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Đặt y = \(x+1=\sqrt[3]{8+2\sqrt{14}}+\sqrt[3]{8-2\sqrt{14}}\)
=> \(y^3=8+2\sqrt{14}+8-2\sqrt{14}+3\sqrt[3]{\left(8+2\sqrt{14}\right)\left(8-2\sqrt{14}\right)}.y\)
<=> \(y^3=16+6y\)
=> \(\left(x+1\right)^3=16+6\left(x+1\right)\)
=> \(x^3+3x^2+3x+1=6x+32\)
<=> \(x^3+3x^2-3x-5=26\)
Ta có:
\(x^6+3x^5-3x^4-2x^3+9x^2-9x+2018\)
= \(x^6+3x^5-3x^4-5x^3+3x^3+9x^2-9x-15+2033\)
= \(\left(x^3+3x^2-3x-5\right)\left(x^3+3\right)+2033\)
= \(26x^3+2111\)
\(=26\left(\sqrt[8]{8+2\sqrt{14}}+\sqrt[8]{8-2\sqrt{14}}-1\right)^3+2033\)
ai nay dung kinh nghiem la chinh
cau a)
ta thay \(10+6\sqrt{3}=\left(1+\sqrt{3}\right)^3\)
\(6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)
khi do \(x=\frac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{5}}\)
\(x=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{1+\sqrt{5}-\sqrt{5}}\)
\(x=\frac{3-1}{1}=2\)
suy ra
x^3-4x+1=1
A=1^2018
A=1
b)
ta thay
\(7+5\sqrt{2}=\left(1+\sqrt{2}\right)^3\)
khi do
\(x=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\frac{1}{\sqrt[3]{\left(1+\sqrt{2}\right)^3}}\)
\(x=1+\sqrt{2}-\frac{1}{1+\sqrt{2}}=\frac{\left(1+\sqrt{2}\right)^2-1}{1+\sqrt{2}}=\frac{2+2\sqrt{2}}{1+\sqrt{2}}\)
x=2
thay vao
x^3+3x-14=0
B=0^2018
B=0
\(x=\dfrac{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}{\sqrt{5}+3-\sqrt{5}}=\dfrac{3}{3}=1\)
\(A=\left(3\cdot1+8\cdot1+2\right)^{2018}=13^{2018}\)
Thêm câu này hộ tớ nx nhé !
e) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right).\left(\sqrt{2}-3\sqrt{0.4}\right)\)
\(a,\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{12}-\sqrt{6}}{2\left(\sqrt{2}-1\right)}-\frac{6\sqrt{6}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}}{2}-\frac{4\sqrt{6}}{2}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\frac{\sqrt{6}-4\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)
\(=\frac{-3\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)
\(=-\frac{3}{2}\)
Đặt \(\sqrt[3]{2\sqrt{14}-8}.\sqrt[3]{2\sqrt{14}-8}=\sqrt[3]{\left(2\sqrt{14}-8\right)\left(2\sqrt{14}+8\right)}=\sqrt[3]{56-64}\)
\(\sqrt[3]{-8}=-2\)
Lời giải:
\(\sqrt[3]{2\sqrt{14}-8}.\sqrt[3]{2\sqrt{14}+8}=\sqrt[3]{(2\sqrt{14}-8)(2\sqrt{14}+8)}\)
\(=\sqrt[3]{(2\sqrt{14})^2-8^2}=\sqrt[3]{-8}=-2\)
Bài này không cần giải phương trình dưới đâu nhé!
Liên hợp ta có:
\(\sqrt{x^2-3x+14}-\sqrt{x^2-3x+8}=2\)
<=> \(\frac{\left(x^2-3x+14\right)-\left(x^2-3x+8\right)}{\sqrt{x^2-3x+14}+\sqrt{x^2-3x+8}}=2\)
<=> \(\frac{6}{\sqrt{x^2-3x+14}+\sqrt{x^2-3x+8}}=2\)
<=> \(\sqrt{x^2-3x+14}+\sqrt{x^2-3x+8}=\frac{6}{2}=3\)
Vậy B = 3.