29⁹ ???

|x-2|+|x-5|=5x

điều kiện 5x>=0<=>x>=0

=>|x-2|+|x-5|=x-2+x-5=5x

=>2x+7=5x

=>7=-2x+5x

=>7=3x

=>x=2.3

                                                      Bài giải

Ta có : \(\left(\frac{1}{2}\right)^4=\frac{1^4}{2^4}=\frac{1}{2^4}\)

\(\left(\frac{1}{4}\right)^4=\frac{1^4}{4^4}=\frac{1}{4^4}\)

Vì \(2^4< 4^4\text{ }\Rightarrow\text{ }\frac{1}{2^4}>\frac{1}{4^4}\)

                                                      Bài giải

Ta có : \(\left(\frac{1}{2}\right)^4=\frac{1^4}{2^4}=\frac{1}{2^4}\)

\(\left(\frac{1}{4}\right)^4=\frac{1^4}{4^4}=\frac{1}{4^4}\)

Vì \(2^4< 4^4\text{ }\Rightarrow\text{ }\frac{1}{2^4}>\frac{1}{4^4}\)

a)\(\frac{9}{4}\cdot\left|x\right|-\frac{5}{2}=\frac{8}{3}\)\(\Rightarrow\frac{9}{4}\cdot\left|x\right|=\frac{8}{3}+\frac{5}{2}\Rightarrow\frac{9}{4}\cdot\left|x\right|=\frac{31}{6}\)

\(\Rightarrow\left|x\right|=\frac{31}{6}:\frac{9}{4}\Rightarrow\left|x\right|=\frac{31}{6}\cdot\frac{4}{9}\Rightarrow\left|x\right|=\frac{62}{27}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{62}{27}\\x=-\frac{62}{27}\end{cases}}\)

b)\(\frac{1}{2}\cdot\left|x\right|+\frac{3}{4}=\frac{2}{3}\Rightarrow\frac{1}{2}\cdot\left|x\right|=\frac{2}{3}-\frac{3}{4}\Rightarrow\frac{1}{2}\cdot\left|x\right|=-\frac{1}{12}\)

\(\Rightarrow\left|x\right|=-\frac{1}{12}:\frac{1}{2}\Rightarrow\left|x\right|=-\frac{1}{12}\cdot2\Rightarrow\left|x\right|=-\frac{1}{6}\)

Ta có\(\left|x\right|\ge0\)mà \(-\frac{1}{6}\le0\)

Do đó ko có giá trị của x thỏa mãn

Ta có : |3x + 2| \(\ge\)0 \(\forall\)x

          |2x + 3| \(\ge\)0 \(\forall\)x

           |x + 1| \(\ge\)0 \(\forall\)x

=> |3x + 2| + |2x + 3| + |x + 1| \(\ge\)0 \(\forall\)x => 10x \(\ge\)0 => x \(\ge\)0

Với : x \(\ge\)0 => 3x + 2 + 2x + 3 + x + 1 = 10x

=> 6x + 6 = 10x

=> 6 = 10x - 6x

=> 4x = 6

=> x = 3/2

Bài 1

a, x=3 hoặc -3

b. X = 1/7 hay -1/7