X= 3\(\frac{7}{10}\)

Tick cho mình nha!!!

cachs giải đâu

Ta có : \(5\frac{8}{17}\div X+\left(-\frac{1}{17}\right)\div X+3\frac{1}{17}\div17\frac{1}{3}=\frac{4}{17}\)

Nên:    \(\left(5\frac{8}{17}+\left(-\frac{1}{17}\right)\right)\div X+\frac{52}{17}\div\frac{52}{3}=\frac{4}{17}\)

           \(5\frac{7}{17}\div X+\frac{52}{17}\times\frac{3}{52}=\frac{4}{17}\)

           \(\frac{92}{17}\div X+\frac{3}{17}=\frac{4}{17}\)

           \(\frac{92}{17}\div X=\frac{4}{17}-\frac{3}{17}\)

           \(\frac{92}{17}\div X=\frac{1}{17}\)

           \(X=\frac{92}{17}\div\frac{1}{17}\)

            \(X=92\)

Vậy \(X=92\)

\(5\frac{8}{17}:x+\left(-\frac{1}{17}\right):x=\frac{52}{51}\)

\(\left(5\frac{8}{17}+-\frac{1}{17}\right):x=\frac{52}{51}\)

\(\frac{92}{17}:x=\frac{52}{51}\)

\(X=\frac{92}{17}:\frac{52}{51}=\frac{69}{13}\)

a) \(\left|x+1\right|+2x=-2\)

\(\Leftrightarrow\left|x+1\right|+2\left(x+1\right)=0\)

\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

b) \(x^2-6x+9=1\)

\(\Leftrightarrow x^2-6x+8=0\)

\(\Leftrightarrow x^2-2x-4x+8=0\)

\(\Leftrightarrow x\left(x-2\right)-4\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x-4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=4\end{array}\right.\)

Bài làm

a)100-7.(X.5)=58

          7.(X.5)=100-58

         7.(X.5)=42

            X.5=42:7

           X.5=6

          X    =6:5

         X=1,2.

#)Giải :

a) \(100-7\times\left(x\times5\right)=58\)

\(\Leftrightarrow7\times\left(x\times5\right)=42\)

\(\Leftrightarrow x\times5=6\)

\(\Leftrightarrow x=1,2\)

b) \(x\times\frac{2}{7}=\frac{242}{363}\times\frac{1616}{2121}\)

\(\Leftrightarrow x\times\frac{2}{7}=\frac{2}{3}\times\frac{16}{21}\)

\(\Leftrightarrow x\times\frac{2}{7}=\frac{32}{63}\)

\(\Leftrightarrow x=\frac{16}{9}\)

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

\(\orbr{\begin{cases}x+2=0\\1-x=0\end{cases}}\)

\(\orbr{\begin{cases}x=-2\\x=1\end{cases}\left(x\text{ ∈}Z\right)}\)

b) \(\left(2x-1\right)^2=9\)

\(\left(2x-1\right)^2=3^2\)

\(2x-1=3\)

\(2x=3+1\)

\(2x=4\)

\(x=2\left(x\text{ ∈}Z\right)\)

c) \(\left(1-5x\right)^3=-27\)

\(\left(1-5x\right)^3=3^3\)

\(1-5x=3\)

\(5x=3+1\)

d, (x - 1)(3 - x) > 0 => (x - 1) và (3 - x) cùng dấu => ta có 2 TH: TH1: (x - 1) và (3 - x) là số nguyên dương => (x - 1) > 0, (3 - x) > 0 => x > 1, 3 > x hay x < 3 => x > 1 và x < 3 => x = 2. TH2: (x - 1) và (3 - x) là số nguyên âm => (x - 1) < 0, (3 - x) < 0 => x < 1, 3 < x hay x > 3 => x < 1, x > 3 (vô lý)(loại). Vậy x = 2

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\(x< y< 0\Rightarrow x,y\)Cùng âm

\(|x|-|y|=100\left(cùngam\right).x< y\Rightarrow x-y=-100\)

a, \(\left(x-2\right)\left(y+1\right)=4\Leftrightarrow x-2;y+1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

tương tự b,c