Bài 4: Liên hệ giữa phép chia và phép khai phương

=>a+b-2căn ab<a-b

=>-2căn ab<0

=>căn ab>0(luôn đúng)

`a.`\(x=\sqrt[3]{-125}+\sqrt[3]{216}\)

\(\Leftrightarrow x=-5+6\)

`<=>x=1` thế vào A, ta được:

\(A=\dfrac{\sqrt{1}-2}{\sqrt{1}+2}=-\dfrac{1}{3}\)

`b.`\(A>\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+2}>\dfrac{1}{3}\)

\(\Leftrightarrow3\sqrt{x}-6>\sqrt{x}+2\)

\(\Leftrightarrow2\sqrt{x}>8\)

\(\Leftrightarrow\sqrt{x}>4\)

\(\Leftrightarrow x>16\)

`c.` \(B=\dfrac{\sqrt{x}+3}{\sqrt{x}+2}+\dfrac{10}{2-\sqrt{x}}+\dfrac{4}{x-4}\)

\(B=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)-10\left(\sqrt{x}+2\right)+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(B=\dfrac{x+\sqrt{x}-6-10\sqrt{x}-20+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(B=\dfrac{x-9\sqrt{x}-22}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(B=\dfrac{\left(\sqrt{x}-11\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(B=\dfrac{\sqrt{x}-11}{\sqrt{x}-2}\) ( sửa đề )

\(B=\dfrac{\sqrt{x}-2-9}{\sqrt{x}-2}=1-\dfrac{9}{\sqrt{x}-2}\)

Để B nguyên thì \(\dfrac{9}{\sqrt{x}-2}\in Z\Rightarrow\sqrt{x}-2\inƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)

Mà \(\sqrt{x}\ge0\Rightarrow\sqrt{x}-2\ge-2\)

`@`\(\sqrt{x}-2=-1\Rightarrow\sqrt{x}=1\Rightarrow x=1\left(tm\right)\)

`@`\(\sqrt{x}-2=1\Rightarrow\sqrt{x}=3\Rightarrow x=9\left(tm\right)\)

`@`\(\sqrt{x}-2=3\Rightarrow\sqrt{x}=5\Rightarrow x=25\left(tm\right)\)

`@`\(\sqrt{x}-2=9\Rightarrow\sqrt{x}=11\Rightarrow x=121\left(tm\right)\)

Vậy \(x\in\left\{1;9;25;121\right\}\) thì B nguyên

`d.`\(P=A.B\)

\(P=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}.\dfrac{\sqrt{x}-11}{\sqrt{x}-2}\)

\(P=\dfrac{\sqrt{x}-11}{\sqrt{x}+2}\)

\(P=\dfrac{\sqrt{x}+2-13}{\sqrt{x}+2}=1-\dfrac{13}{\sqrt{x}+2}\)

Để P nguyên thì \(\dfrac{13}{\sqrt{x}+2}\in Z\Rightarrow\sqrt{x}+2\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)

Mà \(\sqrt{x}+2\ge2\)

`@`\(\sqrt{x}+2=13\Rightarrow\sqrt{x}=11\Rightarrow x=121\left(tm\right)\)

Vậy \(x=121\) thì P nguyên

a) \(\sqrt{25x^2}=100\)

\(25x^2=10000\)

\(x^2=400\)

\(x=\pm20\)

b) \(\left(\sqrt{3}-\sqrt{2}\right)x=\sqrt{27}-\sqrt{18}\)

\(\left(\sqrt{3}-\sqrt{2}\right)x=\sqrt{9.3}-\sqrt{9.2}\)

\(\left(\sqrt{3}-\sqrt{2}\right)x=3\left(\sqrt{3}-\sqrt{2}\right)\)

\(x=3\)

a. \(\sqrt{25x^2}=100\)

\(\Leftrightarrow\sqrt{\left(5x\right)^2}=100\)

\(\Leftrightarrow\left|5x\right|=100\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=100\\5x=-100\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=20\\x=-20\end{matrix}\right.\)

b. \(\left(\sqrt{3}-\sqrt{2}\right)x=\sqrt{27}-\sqrt{18}\)

\(\Leftrightarrow\left(\sqrt{3}-\sqrt{2}\right)x=\sqrt{3^2.3}-\sqrt{3^2.2}\)

\(\Leftrightarrow\left(\sqrt{3}-\sqrt{2}\right)x=3\sqrt{3}-3\sqrt{2}\)

\(\Leftrightarrow\left(\sqrt{3}-\sqrt{2}\right)x=3\left(\sqrt{3}-\sqrt{2}\right)\)

\(\Leftrightarrow x=3\)

Vì \(a\ge0\) và \(b>0\) nên \(\dfrac{\sqrt{a}}{\sqrt{b}}\) xác định và không âm.

Ta có: \(\left(\dfrac{\sqrt{a}}{\sqrt{b}}\right)^2=\dfrac{\left(\sqrt{a}\right)^2}{\left(\sqrt{b}\right)^2}=\dfrac{a}{b}\)

Vậy \(\dfrac{\sqrt{a}}{\sqrt{b}}\) là căn bậc hai số học của \(\dfrac{a}{b}\), tức là: \(\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\)

\(\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\)

\(\Leftrightarrow\sqrt{b}\cdot\sqrt{\dfrac{a}{b}}=\sqrt{a}\)

\(\Leftrightarrow\sqrt{b\cdot\dfrac{a}{b}}=\sqrt{a}\)

\(\Leftrightarrow\sqrt{a}=\sqrt{a}\left(lđ\right)\)

đk a > = 0 

\(\sqrt{25a}+\sqrt{49a}-\sqrt{64a}=5a+7a-8a=4a\)

sai thì thôi nhó :q

`a)Đ k : x ne +- 3`

\(A=\sqrt{\dfrac{1}{x^2-6x+9}}-\dfrac{6}{x^2-9}\)

\(A=\dfrac{\sqrt{1}}{\sqrt{\left(x-3\right)^2}}-\dfrac{6}{\left(x+3\right)\left(x-3\right)}=\dfrac{1}{\left|x-3\right|}-\dfrac{6}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{\left(x+3\right)\left(x-3\right)-\left(x-3\right)6}{\left(x-3\right)^2\left(x+3\right)}=\dfrac{\left(x+3-6\right)\left(x-3\right)}{\left(x-3\right)^2\left(x+3\right)}\)

\(A=\dfrac{\left(x-3\right)^2}{\left(x-3\right)^2\left(x+3\right)}=\dfrac{1}{x+3}\)

\(\sqrt{\dfrac{27\left(x-1\right)^2}{12}}+\dfrac{3}{2}-\left(x-2\right)\sqrt{\dfrac{50x^2}{8\left(x-2\right)^2}}\left(1< x< 2\right)\\ =\dfrac{\sqrt{27}.\sqrt{\left(x-1\right)^2}}{\sqrt{12}}+\dfrac{3}{2}-\left(x-2\right).\dfrac{\sqrt{50}.\sqrt{x^2}}{\sqrt{8}.\sqrt{\left(x-2\right)^2}}\\ =\dfrac{\sqrt{9}.\sqrt{3}.\left|x-1\right|}{\sqrt{4}.\sqrt{3}}+\dfrac{3}{2}-\left(x-2\right).\dfrac{\sqrt{25}.\sqrt{2}.\left|x\right|}{\sqrt{4}.\sqrt{2}.\left|x-2\right|}\\ =\dfrac{3.\left(x-1\right)}{2}+\dfrac{3}{2}-\left(x-2\right).\dfrac{5x}{-2\left(x-2\right)}\\ =\dfrac{3x-3}{2}+\dfrac{3}{2}+\dfrac{5x}{2}\\ =\dfrac{3x-3+3+5x}{2}=\dfrac{8x}{2}=4x\)

a: \(=\dfrac{x^3}{y^2}:\dfrac{x}{y^2}=\dfrac{x^3}{x}=x^2\)

b: \(=\dfrac{3}{2}\left(x-1\right)+\dfrac{3}{2}-\left(x-2\right)\cdot\dfrac{5}{2}\cdot\dfrac{x}{2-x}\)

\(=\dfrac{3}{2}x+\dfrac{5}{2}x=4x\)

\(=\sqrt{\dfrac{1}{125}\cdot\dfrac{32}{35}\cdot\dfrac{225}{56}}=\sqrt{\dfrac{9}{4}\cdot\dfrac{4}{7\cdot35}}=\sqrt{\dfrac{9}{7\cdot35}}\)