\(\left(4x+12\right)\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}4x+12=0\\x+5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}4x=-12\\x=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{-3;-5\right\}\)

(4x+12)(x+5)=0

=>4(x+3)(x+5)=0

=>(x+3)(x+5)=0

=>\(\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}+\dfrac{1}{3^{100}}\\ \Rightarrow3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}+\dfrac{1}{3^{99}}\\ 3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow2B=1-\dfrac{1}{3^{100}}< 1\\ \Rightarrow B< \dfrac{1}{2}< 1\left(DPCM\right)\)

Ta có:

\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\\ \Rightarrow3B=1+\dfrac{1}{3}+...+\dfrac{1}{3^{99}}\\ \Rightarrow3B-B=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\right)\\ \Rightarrow2B=1-\dfrac{1}{3^{100}}\\ \Rightarrow B=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)

Vì \(1-\dfrac{1}{3^{100}}< 1\) nên:

\(\dfrac{1-\dfrac{1}{3^{100}}}{2}< \dfrac{1}{2}< 1\) hay \(B< 1\)

Vậy...

5: a+b=7

=>a=7-b

b+c=9

=>c=9-b

c+a=8

=>7-b+9-b=8

=>16-2b=8

=>2b=16-8=8

=>b=4

=>a=7-4=3;c=9-4=5

1: a+b=17

a+b+c=20

=>c=20-17=3

a+c=15

=>a=15-c=15-3=1

b=17-a=17-1=16

2: \(\left\{{}\begin{matrix}a+b=5\\b+c=9\\a+c=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\c=9-b\\5-b+9-b=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}14-2b=6\\a=5-b\\c=9-b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2b=8\\a=5-b\\c=9-b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=4\\a=5-4=1\\c=9-4=5\end{matrix}\right.\)

3: \(c=\dfrac{abc}{ab}=\dfrac{288}{24}=12\)

bc=96

=>b=96/12=8

\(a=\dfrac{24}{b}=\dfrac{24}{8}=3\)

4: \(\left\{{}\begin{matrix}ab=36\\bc=45\\ca=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{36}{b}\\c=\dfrac{45}{b}\\\dfrac{36}{b}\cdot\dfrac{45}{b}=20\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}b^2=36\cdot45:20=81\\a=\dfrac{36}{b}\\c=\dfrac{45}{b}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b\in\left\{9;-9\right\}\\a=\dfrac{36}{b}\\c=\dfrac{45}{b}\end{matrix}\right.\)

TH1: b=9

\(a=\dfrac{36}{9}=4;c=\dfrac{45}{b}=\dfrac{45}{9}=5\)

TH2: b=-9

=>\(a=\dfrac{36}{-9}=-4;c=\dfrac{45}{-9}=-5\)

`#3107.101107`

`x + 3 - 2x + 5 + 3x - 6 - 4x - 2 + 5x`

`= (x - 2x + 3x - 4x + 5x) + (3 + 5 - 6 - 2)`

`= 3x`

\(x+3-2x+5+3x-6-4x-2+5x\\ =\left(x-2x+3x-4x+5x\right)+\left(3+5-6-2\right)\\ =3x\)

Gọi số bị chia là x

Số chia là 97-x

Thương là 4; số dư là 7 nên ta có:

x=4(97-x)+7

=>x=388-4x+7

=>5x=395

=>x=395:5=79(nhận)

vậy: Số bị chia là 79

Số chia là 97-79=18

\(N=\dfrac{1}{2}+\dfrac{5}{6}+\dfrac{11}{12}+...+\dfrac{89}{90}\)

\(=1-\dfrac{1}{2}+1-\dfrac{1}{6}+...+1-\dfrac{1}{90}\)

\(=\left(1+1+1+...+1\right)-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\)

\(=9-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=9-\left(1-\dfrac{1}{10}\right)=9-1+\dfrac{1}{10}=8+\dfrac{1}{10}=\dfrac{81}{10}\)

Gọi số chia là x

Số bị chia là x+13748

Thương là 3 và dư là 2180 nên ta có:

x+13748=3x+2180

=>3x-x=13748-2180

=>2x=11568

=>x=5784

Vậy: Số chia là 5784

Số bị chia là 5784+13748=19532

\(3^5\cdot x=\left(-3\right)^7\)

=>\(x\cdot3^5=-3^7\)

=>\(x=-\dfrac{3^7}{3^5}=-3^2=-9\)

d: \(\left(315-x\right)+264=327\)

=>315+264-x=327

=>579-x=327

=>x=579-327=252

h: 12x-59=25

=>\(12x=59+25=84\)

=>\(x=\dfrac{84}{12}=7\)

e: 735-(457+x)=124

=>x+457=735-124=611

=>x=611-457=154

k: 36-x:2=16

=>x:2=36-16=20

=>\(x=20\cdot2=40\)

l: 60-3(x-2)=51

=>3(x-2)=60-51=9

=>x-2=3

=>x=3+2=5

m: \(\left(3x-15\right)\cdot7=42\)

=>\(21\left(x-5\right)=42\)

=>x-5=2

=>x=2+5=7

n: \(558-\left(15:x+29\right)\cdot17=14\)

=>\(\left(15:x+29\right)\cdot17=558-14=544\)

=>15:x+29=544:17=32

=>15:x=32-29=3

=>x=15:3=5

p: 30-3x=5*54

=>30-3x=270

=>3x=30-270=-240

=>\(x=-\dfrac{240}{3}=-80\)

q: \(3\left(x-2\right)+2\left(x+5\right)=29\)

=>3x-6+2x+10=29

=>5x+4=29

=>5x=25

=>x=5

v: x+(x+1)+(x+2)+...+(x+30)=1240

=>31x+(1+2+...+30)=1240

=>\(31x+30\cdot\dfrac{31}{2}=1240\)

=>x+15=40

=>x=40-15=25

t: (27-3x)(x-5)=0

=>3(9-x)(x-5)=0

=>(9-x)(x-5)=0

=>(x-9)(x-5)=0

=>\(\left[{}\begin{matrix}x-9=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=5\end{matrix}\right.\)

\(a.187+x=320\\ =>x=320-187=133\\ b.765-x=158\\=> x=765-158=607\\ c.451+\left(x-218\right)=876\\ =>x-218=876-451=425\\ =>x=425+218=643\\ d.\left(315-x\right)+264=327\\ =>315-x=327-264=63\\ =>x=315-63=252\\ e.735-\left(457+x\right)=124\\=>457+x=735-124=611\\ =>x=611-457=154\)

Số chữ số dùng cho trang có 1 chữ số là:

\(\left(9-1+1\right)\cdot1=9\left(chữsố\right)\)

Số chữ số dùng cho trang có 2 chữ số là:

\(\left(99-10+1\right)\cdot2=180\)(chữ số)

Số chữ số còn lại là 282-9-180=93(chữ số)

Số trang sách có 3 chữ số là 93:3=31(trang)

Số trang sách là 99+31=130(trang)