\(\sqrt[3]{26-15\sqrt{3}}=a+b\sqrt{3}.\text{Tính}\) a2-b2
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a)\(A=^3\sqrt{20+14\sqrt{2}}+^3\sqrt{20-14\sqrt{2}}\)
=> \(A^3=\left(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\right)^3\)
\(=20+14\sqrt{2}+20-14\sqrt{2}\)
\(+3\left(\text{}^3\sqrt{20+14\sqrt{2}}+^3\sqrt{20-14\sqrt{2}}\right)\left(^3\sqrt{20+14\sqrt{2}}.^3\sqrt{20-14\sqrt{2}}\right)\)
\(=40+3A.^3\sqrt{\left(20+14\sqrt{2}\right)\left(20+14\sqrt{2}\right)}\)
\(\Rightarrow A^3=40+3.A.2\)
=> \(A^3-6A-40=0\)
<=> \(A^3-16A+10A-40=0\)
<=> \(A\left(A-4\right)\left(A+4\right)+10\left(A-4\right)=0\)
<=> \(\left(A-4\right)\left(A^2+4A+10\right)=0\)
<=> A = 4 ( vì \(A^2+4A+10=\left(A+2\right)^2+6>0\))
Vậy A = 4.
b/ \(B=^3\sqrt{26+15\sqrt{3}}-^3\sqrt{26-15\sqrt{3}}\)
=> \(B^3=\left(^3\sqrt{26+15\sqrt{3}}-^3\sqrt{26-15\sqrt{3}}\right)^3\)
\(=26+15\sqrt{3}-26+15\sqrt{3}\)
\(-3\left(^3\sqrt{26+15\sqrt{3}}-^3\sqrt{26-15\sqrt{3}}\right).^3\sqrt{26+15\sqrt{3}}.^3\sqrt{26-15\sqrt{3}}\)
\(=30\sqrt{3}-3B.1\)
=> \(B^3+3B-30\sqrt{3}=0\)
<=> \(B^3-12B+15B-30\sqrt{3}=0\)
<=> \(B\left(B-2\sqrt{3}\right)\left(B+2\sqrt{3}\right)+15\left(B-2\sqrt{3}\right)=0\)
<=> \(\left(B-2\sqrt{3}\right)\left(B^2+2\sqrt{3}B+15\right)=0\)
<=> \(B-2\sqrt{3}=0\)( vì \(B^2+2\sqrt{3}B+15=\left(B+\sqrt{3}\right)^2+12>0\))
<=> \(B=2\sqrt{3}\)
a, c.Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath
Sửa đề
\(A=\left(2-\sqrt{3}\right)\sqrt[3]{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{26-15\sqrt{3}}\)
\(=\left(2-\sqrt{3}\right)\sqrt[3]{8+12\sqrt{3}+18+3\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{8-12\sqrt{3}+18-3\sqrt{3}}\)
\(=\left(2-\sqrt{3}\right)\sqrt[3]{\left(2+\sqrt{3}\right)^3}-\left(2+\sqrt{3}\right)\sqrt[3]{\left(2-\sqrt{3}\right)^3}\)
\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)-\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=0\)
1.
Sửa đề: \(S=\dfrac{1}{6}\left(ch_a+bh_c+ah_b\right)\)
\(a.h_a=b.h_b=c.h_c=2S\Rightarrow\left\{{}\begin{matrix}h_a=\dfrac{2S}{a}\\h_b=\dfrac{2S}{b}\\h_c=\dfrac{2S}{c}\end{matrix}\right.\)
\(\Rightarrow6S=\dfrac{2Sc}{a}+\dfrac{2Sb}{c}+\dfrac{2Sa}{b}\)
\(\Leftrightarrow\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=3\)
Mặt khác theo AM-GM: \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge3\sqrt[3]{\dfrac{abc}{abc}}=3\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c\)
\(\Leftrightarrow\) Tam giác đã cho đều
2.
Bạn coi lại đề, biểu thức câu này rất kì quặc (2 vế không đồng bậc)
Ở vế trái là \(2\left(a^2+b^2+c^2\right)\) hay \(2\left(a^3+b^3+c^3\right)\) nhỉ?
3.
Theo câu a, ta có:
\(VT=\dfrac{2S}{a}+\dfrac{2S}{b}+\dfrac{2S}{c}\ge\dfrac{18S}{a+b+c}=\dfrac{18.pr}{a+b+c}=9r\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c\)
Hay tam giác đã cho đều
a = \(\sqrt[3]{26+15\sqrt{3}}\)+\(\sqrt[3]{26-15\sqrt{3}}\)=\(\sqrt[3]{8+2.3.3+3.4.\sqrt{3}+3\sqrt{3}}+\sqrt[3]{8-3.4.\sqrt{3}+2.3.3-3\sqrt{3}}\)
=\(\sqrt[3]{2+\sqrt{3}}^3\)+\(\sqrt[3]{2-\sqrt{3}}^3\)
=2+\(\sqrt{3}\)+2-\(\sqrt{3}\)
=4=\(2^2\)
Ta có \(a=\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}=\sqrt[3]{8+12\sqrt{3}+18+3\sqrt{3}}+\sqrt[3]{8-12\sqrt{3}+18-3\sqrt{3}}=\sqrt[3]{2^3+3.2^2.\sqrt{3}+3.2.\left(\sqrt{3}\right)^2+\left(\sqrt{3}\right)^3}+\sqrt[3]{2^3-3.2^2.\sqrt{3}+3.2.\left(\sqrt{3}\right)^2-\left(\sqrt{3}\right)^3}=\sqrt[3]{\left(2+\sqrt{3}\right)^3}+\sqrt[3]{\left(2-\sqrt{3}\right)^3}=2+\sqrt{3}+2-\sqrt{3}=4=2^2\)
Vậy a là bình phương của một số nguyên
Nháp:
\(9\pm\sqrt{80}=9\pm4\sqrt{5}=\dfrac{72\pm32\sqrt{5}}{8}=\left(\dfrac{3\pm\sqrt{5}}{2}\right)^3\)
\(\Rightarrow \sqrt[3]{9+\sqrt{80}}=\dfrac{3+\sqrt{5}}{2}\); \(\Rightarrow \sqrt[3]{9-\sqrt{80}}=\dfrac{3-\sqrt{5}}{2}\)
\(S=\sqrt[3]{26+15\sqrt{3}}\left(2-\sqrt{3}\right)+\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\\ S=\sqrt[3]{\left(2+\sqrt{3}\right)^3}+\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\\ S=2+\sqrt{3}+\dfrac{3+\sqrt{5}}{2}+\dfrac{3-\sqrt{5}}{2}\\ S=2+\sqrt{3}+3\\ S=5+\sqrt{3}\)
\(\sqrt[3]{26-15\sqrt[]{3}}=\sqrt[3]{8-3.4.\sqrt[]{3}+3.2.3-3\sqrt[]{3}}\)
\(=\sqrt[3]{2^3-3.2^2.\sqrt[]{3}+3.2.\left(\sqrt[]{3}\right)^2-\left(\sqrt[]{3}\right)^3}\)
\(=\sqrt[3]{\left(2-\sqrt[]{3}\right)^3}=2-\sqrt[]{3}\)
\(\Rightarrow\left\{{}\begin{matrix}a=2\\b=-1\end{matrix}\right.\) \(\Rightarrow a^2-b^2=3\)