khó quá

khó quá e hông hỉu  e moi hc lop 5 thui

1)(x-3)²=0

=>x-3=0

x=0+3

x=3

2)(2x+1)²=4=2²

=>2x+1=2

2x=2-1

2x=1

x=1/2

3)(2x-3)³=8=2³

=>2x-3=2

2x=2+3

2x=5

x=5/2

4)(x+1/2)⁴=1/16=(1/4)⁴

=>x+1/2=1/4

x=1/4-1/2=1/4-2/4

x=-1/4

5)9,27<=3^x<=243

3²<=3^x<=3⁵ (vì x thuộc Z nên làm tròn số 9,27)

=>x thuộc{3;4}

\(\left(2\frac{1}{3}+3\frac{1}{2}\right):\left(-4\frac{1}{6}+3\frac{1}{7}\right)+7\frac{1}{2}=5\frac{5}{6}:-1\frac{1}{42}+7\frac{1}{2}=-5\frac{30}{43}+7\frac{1}{2}=\frac{155}{86}=1\frac{69}{86}\)

\(VT=\left|x-3\right|+\left|5-x\right|+2\left|x-4\right|\ge\left|x-3+5-x\right|+2.0=2\)

Dấu "=" xảy ra khi: \(\left(x-3\right)\left(5-x\right)\ge0\text{ và }x-4=0\)\(\Leftrightarrow x=4\)

Vậy x = 4.

\(\frac{x+1}{2014}+\frac{x+2}{2013}=\frac{x+3}{2012}+\frac{x+4}{2011}\)

\(\frac{x+1}{2014}+1+\frac{x+2}{2013}+1=\frac{x+3}{2012}+1+\frac{x+4}{2011}+1\)

\(\frac{x+2015}{2014}+\frac{x+2015}{2013}=\frac{x+2015}{2012}+\frac{x+2015}{2011}\)

\(\frac{x+2015}{2014}+\frac{x+2015}{2013}-\frac{x+2015}{2012}-\frac{x+2015}{2011}=0\)

\(\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)=0\)

Vì \(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\ne0\) nên x + 2015 = 0

x = 0 - 2015

x = -2015

\(\frac{x+1}{2014}+\frac{x+2}{2013}=\frac{x+3}{2012}+\frac{x+4}{2011}\)

\(1+\frac{x+1}{2014}+1+\frac{x+2}{2013}=1+\frac{x+3}{2012}+1+\frac{x+4}{2011}\)

\(\frac{x+1+2014}{2014}+\frac{x+2+2013}{2013}=\frac{x+3+2012}{2012}+\frac{x+4+2011}{2011}\)

\(\frac{x+2015}{2014}+\frac{x+2015}{2013}=\frac{x+2015}{2012}+\frac{x+2015}{2011}\)

\(\Rightarrow\frac{x+2015}{2014}+\frac{x+2015}{2013}-\frac{x+2015}{2012}-\frac{x+2015}{2011}=0\)

\(\Rightarrow\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)=0\)

=> x + 2015 = 0 ( vì 1/2014 + 1/2013 - 1/2012  - 1/2011 khác 0)

=> x = -2015

áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{x+2}{x+6}=\frac{x+4}{x+5}=\frac{\left(x+2\right)-\left(x+4\right)}{\left(x+6\right)-\left(x+5\right)}=\frac{x+2-x-4}{x+6-x-5}=\frac{-2}{1}=-2\)

suy ra:

\(\frac{x+2}{x+6}=-2\Rightarrow x+2=-2.\left(x+6\right)\)

\(x+2=-2x-12\)

\(x+2x=-12-2\)

\(3x=-14\)

\(x=\frac{-14}{3}\)

Để D>0 thì \(\frac{x+1}{x+9}>0\)

=> \(x+1>0\) \(;x+9>0\) hoặc \(x+1