d: =>4x+6=15x-12

=>4x-15x=-12-6=-18

=>-11x=-18

hay x=18/11

e: =>\(45x+27=12+24x\)

=>21x=-15

hay x=-5/7

f: =>35x-5=96-6x

=>41x=101

hay x=101/41

g: =>3(x-3)=90-5(1-2x)

=>3x-9=90-5+10x

=>3x-9=10x+85

=>-7x=94

hay x=-94/7

làm rõ ra giúp với ạ, ghi v k hỉu j hết ;-;

d,

\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)

e,

\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)

\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)

\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)

Vậy không tồn tại $x$ thỏa mãn đề bài.

f, 

\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)

\(\Leftrightarrow 6x-3=10+6x\)

\(\Leftrightarrow 13=0\) (vô lý)

Vậy không tồn tại $x$ thỏa mãn đề bài.

a,

$0-|x+1|=5$

$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)

Do đó không tồn tại $x$ thỏa mãn điều kiện đề.

b,

\(2-|\frac{3}{4}-x|=\frac{7}{12}\)

\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)

\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)

c, 

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

\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)

\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)

\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)

a: 0-|x+1|=5

=>|x+1|=0-5=-5<0(vô lý)

=>x∈∅

b: \(2-\left|\frac34-x\right|=\frac{7}{12}\)

=>\(\left|x-\frac34\right|=2-\frac{7}{12}=\frac{17}{12}\)

=>\(\left[\begin{array}{l}x-\frac34=\frac{17}{12}\\ x-\frac34=-\frac{17}{12}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{17}{12}+\frac{9}{12}=\frac{26}{12}=\frac{13}{6}\\ x=-\frac{17}{12}+\frac34=-\frac{17}{12}+\frac{9}{12}=-\frac{8}{12}=-\frac23\end{array}\right.\)

c: \(2\left|\frac12x-\frac13\right|-\frac32=\frac14\)

=>\(\left|2\left(\frac12x-\frac13\right)\right|=\frac32+\frac14=\frac74\)

=>\(\left|x-\frac23\right|=\frac74\)

=>\(\left[\begin{array}{l}x-\frac23=\frac74\\ x-\frac23=-\frac74\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac74+\frac23=\frac{21+8}{12}=\frac{29}{12}\\ x=-\frac74+\frac23=\frac{-21+8}{12}=-\frac{13}{12}\end{array}\right.\)

d: \(\left|x-\frac13\right|=\frac56\)

=>\(\left[\begin{array}{l}x-\frac13=\frac56\\ x-\frac13=-\frac56\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac56+\frac13=\frac76\\ x=-\frac56+\frac13=-\frac36=-\frac12\end{array}\right.\)

e: \(\frac34-2\left|2x-\frac23\right|=2\)

=>\(2\left|2x-\frac23\right|=\frac34-2=-\frac54<0\) (vô lý)

=>x∈∅

f: \(\frac{2x-1}{2}=\frac{5+3x}{3}\)

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

=>6x-3=6x+10

=>-3=10(vô lý)

=>x∈∅

\(\frac{x}{2^2}+\frac{x}{3^2}+\frac{x}{4^2}=\frac{x}{2^3}+\frac{x}{3^3}+\frac{x}{4^3}\)

\(\Rightarrow x.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}\right)=x.\left(\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}\right)\)

Mà \(\frac{1}{2^2}>\frac{1}{2^3};\frac{1}{3^2}>\frac{1}{3^3};\frac{1}{4^2}>\frac{1}{4^3}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}\ne\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}\)

=> x = 0

Vậy x = 0

giải toán violympic cần nhanh, chính xác

= x( 1/2² + .....- 1/4³) = 0

x = 0

\(\frac{x}{2^2}+\frac{x}{2^3}+\frac{x}{2^4}=\frac{x}{3^2}+\frac{x}{3^3}+\frac{x}{3^4}\)

\(\Rightarrow\frac{x}{2^2}+\frac{x}{2^3}+\frac{x}{2^4}-\frac{x}{3^2}-\frac{x}{3^3}-\frac{x}{3^4}=0\)

\(x\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{3^2}-\frac{1}{3^3}-\frac{1}{3^4}\right)=0\)

Mà \(\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{3^2}-\frac{1}{3^3}-\frac{1}{3^4}\right)\ne0\)

\(\Rightarrow x=0\)

\(\frac{x}{2^2}+\frac{x}{2^3}+\frac{x}{2^4}=\frac{x}{3^2}+\frac{x}{3^3}+\frac{x}{3^4}\)

\(\Leftrightarrow\frac{x}{2^2}+\frac{x}{2^3}+\frac{x}{2^4}-\frac{x}{3^2}-\frac{x}{3^3}-\frac{x}{3^4}=0\)

\(\Leftrightarrow x\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{3^2}-\frac{1}{3^3}-\frac{1}{3^4}\right)=0\)

\(\Leftrightarrow x=0\). Do \(\Leftrightarrow x=0\)

a) \(\left(x+1\right).\left(x-2\right)< 0\)

\(\Rightarrow x+1\) và \(x-2\) trái dấu.

Ta có 2 trường hợp:

TH1: \(\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\left(loại\right)\)

TH2: \(\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\Rightarrow-1< x< 2.\)

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

Vậy \(x\in\left\{0;1\right\}\) thì \(\left(x+1\right).\left(x-2\right)< 0.\)

Chúc bạn học tốt!

Bạn ơi câu tính giá trị bthuc thì thừa số thứ nhất có bị sai đề ko vậy??

a, 3/4 + 1/4.x=2

    1/4.x        = 2-3/4

     1/4.x      =5/4

     x           = 5/4:1/4

     x           = 5

b, x-2/3.9/4=2,5-1/2

    x-2/3.9/4=2

    x-2/3     =2:9/4

    x-2/3   =8/9

     x        = 8/9+2/3

     x         = 14/9

Ta có : \(\frac{x}{2^2}+\frac{x}{2^3}+\frac{x}{2^4}=\frac{x}{3^2}+\frac{x}{3^3}+\frac{x}{3^4}\)

<=> \(\frac{x}{2^2}+\frac{x}{2^3}+\frac{x}{2^4}-\frac{x}{3^2}-\frac{x}{3^3}-\frac{x}{3^4}=0\)

<=> \(x\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{3^2}-\frac{1}{3^3}-\frac{1}{3^4}\right)=0\)

Mà \(\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{3^2}-\frac{1}{3^3}-\frac{1}{3^4}\right)\ne0\)

Vậy : x = 0

\(\Rightarrow x.\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}\right)=x.\left(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)\)

\(\Rightarrow x.\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}\right)-x.\left(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)=0\)

\(\Rightarrow x=0\)

Vậy x=0 nha