a)\(2\left|2x-3\right|=\frac{1}{2}\)

\(\Leftrightarrow\left|2x-3\right|=\frac{1}{4}\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=\frac{1}{4}\\2x-3=-\frac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{13}{8}\\x=\frac{11}{8}\end{matrix}\right.\)

Vậy....

b)\(7,5-3\left|5-2x\right|=-4,5\)

\(\Leftrightarrow\left|5-2x\right|=4\)

\(\Rightarrow\left[{}\begin{matrix}5-2x=4\\5-2x=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{9}{2}\end{matrix}\right.\)

VẬy...

c)\(\left|3x-4\right|+\left|5-2x\right|=0\)

Có: \(\left|3x-4\right|\ge0với\forall x\\ \left|5-2x\right|\ge0với\forall x\)

\(\Rightarrow\left[{}\begin{matrix}3x-4=0\\5-2x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x=\frac{5}{2}\end{matrix}\right.\)

\(\Rightarrow x\in\varnothing\)

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

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

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

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{13}{8}\\x=\dfrac{11}{8}\end{matrix}\right.\)

b, \(7,5-3\left|5-2x\right|=-4,5\)

\(\Rightarrow3\left|5-2x\right|=12\)

\(\Rightarrow\left|5-2x\right|=4\)

\(\Rightarrow\left\{{}\begin{matrix}5-2x=4\\5-2x=-4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\)

c, \(\left|3x-4\right|+\left|3y+5\right|=0\)

Với mọi giá trị của \(x;y\in R\) ta có:

\(\left|3x-4\right|\ge0;\left|3y+5\right|\ge0\)

\(\Rightarrow\left|3x-4\right|+\left|3y+5\right|\ge0\) với mọi giá trị của \(x;y\in R\).

Để \(\left|3x-4\right|+\left|3y+5\right|=0\) thì

\(\left\{{}\begin{matrix}\left|3x-4\right|=0\\\left|3y+5\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x-4=0\\3y+5=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3x=4\\3y=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=-\dfrac{5}{3}\end{matrix}\right.\)

Vậy.............

Chúc bạn học tốt!!!

a) (x-3)x(4-5x x)=0

=> x-3=0 hoặc 4-5x x=0

=>x=3 hoặc x=0,8

b) x²-2=0

=>x²=2

=>x=\(\sqrt{2}\)

c) x²+\(\sqrt{3}\)=0

=>x²= -\(\sqrt{3}\)

=> Vô nghiệm

d) x²+2x x=0

=> x x(x+2)=0

=> x=0 hoặc x+2=0

=>x=0 hoặc x=-2

e) x² + 2x x-3=0

=>x²- x+ 3x -3=0

=>(x²-x)+ (3x - 3)=0

=> x(x-1)+ 3(x-1)=0

=>(x-1) x (x+3)=0

=> x-1 =0 hoặc x+3=0

=> x= 1 hoặc x=-3

a) (x-3)x(4-5x x)=0

=> x-3=0 hoặc 4-5x x=0

=>x=3 hoặc x=0,8

b) x²-2=0

=>x²=2

=>x=\(\sqrt{2}\)

c) x²+\(\sqrt{3}\)=0

=>x²= -\(\sqrt{3}\)

=> Vô nghiệm

d) x²+2x x=0

=> x x(x+2)=0

=> x=0 hoặc x+2=0

=>x=0 hoặc x=-2

e) x² + 2x x-3=0

=>x²- x+ 3x -3=0

=>(x²-x)+ (3x - 3)=0

=> x(x-1)+ 3(x-1)=0

=>(x-1) x (x+3)=0

=> x-1 =0 hoặc x+3=0

=> x= 1 hoặc x=-3

a) (x-3)x(4-5x x)=0

=> x-3=0 hoặc 4-5x x=0

=>x=3 hoặc x=0,8

b) x²-2=0

=>x²=2

=>x=\(\sqrt{2}\)

c) x²+\(\sqrt{3}\)=0

=>x²= -\(\sqrt{3}\)

=> Vô nghiệm

d) x²+2x x=0

=> x x(x+2)=0

=> x=0 hoặc x+2=0

=>x=0 hoặc x=-2

e) x² + 2x x-3=0

=>x²- x+ 3x -3=0

=>(x²-x)+ (3x - 3)=0

=> x(x-1)+ 3(x-1)=0

=>(x-1) x (x+3)=0

=> x-1 =0 hoặc x+3=0

=> x= 1 hoặc x=-3

c.\(\dfrac{3}{7}+\dfrac{5}{7}:x=\dfrac{1}{3}\)

\(\dfrac{5}{7}:x=\dfrac{1}{3}-\dfrac{3}{7}\)

\(\dfrac{5}{7}:x=-\dfrac{2}{21}\)

\(x=\dfrac{5}{7}:-\dfrac{2}{21}\)

\(x=-\dfrac{15}{2}\)

d.\(3\dfrac{1}{4}:\left|2x-\dfrac{5}{12}\right|=\dfrac{39}{16}\)

\(\left|2x-\dfrac{5}{12}\right|=3\dfrac{1}{4}:\dfrac{39}{16}\)

\(\left|2x-\dfrac{5}{12}\right|=\dfrac{4}{3}\)

\(\rightarrow\left[{}\begin{matrix}2x-\dfrac{5}{12}=\dfrac{4}{3}\\2x-\dfrac{4}{12}=-\dfrac{4}{3}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}2x=\dfrac{7}{4}\\2x=-\dfrac{11}{12}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=-\dfrac{11}{24}\end{matrix}\right.\)

A, \(\dfrac{4}{9}+x=\dfrac{5}{3}\)

\(x\)\(=\dfrac{5}{3}-\dfrac{4}{9}\)

\(x\)\(=\dfrac{11}{9}\)

B,\(\dfrac{3}{4}.x=\dfrac{-1}{2}\)

\(x=\dfrac{-1}{2}:\dfrac{3}{4}\)

\(x=\)\(\dfrac{-2}{3}\)

a)

\(\frac{4}{9} + x = \frac{5}{3}\)

=> \(x = \frac{5}{3}-\frac{4}{9}\)

=> \(x = \) \(\frac{11}{9}\)

Vậy \(x = \dfrac{11}{9}\)

b) 

\(\dfrac{3}{4} .x = \dfrac{-1}{2}\)

=> \(x = \dfrac{-1}{2} : \dfrac{3}{4}\)

=> \(x = \dfrac{-2}{3}\)

Vậy \(x = \dfrac{-2}{3}\)

c)

\( \dfrac{3}{7}+ \dfrac{5}{7}:x = \dfrac{1}{3}\)

=> \(\dfrac{5}{7}:x = \dfrac{1}{3}-\) \( \dfrac{3}{7}\)

=> \(\dfrac{5}{7}:x = \dfrac{-2}{21}\)

=> \(x = \dfrac{5}{7}:\dfrac{-2}{21}\)

=> \(x = \dfrac{-15}{2}\)

Vậy \(x = \dfrac{-15}{2}\)

d) 

\(3\dfrac{1}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)

=> \(\dfrac{13}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)

=> \( |2x - \dfrac{5}{12} | =\dfrac{13}{4} : \dfrac{39}{16}\)

=> \(|2x-\dfrac{5}{12} |= \dfrac{4}{3}\)

=> \(\left[\begin{matrix} 2x - \dfrac{5}{12} = \dfrac{4}{3}\\ 2x - \dfrac{5}{12} = \dfrac{4}{3}\end{matrix}\right.\)

=> \(\left[\begin{matrix} 2x = \dfrac{-4}{3}+\dfrac{5}{12}\\ 2x = \dfrac{-4}{3}+\dfrac{5}{12} \end{matrix}\right.\)

=> \(\left[\begin{matrix} 2x = \dfrac{7}{4}\\ 2x = \dfrac{-11}{12} \end{matrix}\right.\)

=> \(\left[\begin{matrix} x = \dfrac{7}{8}\\ x = \dfrac{-11}{24} \end{matrix}\right.\)

Vậy \(x \in \) { \(\dfrac{7}{8} ; \dfrac{-11}{24}\) }

\(E=\frac{4}{\left(2x-3\right)^2+5}\)

\(E\le\frac{4}{5}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow2x-3=0\Leftrightarrow x=\frac{3}{2}\)

\(E=\frac{4}{\left(2x-3\right)^2+5}\)

\(E\le\frac{4}{5}\forall x\)

Dấu " = " xảy ra <=> 2x - 3 = 0 <=> x = 3/2

\(E=\frac{4}{\left(2x-3\right)^2+5}\le\frac{4}{0+5}=\frac{4}{5}\)

Dấu "=" \(\Leftrightarrow2x-3=0\)

               \(\Leftrightarrow x=\frac{3}{2}\)

Vậy ............

Câu 2:

a: 10km=10000m

10000m dây đồng có cân nặng là:

\(47:5\cdot10000=94000\left(g\right)\)

b: 300g=0,3kg=0,003 tạ

0,003 tạ nặng:

\(2,5:1\cdot0,003=\dfrac{3}{400}\left(kg\right)\)

Câu 1:

a:

\(\left|1-2x\right|>=0\forall x\)

=>\(3\left|1-2x\right|>=0\forall x\)

=>\(3\left|1-2x\right|-5>=-5\forall x\)

=>\(A>=-5\forall x\)

Dấu '=' xảy ra khi 1-2x=0

=>2x=1

=>x=1/2

Vậy: \(A_{Min}=-5\) khi x=1/2

b: \(2x^2>=0\forall x\)

=>\(2x^2+1>=1\forall x\)

=>\(\left(2x^2+1\right)^4>=1^4=1\forall x\)

=>\(\left(2x^2+1\right)^4-3>=1-3=-2\forall x\)

=>B>=-2\(\forall\)x

Dấu '=' xảy ra khi x=0

c: \(\left|x-\dfrac{1}{2}\right|>=0\forall x\)

\(\left(y+2\right)^2>=0\forall y\)

Do đó: \(\left|x-\dfrac{1}{2}\right|+\left(y+2\right)^2>=0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+2=0\end{matrix}\right.\)

=>x=1/2 và y=-2

\(=\frac{\left(2^2\right)^5.2.5.5^6+2^8.\left(5^2\right)^5}{2^8.5^4+2^5.5^7}=\frac{2^{10}.2.5^7+2^8.5^{10}}{2^5.5^4\left(2^3+5^3\right)}=\frac{2^{11}.5^7+2^8.5^{10}}{2^5.5^4.\left(2^3+5^3\right)}=\frac{2^8.5^7\left(2^3+5^3\right)}{2^5.5^4\left(2^3+5^3\right)}=\frac{2^3.5^3.1}{1.1.1}=1000\)

|2x-1|=1,5

TH(1)2x-1=1,5

2x =1,5+1

2x =2,5

x =2,5 :2

x =1,25

TH(2) 2x-1=-1,5

2x =-1,5+1

2x =-0,5

x =-0,5:2

x =-0,25

các câu khác cứ tương tự bạn nhé

b) \(7,5-\left|5-2x\right|=-4,5\)

\(\left|5-2x\right|=7,5+4,7\)

\(\left|5-2x\right|=12\)

th1 :\(5-2x=12\)

\(2x=5-12\)

\(2x=-7\)

\(x=-7:2\)

\(x=-3,5\)

th2: \(5-2x=-12\)

\(2x=5+12\)

\(2x=17\)

\(x=17:2\)

\(x=8,5\)

c) \(-3+\left|x\right|=-1\)

\(\left|x\right|=-1+3\)

\(\left|x\right|=2\)

th1: \(x=-2\)

th2 : \(x=2\)

d)\(\left|2\dfrac{1}{3}-x\right|=\dfrac{1}{6}\)

\(\left|\dfrac{7}{3}-x\right|=\dfrac{1}{6}\)

th1 :\(\dfrac{7}{3}-x=\dfrac{1}{6}\)

\(x=\dfrac{7}{3}-\dfrac{1}{2}\)

\(x=\dfrac{11}{6}\)

th2: \(\dfrac{7}{3}-x=\dfrac{-1}{6}\)

\(x=\dfrac{7}{3}+\dfrac{1}{6}\)

\(x=\dfrac{-5}{2}\)

e) \(\dfrac{5}{7}-\left|x+1\right|=\dfrac{1}{14}\)

\(\left|x+1\right|=\dfrac{5}{7}-\dfrac{1}{14}\)

\(\left|x+1\right|=\dfrac{9}{14}\)

th1 :\(x+1=\dfrac{9}{14}\)

\(x=\dfrac{9}{14}-1\)

\(x=\dfrac{-5}{14}\)

th2 : \(x+1=\dfrac{-9}{14}\)

\(x=\dfrac{-9}{14}-1\)

\(x=\dfrac{-5}{14}\)