac{1}{-2024}< \dfrac{-2023}{-45}\)

x: Đặt \(A=\dfrac{2024^{2025}+1}{2024^{2026}+1};B=\dfrac{2024^{2000}+1}{2024^{2001}+1}\)

\(2024A=\dfrac{2024^{2026}+2024}{2024^{2026}+1}=1+\dfrac{2023}{2024^{2026}+1}\)

\(2024B=\dfrac{2024^{2001}+2024}{2024^{2001}+1}=1+\dfrac{2023}{2024^{2001}+1}\)

\(2024^{2026}+1>2024^{2001}+1\)

=>\(\dfrac{2023}{2024^{2026}+1}< \dfrac{2023}{2024^{2001}+1}\)

=>2024A<2024B

=>A<B

AH
Akai Haruma
Giáo viên
26 tháng 11 2023

1/

$C=5+(5^2+5^3)+(5^4+5^5)+.....+(5^{2022}+5^{2023})$

$=5+5^2(1+5)+5^4(1+5)+....+5^{2022}(1+5)$

$=5+(1+5)(5^2+5^4+....+5^{2022})$
$=5+6(5^2+5^4+....+5^{2022})$

$\Rightarrow C$ chia $6$ dư $5$

$\Rightarrow C\not\vdots 6$

AH
Akai Haruma
Giáo viên
26 tháng 11 2023

2/

$D=(1+2+2^2)+(2^3+2^4+2^5)+....+(2^{2019}+2^{2020}+2^{2021})$

$=(1+2+2^2)+2^3(1+2+2^2)+....+2^{2019}(1+2+2^2)$

$=(1+2+2^2)(1+2^3+...+2^{2019})$

$=7(1+2^3+...+2^{2019})\vdots 7$ 

Ta có đpcm.

24 tháng 10 2021

a: Ư(8)={1;2;4;8}

Ư(12)={1;2;3;4;6;12}

UC(8;12)={1;2;4}

b: B(16)={0;16;32;...}

B(24)={0;24;48;...}

BC(16,24)={0;48;96;...}

24 tháng 10 2021

chx lm hết ạ