làm phần 2 thôi nhá :v

1+2+4=7=(4.2)-1

1+2+4+8=15=(8.2)-1

......

1+2+..+8192=(8192.2)-1=16383

\(18+\dfrac{1}{11}\times\left(x-18\right)=36+\dfrac{1}{11}\times\left[\dfrac{10}{11}\times\left(x-18\right)-36\right]\)

\(\Leftrightarrow\dfrac{198}{11}+\dfrac{1}{11}\times\left(x-18\right)=36+\dfrac{1}{11}\times\left[\dfrac{10}{11}\times\left(x-18\right)-\dfrac{396}{11}\right]\)

\(\Leftrightarrow\dfrac{198+x-18}{11}=36+\dfrac{1}{11}\times\dfrac{10x-180-396}{11}\)

\(\Leftrightarrow\dfrac{180+x}{11}=36+\dfrac{10x-576}{121}\)

\(\Leftrightarrow\dfrac{1980+11x}{121}=\dfrac{4356}{121}+\dfrac{10x-576}{121}\)

\(\Leftrightarrow1980+11x=4356+10x-576\)

\(\Leftrightarrow11x-10x=4356-1980-576\)

\(\Leftrightarrow x=1800\)

`(5/6 -x+7/12) : ( 11/24 -1/8)=11/36`

`=>(5/6 -x+7/12) : (11/24 - 3/24)=11/36`

`=>(5/6 -x+7/12) : 8=11/36`

`=>5/6 -x+7/12=11/36 xx 8`

`=>5/6 -x+7/12=22/9`

`=> x+7/12=5/6-22/9`

`=> x+7/12=-29/18`

`=>x=-29/18 -7/12`

`=>x=-79/36`

79/36 nhé

Bài 1:

a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)

b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)

c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)

d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)

hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)

e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)

hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)

dad

a) 26 x 11 = 26 x (10 + 1)

= 26 x 10 + 26 x 1 = 260 + 26 = 286

35 x 101 = 35 x (100 + 1)

= 35 x 100 + 35 x 1 = 3500 + 35 = 3535

b) 213 x 11 = 213 x (10 +1)

= 213 x 100 + 213 x 1 = 2130 + 213 = 2343

123 x 101 = 123 x (100 + 1)

= 123 x 100 + 123 x 1

= 12300 + 123 = 12423

Nói thêm: Muốn nhân một số với 11 (hoặc 101) ta nhân số đó với 10 (hoặc 100) rồi cộng thêm chính số đó.

1: ĐKXĐ: x>=8/3

\(\sqrt{3x-8}-\sqrt{x+1}=\frac{2x-11}{5}\)

=>\(\sqrt{3x-8}-1+2-\sqrt{x+1}=\frac{2x-11}{5}+1\)

=>\(\frac{3x-8-1}{\sqrt{3x-8}+1}+\frac{4-x-1}{2+\sqrt{x+1}}=\frac{2x-11+5}{5}\)

=>\(\left(x-3\right)\left(\frac{3}{\sqrt{3x-8}+1}-\frac{1}{2+\sqrt{x+1}}-\frac25\right)=0\)

=>x-3=0

=>x=3(nhận)

3: ĐKXĐ: -5/2<=x<=5/2

Đặt \(a=\sqrt{5+2x};b=\sqrt{5-2x}\)

=>\(ab=\sqrt{\left(5+2x\right)\left(5-2x\right)}=\sqrt{25-4x^2}\)

Theo đề, ta có: a+b+5=3ab

=>3ab-a-b-5=0

=>a(3b-1)-b+1/3-16/3=0

=>\(3a\left(b-\frac13\right)-\left(b-\frac13\right)=\frac{16}{3}\)

=>\(\left(b-\frac13\right)\left(3a-1\right)=\frac{16}{3}\)

=>(3a-1)(3b-1)=16

=>(3a-1;3b-1)∈{(1;16);(16;1);(2;8);(8;2);(4;4)}

=>(3a;3b)∈{(2;17);(17;2);(3;9);(9;3);(5;5)}

=>(a;b)∈{(2/3;17/3);(17/3;2/3);(1;3);(3;1);(5/3;5/3)}

mà a<>b

nên (a;b)∈{(2/3;17/3);(17/3;2/3);(1;3);(3;1)}

TH1: a=2/3 và b=17/3

=>\(\begin{cases}5+2x=\frac49\\ 5-2x=\frac{289}{9}\end{cases}\Rightarrow\begin{cases}2x=\frac49-5=\frac49-\frac{45}{9}=-\frac{41}{9}\\ 2x=5-\frac{289}{9}=-\frac{244}{9}\end{cases}\)

=>x∈∅

TH2: a=17/3 và b=2/3

=>\(\begin{cases}5+2x=\frac{289}{9}\\ 5-2x=\frac49\end{cases}\Rightarrow\begin{cases}2x=\frac{289}{9}-5=\frac{244}{9}\\ 2x=5-\frac49=\frac{41}{9}\end{cases}\)

=>x∈∅

TH3: a=1 và b=3

=>5+2x=1 và 5-2x=9

=>2x=-4 và 2x=5-9=-4

=>x=-2(nhận)

TH4: a=3 và b=1

=>5+2x=9 và 5-2x=1

=>2x=4 và 2x=4

=>x=2(nhận)

\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+.....+\frac{3}{x\left(x+3\right)}\right)=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+....+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{1}{18}\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{1}{18}:\frac{1}{3}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{1}{6}=\frac{1}{30}\)

<=>x+3=30

<=>x=27

Vậy x=27