\(lim\left(\sqrt[3]{n^3+4}-\sqrt[3]{n^3-1}\right)\)

\(=lim\left(\sqrt[3]{1+\dfrac{4}{n^3}}-\sqrt[3]{1-\dfrac{1}{n^3}}\right)=\sqrt[3]{1}-\sqrt[3]{1}=0\)

ĐKXĐ: \(0\le x\le9\)

Bình phương 2 vế ta được:

\(x+9-x+2\sqrt{x\left(9-x\right)}=-x^2+9x+9\)

\(\Leftrightarrow-x^2+9x-2\sqrt{-x^2+9x}=0\)

\(\Leftrightarrow\sqrt{-x^2+9x}\left(\sqrt{-x^2+9x}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{-x^2+9x}=0\\\sqrt{-x^2+9x}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-x^2+9x=0\\-x^2+9x-4=0\end{matrix}\right.\)

Tới đây em tự hoàn thành nốt

3/7

=6/14=3/7

\(\dfrac{1}{2}.\dfrac{6}{7}=\dfrac{6}{14}\)

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

\(\Leftrightarrow\left(3x-2\right)^2-\left(x-2\right)\left(2x-3\right)=0\)

\(\Leftrightarrow9x^2-12x+4-\left(2x^2-7x+6\right)=0\)

\(\Leftrightarrow9x^2-12x+4-2x^2+7x-6=0\)

\(\Leftrightarrow7x^2-5x-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{7}\end{matrix}\right.\)

Vậy \(S=\left\{1;-\dfrac{2}{7}\right\}\)

`(3-2x)^2=(x-2)(2x-3)`

`<=>(2x-3)^2 -(x-2)(2x-3)=0`

`<=> (2x-3)(2x-3-x+2)=0`

`<=> (2x-3)(x-1)=0`

\(< =>\left[{}\begin{matrix}2x-3=0\\x-1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=1\end{matrix}\right.\)

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4/1 x 13/15

= 52/15

52/15

mình có câu trl rồi nhé:

4 x 13/15 = 4 x 13/15 = 52/15 nhé!