Bài 3: Liên hệ giữa phép nhân và phép khai phương

=x-3-4x+25+3x-3căn x

=-3căn x+22

\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}-\sqrt{3}\right)\)

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

\(=\sqrt{3}+2+2\sqrt{2}-2+2-\sqrt{2}-\sqrt{2}+\sqrt{3}\)

\(=2\sqrt{3}+2\)

`[3+2\sqrt{3}]/\sqrt{3}+[2+\sqrt{2}]/[\sqrt{2}+1]-(\sqrt{2}-\sqrt{3})`

`=[\sqrt{3}(\sqrt{3}+2)]/\sqrt{3}+[\sqrt{2}(\sqrt{2}+1)]/[\sqrt{2}+1]-\sqrt{2}+\sqrt{3}`

`=\sqrt{3}+2+\sqrt{2}-\sqrt{2}+\sqrt{3}`

`=2\sqrt{3}+2`

`\sqrt{-x^2-2x+4}=x-3`              `ĐK:x >= 3`

`<=>-x^2-2x+4=x^2-6x+9`

`<=>2x^2-4x-5=0`

Ptr có: `\Delta'=(-2)^2-2.(-5)=14 > 0`

`=>` Ptr có `2` nghiệm pb

`x_1=[-b'+\sqrt{\Delta'}]/a=[2+\sqrt{14}]/2` (ko t/m)

`x_2=[-b'-\sqrt{\Delta'}]/a=[2-\sqrt{14}]/2` (ko t/m)

Vậy ptr vô nghiệm

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`\sqrt{x-3}-2\sqrt{x^2-9}=0`         `ĐK: x >= 3`

`<=>\sqrt{x-3}-2\sqrt{(x-3)(x+3)}=0`

`<=>\sqrt{x-3}(1-2\sqrt{x+3})=0`

`<=>` $\left[\begin{matrix} \sqrt{x-3}=0\\ 1-2\sqrt{x+3}=0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x-3=0\\ \sqrt{x+3}=\dfrac{1}{2}\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=3(t/m)\\ x+3=\dfrac{1}{4}=\dfrac{-11}{4} (ko t/m)\end{matrix}\right.$

Vậy `S={3}`

Bạn thêm dấu tương đương trước mỗi phép biến đổi giúp mình nhé

\(a.\sqrt{\left(x-2\right)^{^2}}=2x-2\)

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

\(x-2=2x-2\)

\(x=0\)

ĐK: x >= 2

\(b.\sqrt{x^{^2}-2x}=\sqrt{2-3x}\)

\(x^{^2}-2x=2-3x\)

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

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

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

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

\(x-1=0hoacx+2=0\)

\(x=1hoacx=-2\)

`a)`\(\sqrt{\sqrt{x^4+4}-x^2}.\sqrt{\sqrt{x^4+4}+x^2}\)

\(=\sqrt{\left(\sqrt{x^4+4}-x^2\right).\left(\sqrt{x^4+4}+x^2\right)}\)

\(=\sqrt{\left(\sqrt{x^4+4}\right)^2-\left(x^2\right)^2}\)

\(=\sqrt{x^4+4-x^4}\)

\(=\sqrt{4}=2\)

`b)`\(\dfrac{x\sqrt{y}+y\sqrt{x}}{x+2\sqrt{xy}+y}=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)^2}=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

`c)`\(\dfrac{3\sqrt{a}-2a-1}{4a-4\sqrt{a}+1}=\dfrac{\left(\sqrt{a}-1\right)\left(2\sqrt{a}-1\right)}{\left(2\sqrt{a}-1\right)^2}=\dfrac{\sqrt{a}-1}{2\sqrt{a}-1}\)

`d)`\(\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)

\(=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)

\(=\sqrt{a}+2-\left(\sqrt{a}+2\right)\)

\(=0\)

a: \(=\left|x\right|\cdot\left|x+1\right|=x\left(x+1\right)=x^2+x\)

b: \(=9\cdot1=9\)

a.

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

\(M=\dfrac{2x-\sqrt{x}+2}{x-4}+\dfrac{\sqrt{x}-2}{x-4}-\dfrac{x+2\sqrt{x}}{x-4}\)

\(M=\dfrac{2x-\sqrt{x}+2+\sqrt{x}-2-x-2\sqrt{x}}{x-4}\)

\(M=\dfrac{x-2\sqrt{x}}{x-4}\)

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

b. hình như thiếu dữ kiện đk ạ?

c. \(M=\dfrac{\sqrt{x}}{\sqrt{x}+2}< \dfrac{1}{2}\)

đkxđ: \(x\ge0\)

\(\Leftrightarrow2\sqrt{x}< \sqrt{x}+2\)

\(\Leftrightarrow2\sqrt{x}-\sqrt{x}-2< 0\)

\(\Leftrightarrow\sqrt{x}-2< 0\)

\(\Leftrightarrow\sqrt{x}< 2\)

\(\Leftrightarrow x< 4\)

Kết hợp vs đk, ta được giá trị của x là: \(\left\{0\le x< 4\right\}\)

\(Q=2+\sqrt{3}+\sqrt{2}-\sqrt{2}-\sqrt{3}=2\)