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4 tháng 1 2018

\(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot...\cdot\dfrac{14}{30}.\dfrac{15}{32}=\dfrac{1}{2^x}\)

\(\Rightarrow\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot14\cdot15}{4\cdot6\cdot8\cdot10\cdot...\cdot30\cdot32}=\dfrac{1}{2^x}\)

\(\Rightarrow\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot14\cdot15}{2\cdot4\cdot6\cdot8\cdot10\cdot...\cdot30\cdot32}=\dfrac{1}{2^{x+1}}\)

\(\Rightarrow\dfrac{1}{2^{15}\cdot32}=\dfrac{1}{2^{x+1}}\)

\(\Rightarrow2^{15}.2^5=2^{x+1}\)

\(\Rightarrow2^{20}=2^{x+1}\)

\(\Rightarrow x+1=20\Rightarrow x=19\)

Vậy x = 19.

18 tháng 7 2018

\(a,x^2=16\)

\(x^2=4^2=\left(-4\right)^2\)

\(x=2\) hoặc \(x=-2\)

\(b,x^3=-8\)

\(x^3=\left(-2\right)^3\)

\(x=-2\)

\(c,\left(x+2\right)^2=4\)

\(\left(x+2\right)^2=2^2=\left(-2\right)^2\)

\(x+2=2\Rightarrow x=0\) hoặc \(x+2=-2\Rightarrow x=-4\)

\(d,\left(1-x\right)^3=1\)

\(1-x=1\)

\(x=0\)

e,phần này mk chưa nghĩ ra,sorry bn nha!

28 tháng 10 2017

\(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot...\cdot\dfrac{15}{32}=2^x\)

\(\Rightarrow\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot15}{4\cdot6\cdot8\cdot10\cdot...\cdot32}=2^x\Rightarrow\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot15}{\left(2\cdot2\right)\left(2\cdot3\right)\left(2\cdot4\right)\cdot...\cdot\left(2.16\right)}=2^x\)

\(\Rightarrow\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot15}{2^{15}\cdot\left(2\cdot3\cdot4\cdot...\cdot16\right)}=2^x\Rightarrow\dfrac{1}{2^{19}}=2^x\)

\(\Rightarrow2^{-19}=2^x\Rightarrow x=-19\)

28 tháng 10 2017

\(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot...\cdot\dfrac{15}{32}\\ =\dfrac{1\cdot2\cdot3\cdot...\cdot15}{4\cdot6\cdot8\cdot...\cdot32}=\dfrac{1}{2^{14}\cdot2^5}=\dfrac{1}{2^{19}}=2^{-19}=2^x\\ \text{Vì }2\ne0;2\ne\pm1\text{ nên }x=-19\)

30 tháng 8 2019

1) -2/3

1: \(\Leftrightarrow3x+4=2\)

=>3x=-2

=>x=-2/3

2: \(\Leftrightarrow7x-7=6x-30\)

=>x=-23

3: =>\(5x-5=3x+9\)

=>2x=14

=>x=7

4: =>9x+15=14x+7

=>-5x=-8

=>x=8/5

a: \(\Leftrightarrow\dfrac{x+1}{2x+1}=\dfrac{x+4}{2x+6}\)

=>(x+1)(2x+6)=(2x+1)(x+4)

\(\Leftrightarrow2x^2+6x+2x+6=2x^2+8x+x+4\)

=>9x+4=8x+6

=>x=2

b: \(x^2+5x=0\)

=>x(x+5)=0

=>x=0 hoặc x=-5

AH
Akai Haruma
Giáo viên
15 tháng 11 2017

Lời giải:

Ta có:

\(\text{VT}=\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.....\frac{30}{62}.\frac{31}{64}=\frac{1.2.3....31}{2.4.6.8...64}\)

Xét mẫu số:

\(2.4.6.8.....62.64=(2.1)(2.2)(2.3)(2.4)....(2.31)(2.32)\)

\(=2^{32}(1.2.3....31.32)\)

Suy ra:

\(\text{VT}=\frac{1.2.3....31}{2^{32}.(1.2.3...31.32)}=\frac{1}{2^{32}.32}=\frac{1}{2^{37}}\)

Do đó \(4^x=\frac{1}{2^{37}}\Leftrightarrow 2^{2x}=\frac{1}{2^{37}}\Leftrightarrow 2^{2x+37}=1\)

\(\Leftrightarrow 2x+37=0\Leftrightarrow x=-\frac{37}{2}\)

Vậy \(x=\frac{-37}{2}\)

18 tháng 11 2017

Số 2 nó ở đâu chui ra v Violympic toán 7

=>\(1\cdot\dfrac{2}{4}\cdot\dfrac{3}{6}\cdot...\cdot\dfrac{31}{62}\cdot\dfrac{1}{64}=2^x\)

=>\(2^x=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot...\cdot\dfrac{1}{2}\cdot\dfrac{1}{64}=\left(\dfrac{1}{2}\right)^{30}\cdot\left(\dfrac{1}{2}\right)^6=\dfrac{1}{2^{36}}\)

=>x=-36

10 tháng 2 2018

a) \(-5\cdot\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}\cdot\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}\cdot x-\dfrac{5}{6}\\ -5\cdot x+1-\dfrac{1}{2}\cdot x-\dfrac{1}{3}=\dfrac{3}{2}\cdot x-\dfrac{5}{6}\\ x\cdot\left(-5-\dfrac{1}{2}\right)+\dfrac{1}{3}+\dfrac{5}{6}=\dfrac{3}{2}\cdot x\\ x\cdot\dfrac{-11}{2}+\dfrac{7}{6}=\dfrac{3}{2}\cdot x\\ \dfrac{3}{2}\cdot x-\dfrac{-11}{2}\cdot x=\dfrac{7}{6}\\ x\cdot\left(\dfrac{3}{2}-\dfrac{-11}{2}\right)=\dfrac{7}{6}\\ x\cdot7=\dfrac{7}{6}\\ x=\dfrac{7}{6}:7\\ x=\dfrac{1}{6}\)

Vậy x = \(\dfrac{1}{6}\)

11 tháng 2 2018

b, \(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2^x\\ \dfrac{1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot30\cdot31}{2^{30}\cdot\left(1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot30\cdot31\right)\cdot64}=2^x\\ \dfrac{1}{2^{30}\cdot2^6}=2^x\\ \dfrac{1}{2^{36}}=2^x\\ 2^{-36}=2^x\\ \Rightarrow x=-36\)

a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< =x< =\dfrac{-13}{5}:\dfrac{21}{15}\)

=>-10<=x<=-13/7

hay \(x\in\left\{-10;-9;...;-2\right\}\)

b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< =x< =-\dfrac{2}{3}\cdot\dfrac{-11}{12}\)

=>-13/9<=x<=11/18

hay \(x\in\left\{-1;0\right\}\)