Đặt \(A=\sqrt[3]{22\sqrt{2}+25}-\sqrt[3]{22\sqrt{2}-25}\)

\(\Rightarrow A^3=50-3\sqrt[3]{\left(22\sqrt{2}+25\right)\left(22\sqrt{2}-25\right)}\left(\sqrt[3]{22\sqrt{2}+25}-\sqrt[3]{22\sqrt{2}-25}\right)\)

\(\Rightarrow A^3=50-3\sqrt[3]{\left(22\sqrt{2}+25\right)\left(22\sqrt{2}-25\right)}\cdot A\)

\(\Rightarrow A^3=50-3A\sqrt[3]{343}=50-21A\)

\(\Rightarrow A^3+21A-50=0\Leftrightarrow A^3-4A+25A-50=0\)

\(\Leftrightarrow\left(A-2\right)\left(A^2+2A+25\right)=0\)

\(\Leftrightarrow A=2\left(A^2+2A+25>0,\forall A\right)\)

\(\Rightarrow\sqrt[3]{22\sqrt{2}+25}-\sqrt[3]{22\sqrt{2}-25}=2\)

Tick nha bạn 😘

a, x=2=35/2

x=log(35/2)

x=log(35)-log(20)

x=log(35)-1

b) \(\left(\sqrt{9}+\sqrt{4}\right).\sqrt{x}=10\)

\(\left(3+2\right).\sqrt{x}=10\)

\(5.\sqrt{x}=10\)

\(\sqrt{x}=2\)

\(x=\sqrt{2}\)

a) Ta có: \(\sqrt{x-2\sqrt{x-1}}-\sqrt{x-1}=1\)

\(\Leftrightarrow\left|\sqrt{x-1}-1\right|=\sqrt{x-1}+1\)

\(\Leftrightarrow\sqrt{x-1}=\sqrt{x-1}+1+1\)(Vô lý)

Vậy: \(S=\varnothing\)

b) Ta có: \(\sqrt{x^4+2x^2+1}=\sqrt{x^2+10x+25}-10x+22\)

\(\Leftrightarrow x^2+1=\left|x+5\right|-10x+22\)

\(\Leftrightarrow\left|x+5\right|=x^2+1+10x-22=x^2+10x-21\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=x^2+10x-21\left(x\ge-5\right)\\-x-5=x^2+10x-21\left(x< -5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+10x-21-x-5=0\\x^2+10x-21+x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+9x-26=0\\x^2+11x-16=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-9+\sqrt{185}}{2}\\x=\dfrac{-11-\sqrt{185}}{2}\end{matrix}\right.\)

\(A=\dfrac{1}{\sqrt{25}+\sqrt{24}}+\dfrac{1}{\sqrt{24}+\sqrt{23}}+....+\dfrac{1}{\sqrt{2}+1}\)

\(A=\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+......+\sqrt{2}-1=\sqrt{25}-1=4\)

a, \(\sqrt{4x^2+20x+25}\) + \(\sqrt{x^2-8x+16}\) = \(\sqrt{x^2+18x+81}\)

⇔ 4x² + 20x + 25 + \(2\sqrt{\left(4x^2+20x+25\right)\left(x^2-8x+16\right)}\) = x² + 18x + 81

⇔ 4x² + 20x + 25 - x² - 18x - 81 + \(2\sqrt{\left(2x+5\right)^2.\left(x-4\right)^2}\) = 0

⇔ 3x² + 2x - 56 + 2.(2x + 5) . (x - 4) = 0

⇔ 3x² + 2x - 56 + (4x + 10) . (x - 4) = 0

⇔ 3x² + 2x - 56 + 4x² - 16x + 10x - 40 = 0

⇔ 7x² - 4x - 96 = 0

x₁ = 4 ( nhận )

x₂ = \(\frac{-24}{7}\) ( nhận )

Vậy: S = {4; \(\frac{-24}{7}\)}