\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Leftrightarrow\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}=\dfrac{\left(x-2y+3z\right)-\left(1+2+3\right)}{2-6+12}=-\dfrac{106}{8}=\dfrac{53}{4}\)

- \(\dfrac{53}{4}=\dfrac{x-1}{2}\Rightarrow x-1=\dfrac{53}{2}\Rightarrow x=\dfrac{51}{2}\)

- \(\dfrac{53}{4}=\dfrac{y-2}{3}\Rightarrow y-2=\dfrac{159}{4}\Rightarrow y=\dfrac{155}{4}\)

- \(\dfrac{53}{4}=\dfrac{z-3}{4}\Rightarrow z-3=53\Rightarrow z=50\)

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\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\Leftrightarrow\dfrac{x}{3}=\dfrac{2y}{8}=\dfrac{3z}{15}=\dfrac{x-2y+3z}{3-8+15}=\dfrac{35}{10}\)

- \(\dfrac{35}{10}=\dfrac{x}{3}\Rightarrow x=\dfrac{21}{2}\)

- \(\dfrac{35}{10}=\dfrac{y}{4}\Rightarrow y=14\)

-\(\dfrac{35}{10}=\dfrac{z}{5}\Rightarrow z=\dfrac{35}{2}\)

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D=(6x−5)(x+8)−(3x−1)(2x+3)−9(4x−3)

\(\Rightarrow D=\left(6x^2+48x-5x-40\right)-\left(6x^2+9x-2x-3\right)+\left(-36x+27\right)\)

\(\Rightarrow D=6x^2+48x-5x-40-6x^2-9x+2x+3-36x+27\)

\(\Rightarrow D=\left(6x^2-6x^2\right)+\left(48x-5x-9x-36x+2x\right)-40+3+27\)

\(\Rightarrow D=-40+3+27=-10\)

Vậy biểu thức D không phụ thuộc vào giá trị của biến x.(đpcm)

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\(x^2-1=\left(x-1\right)^2⋮x-1\)

\(x^2-4⋮x-1\)

Ta có : \(x^2-4=\left(x^2-1\right)-3\)

Vì \(x^2-1⋮x-1\) nên \(4⋮x-1\\ \Rightarrow x-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\Rightarrow x-1\in\left\{2;0;3;-1;5;-3\right\}\)

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Ai mà biết được

Bài 3 :

a. \(-x+\dfrac{4}{9}=\dfrac{2}{27}\\ \Rightarrow-x=\dfrac{2}{27}-\dfrac{4}{9}=-\dfrac{10}{27}\\ x=\dfrac{10}{27}\)

Bài 2 :

a. \(\dfrac{2}{5}\cdot17\cdot\dfrac{1}{3}-\dfrac{2}{5}\cdot47\cdot\dfrac{1}{3}=\dfrac{1}{3}\left(\dfrac{2}{5}\left(17-47\right)\right)=\dfrac{1}{3}\left(\dfrac{2}{5}\cdot\left(-30\right)\right)=\dfrac{1}{3}\left(-12\right)=-4\)

b. \(\left(\dfrac{6}{5}-\dfrac{2}{7}\right)+15:\dfrac{35}{2^2}=\dfrac{32}{35}+\dfrac{12}{7}=\dfrac{92}{35}\)

Bài 1 :

a. \(\left(a^m\right)^n=a^{m\cdot n}\)

     \(a^m:a^n=a^{m-n}\)

b.\(\left(\left(-\dfrac{1}{2}^3\right)^2\right):\dfrac{\left(\dfrac{2}{3}\right)^7}{\left(\dfrac{2}{3}\right)^5}=-\dfrac{1}{2}^6:\dfrac{2}{3}^2=\dfrac{1}{64}:\dfrac{4}{9}=\dfrac{1\cdot9}{64\cdot4}=\dfrac{9}{256}\)