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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

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

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

=>x=-36

14 tháng 6 2017

a.

| x | = 5,6

=>\(\left[{}\begin{matrix}x=5,6\\x=-5,6\end{matrix}\right.\)

Vậy \(x\in\left\{-5,6;5,6\right\}\)

b, \(\left|x-3,5\right|=5\)

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

=>\(\left[{}\begin{matrix}x=8,5\\x=-1,5\end{matrix}\right.\)

Vậy \(x\in\left\{-1,5;8,5\right\}\)

c,\(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)

=> \(\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)

=>\(\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{1}{4};\dfrac{5}{4}\right\}\)

d,\(\left|4x\right|-\left(\left|-13,5\right|\right)=\left|\dfrac{1}{4}\right|\)

=> \(\left|4x\right|-13,5=\dfrac{1}{4}\)

=> \(\left|4x\right|=13,75\)

=>\(\left[{}\begin{matrix}4x=13,75\\4x=-13,75\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=3,4375\\x=-3,4375\end{matrix}\right.\)

Vậy \(x\in\left\{-3,4375;3,4375\right\}\)

14 tháng 6 2017

e, ( x - 1 ) 3 = 27

=> x - 1 = 3

=> x = 4

Vậy x = 4

f, ( 2x - 3)2 = 36

=> \(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=4,5\\x=-1,5\end{matrix}\right.\)

Vậy x\(\in\left\{-1,5;4,5\right\}\)

g, \(5^{x+2}=625\)

=> \(5^{x+2}=5^4\)

=> x + 2 = 4

=> x = 2

Vậy x = 2

h, ( 2x - 1)3 = -8

=> 2x - 1 = -2

=> x = \(\dfrac{-1}{2}\)

Vậy x = \(\dfrac{-1}{2}\)

i, \(\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}...\dfrac{30}{62}.\dfrac{31}{64}=2^x\)

=> \(\dfrac{1.2.3.4.5...30.31}{4.6.8.10.12...62.64}=2^x\)

=>\(\dfrac{1.2.3.4.5...30.31}{\left(2.3.4.5...30.31.32\right)\left(2.2.2.2...2.2_{ }\right)}=2^x\)(có 31 số 2)

=> \(\dfrac{1}{32.2^{31}}=2^x\)

=> \(\dfrac{1}{2^{36}}=2^x\)

=> x = -36

Vậy x = -36

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

a) Ta có: \(\dfrac{4}{5}-3\left|x\right|=\dfrac{1}{5}\)

\(\Leftrightarrow3\left|x\right|=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\)

\(\Leftrightarrow\left|x\right|=\dfrac{1}{5}\)

hay \(x\in\left\{\dfrac{1}{5};-\dfrac{1}{5}\right\}\)

b) Ta có: \(4x-\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{4}{5}\)

nên \(\dfrac{41}{10}x=\dfrac{4}{5}\)

hay \(x=\dfrac{8}{41}\)

c) Ta có: \(\left(2x-8\right)\left(10-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-8=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

d) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}\)

\(\Leftrightarrow\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}-\dfrac{3}{4}=\dfrac{14}{4}-\dfrac{3}{4}=\dfrac{11}{4}\)

\(\Leftrightarrow\left|2x-1\right|=11\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=11\\2x-1=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=12\\2x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)

\(\dfrac{1}{2.2}.\dfrac{2}{2.3}.....\dfrac{31}{64}=2^x\\ =>\dfrac{1}{2.2.2.....2.64}=2^x\\ \dfrac{1}{2^{30}.26}=2^x\\ =>\dfrac{1}{2^{36}}=2^x\\ =>2^{-36}=2^x\\ =>x=-36\)

6 tháng 8 2021

\(n=-36\)

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

20 tháng 6 2018

a)x=1;2;-2(bạn nên tự giải)

b)=>\(\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}\)=2x

=>\(\dfrac{2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31}{60\left(2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31\right)\cdot64}=2x\)

=>\(\dfrac{1}{60\cdot64}=2x\)=> 1/3840 =2x

=>x = 1/7680

c)=>4x - 2x = 6x - 3x

=>2x (2x-1)= 3x(2x-1)

=> 2x = 3x

=>x = 0

21 tháng 6 2018

ak mình nhầm

10 tháng 8 2018

\(xy-3x-y=6\)

\(=>xy+3x-y-3=6-3\)

\(=>x\left(y+3\right)-\left(y+3\right)=3\)

\(=>\left(y+3\right)\left(x-1\right)=3\)

y+3 -1 3 1 -3
x-1 -3 1 3 -1

y+3 -1 3 -3 1
y -4 -1 -7 -3

x-1 -3 1 3 -1
x -2 2 4 0