Cho `2x^3=3y^3=4z^3`
`CMR:(\root{3}{2x^2+3y^2+4z^2})/(\root{3}{2}+\root{3}{3}+\root{3}{4})=1`
Giúp!
root(2x + 2, 3) + root(2x + 1, 3) = root(2x ^ 2, 3) + root(2x ^ 2 + 1, 3)
\(2x^3=3y^3=4z^3\).CMR:\(\frac{\sqrt[3]{2x^2+3y^2+4z^2}}{\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{4}}\)=1
$x=\root(3)(22\sqrt(2+)25-\root(3)(22\sqrt(2))- 25)$
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 😘
Rút gọn D = root(3, 27) - root(3, - 8) - root(3, 125)
Cho 2x3=3y3=4z3. Chứng minh rằng \(\frac{\sqrt[3]{2x^2+3y^2+4z^2}}{\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{4}}=1\)
Mong các cao nhân lm giúp e vs ạ
Cho \(2x^3=3y^3=4z^3\) . Chứng minh: \(\frac{\sqrt[3]{2x^2+3y^2+4z^2}}{\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{4}}=1\)
Cho \(2x^3=3y^3=4z^3\). Chứng minh rằng: \(\frac{\sqrt[3]{2x^2+3y^2+4z^2}}{\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{4}}=1\)
Cho 2x3 = 3y3 = 4z3
Chứng minh rằng : \(\dfrac{\sqrt[3]{2x^2+3y^2+4z^2}}{\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{4}}=1\)
Bài này hay phết: Theo mik bạn nên thêm ĐK: x;y;z đồng thời khác 0.
Có \(2x^3=3y^3=4z^3\\ \)
Hình như thiếu ĐK\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\\ \)
Cho \(2x^3=3y^3=4z^3\)
CHứng minh rằng :\(\frac{\sqrt[3]{2x^2+3y^2+4z^2}}{\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{4}}=1\)