Là những từ như Often, Alway, Usualy, .... dùng để nói về việc bạn làm một điều gì đó thường xuyên như thế nào.

@Cỏ

#Forever

\(g,ĐK:x\ge0\\ PT\Leftrightarrow10\sqrt{x}+8\sqrt{x}-11\sqrt{x}=21\\ \Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\left(tm\right)\\ h,ĐK:x\ge0\\ PT\Leftrightarrow6\sqrt{3x}+2\sqrt{3x}-3\sqrt{3x}=15\\ \Leftrightarrow\sqrt{3x}=5\Leftrightarrow3x=25\Leftrightarrow x=\dfrac{25}{3}\left(tm\right)\\ i,ĐK:x\ge0\\ PT\Leftrightarrow12\sqrt{x}-21-2\sqrt{x}+10=6\sqrt{x}-12\\ \Leftrightarrow4\sqrt{x}=-1\Leftrightarrow\sqrt{x}=-\dfrac{1}{4}\Leftrightarrow x\in\varnothing\\ j,ĐK:x\ge2\\ PT\Leftrightarrow6\sqrt{x-2}-15\cdot\dfrac{1}{5}\sqrt{x-2}=20+4\sqrt{x-2}\\ \Leftrightarrow\sqrt{x-2}=-20\Leftrightarrow x\in\varnothing\)

\(k,ĐK:x\ge3\\ PT\Leftrightarrow6\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=20\\ \Leftrightarrow4\sqrt{x-3}=20\Leftrightarrow\sqrt{x-3}=5\\ \Leftrightarrow x-3=25\Leftrightarrow x=28\left(tm\right)\\ l,ĐK:x\ge5\\ PT\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

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\(P=\left[\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(2-x\right)\left(x^2+4\right)}\right]\cdot\dfrac{x^2-x-2}{x^2}\\ P=\dfrac{-x\left(x-2\right)^2-4x^2}{2\left(x^2+4\right)\left(2-x\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\\ P=\dfrac{x^3+4x}{2\left(x^2+4\right)\left(x-2\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\\ P=\dfrac{x\left(x^2+4\right)\left(x+1\right)}{2x^2\left(x^2+4\right)}=\dfrac{x+1}{2x}\)

\(\widehat{XAB}\) + \(\widehat{ABZ}\) = 130⁰ + 50⁰ = 180⁰

Vì góc XAB và góc ABZ là hai góc trong cùng phía nên 

Ax // BZ

BZ // Cy ⇔ \(x\) + \(\widehat{yCB}\)  =180⁰

             ⇒ \(x\)              = 180⁰ - 145⁰ = 35⁰

Bài 2: 

Áp dụng tính chất của dãy tỉ số bằng nhau,ta được

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{a+b+c}{2+3+4}=\dfrac{45}{9}=5\)

Do đó: a=10; b=15;c=20