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2 tháng 10 2016

b) \(3.2^{x+1}=12\)

\(2^{x+1}=12:3\)

\(2^{x+1}=4\)

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

\(x+1=2\)

\(x=2-1\)

\(x=1\)

Vậy \(x=1\)

c) \(2^{x-1}=2^3+2^4-2^3\)

\(2^{x-1}=8+16-8\)

\(2^{x-1}=16\)

\(2^{x-1}=2^4\)

\(x-1=4\)

\(x=5\)

Vậy \(x=5\)

d) \(x^{50}=x\)

\(x^{50}-x=0\)

\(\Rightarrow x\in\left\{0;1\right\}\)

Vậy \(x\in\left\{0;1\right\}\)

2 tháng 10 2016

\(b.3.2^{x+1}=12\\ \Rightarrow2^{x+1}=4\\ \Rightarrow2^{x+1}=2^2\\ \Rightarrow x=1\\ \)

c) \(2^{x-1}=2^3-2^3+2^4\\ \Rightarrow2^{x-1}=0+16\\ \Rightarrow2^{x-1}=16\\ \Rightarrow2^{x-1}=2^4\\ \Rightarrow x-1=4\\ \Rightarrow x=5\)

d) \(x^{50}=x\\ \Rightarrow x=0;1\)

e) \(2\left(2x-1\right)^4=32\\ \Rightarrow\left(2x-1\right)^4=16\\ \Rightarrow\left(2x-1\right)^4=2^4\\ \Rightarrow2x-1=2\\ \Rightarrow2x=3\\ \Rightarrow x=\frac{3}{2}\)

g) Bí 

7 tháng 1 2016

bạn giúp mình giải 3 câu này nhé

 

a) \(3x-10=2x+13\Leftrightarrow x=13+10=23\)

b) \(x+12=-5-x\Leftrightarrow2x=-17\Leftrightarrow x=-\frac{17}{2}\)

c) \(x+5=10-x\Leftrightarrow2x=5\Leftrightarrow x=\frac{5}{2}\)

d) \(6x+23=2x-12\Leftrightarrow4x=-35\Leftrightarrow x=-\frac{35}{4}\)

e) \(12-x=x+1\Leftrightarrow2x=11\Leftrightarrow x=\frac{11}{2}\)

f) \(14+4x=3x+20\Leftrightarrow x=20-14=6\)

g) \(2\left(x-1\right)+3\left(x-2\right)=x-4\Leftrightarrow2x-2+3x-6=x-4\)

\(\Leftrightarrow5x-8=x-4\Leftrightarrow4x=4\Leftrightarrow x=1\)

h ) \(3\left(4-x\right)-2\left(x-1\right)=x+20\Leftrightarrow12-3x-2x+2=x+20\)

\(\Leftrightarrow14-5x=x+20\Leftrightarrow6x=-6\Leftrightarrow x=-1\)

i ) \(4\left(2x+7\right)-3\left(3x-2\right)=24\Leftrightarrow8x+28-9x+6=24\)

\(\Leftrightarrow34-x=24\Leftrightarrow x=34-24=10\)

k) \(3\left(x-2\right)+2x=10\Leftrightarrow3x-6+2x=5x-6=10\)

\(\Leftrightarrow5x=10+6=16\Leftrightarrow x=\frac{16}{5}\)

Tự KL cho các phần

2 tháng 12 2020

a, \(2\left(x+3\right)\left(x-4\right)=\left(2x-1\right)\left(x+2\right)-27\)

\(\Leftrightarrow2\left(x^2-4x+3x-12\right)=2x^2+4x-x-2-27\)

\(\Leftrightarrow2x^2-2x-24=2x^2+3x-29\Leftrightarrow-5x+5=0\Leftrightarrow x=1\)

b, \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)

\(\Leftrightarrow x^3-8-x\left(x^2-9\right)=26\Leftrightarrow-8+9x=26\)

\(\Leftrightarrow9x=18\Leftrightarrow x=2\)

a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

1 tháng 10 2021

giải phần còn lại giúp mình được ko?

27 tháng 4 2023

Cậu tách ra `2->3` câu thôi nhe

 

a: =>17x-5x-15-2x-5=0

=>10x-20=0

=>x=2

b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)

=>11x+23=-2x-16

=>13x=-39

=>x=-3(nhận)

c: =>5x+7>=3x-3

=>2x>=-10

=>x>=-5

d: =>5(3x-1)=-2(x+1)

=>15x-5=-2x-2

=>17x=3

=>x=3/17

e: =>4x^2-1-4x^2-3x-2=0

=>-3x-3=0

=>x=-1

g: =>7x-5-8x+2-7<0

=>-x-10<0

=>x+10>0

=>x>-10

a: =>17x-5x-15-2x-5=0

=>10x-20=0

=>x=2

b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)

=>11x+23=-2x-16

=>13x=-39

=>x=-3(nhận)

c: =>5x+7>=3x-3

=>2x>=-10

=>x>=-5

d: =>5(3x-1)=-2(x+1)

=>15x-5=-2x-2

=>17x=3

=>x=3/17

e: =>4x^2-1-4x^2-3x-2=0

=>-3x-3=0

=>x=-1

g: =>7x-5-8x+2-7<0

=>-x-10<0

=>x+10>0

=>x>-10

22 tháng 12 2020

a) Ta có: \(\dfrac{9-3x}{x^2+3x+4}-\dfrac{3x-23}{\left(1-x\right)\left(x+4\right)}\)

\(=\dfrac{9-3x}{x^2+3x+4}+\dfrac{3x-23}{x^2+3x-4}\)

\(=\dfrac{\left(9-3x\right)\left(x^2+3x-4\right)}{\left(x^2+3x+4\right)\left(x^2+3x-4\right)}+\dfrac{\left(3x-23\right)\left(x^2+3x+4\right)}{\left(x^2+3x-4\right)\left(x^2+3x+4\right)}\)

\(=\dfrac{9x^2+27x-36-3x^3-9x^2+12x+3x^3+9x^2+12x-23x^2-69x-92}{\left(x^2+3x-4\right)\left(x^2+3x+4\right)}\)

\(=\dfrac{-14x^2-18x-128}{\left(x^2+3x-4\right)\left(x^2+3x+4\right)}\)

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

\(=\dfrac{4-x}{x\left(x^2+2\right)}-\dfrac{x+5}{x^2\left(x-1\right)+2\left(x-1\right)}\)

\(=\dfrac{4-x}{x\left(x^2+2\right)}-\dfrac{x+5}{\left(x-1\right)\left(x^2+2\right)}\)

\(=\dfrac{\left(4-x\right)\left(x-1\right)}{x\left(x-1\right)\left(x^2+2\right)}-\dfrac{x\left(x+5\right)}{x\left(x-1\right)\left(x^2+2\right)}\)

\(=\dfrac{4x-4-x^2+x-x^2-5x}{x\left(x-1\right)\left(x^2+2\right)}\)

\(=\dfrac{-2x^2-4}{x\left(x-1\right)\left(x^2+2\right)}\)

\(=\dfrac{-2\left(x^2+2\right)}{x\left(x-1\right)\left(x^2+2\right)}\)

\(=\dfrac{-2}{x\left(x-1\right)}\)