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k nguyên dương => \(k\ge1\)\(\Leftrightarrow\)\(a^k\ge a\)\(\Leftrightarrow\)\(\frac{a^k}{b+c}\ge\frac{a}{b+c}\)
Tương tự với 2 phân thức còn lại, cộng 3 bđt ta thu đc bđt Nesbit 3 ẩn => đpcm
Chứng minh:
\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
Ta có: \(a+b\in Z\)
và \(a^2+b^2=\left(a+b\right)^2-2ab\in Z\Rightarrow2ab\in Z\)
\(a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2\in Z\Rightarrow2a^2b^2\in Z\)
Đặt 2ab=k , k thuộc Z => \(4a^2b^2=k^2\Rightarrow2a^2b^2=\frac{k^2}{2}\in Z\Rightarrow\frac{k}{2}\in Z\)=> ab thuộc Z
=> \(a^3+b^3\in Z\)
Em chưa hiểu chỗ này: \(\frac{k^2}{2}\inℤ\Rightarrow\frac{k}{2}\inℤ\)
câu này khá khó mình ko biết làm có đúng ko nữa
để \(\left(d1\right)\perp\left(d2\right)\)
\(\Leftrightarrow\)\(\left(k-3\right).\left(2k+1\right)=-1\)
\(\Leftrightarrow2k^2+k-6k-3+1=0\)
\(\Leftrightarrow2k^2-5k-2=0\)
\(\Leftrightarrow k^2-\frac{5}{2}k-1=0\)
\(\Leftrightarrow\)\(k^2-2.k.\frac{5}{4}+\frac{25}{16}-\frac{25}{16}-1=0\)
\(\Leftrightarrow\left(k-\frac{5}{4}\right)^2-\frac{41}{16}=0\)
\(\Leftrightarrow\left(k-\frac{5}{4}-\frac{\sqrt{41}}{4}\right)\left(k-\frac{5}{4}+\frac{\sqrt{41}}{4}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}k-\frac{5}{4}-\frac{\sqrt{41}}{4}=0\\k-\frac{5}{4}+\frac{\sqrt{41}}{4}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}k=\frac{5+\sqrt{41}}{4}\\k=\frac{5-\sqrt{41}}{4}\end{cases}}\) ( Thỏa mãn \(k\ne3;k\ne\frac{-1}{2}\))
vậy \(k=\frac{5-\sqrt{41}}{4}\) ; \(k=\frac{5+\sqrt{41}}{4}\)
a: \(=\sqrt{11}-1\)
b: \(=3\sqrt{3}+1\)
c: \(=\sqrt{3}+\sqrt{2}\)
d: \(=\sqrt{3}-\sqrt{2}\)
e: \(=\sqrt{3}-1\)
g: \(=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)
a) Ta có: \(\sqrt{11-2\sqrt{10}}\)
\(=\sqrt{10-2\cdot\sqrt{10}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{10}-1\right)^2}\)
\(=\left|\sqrt{10}-1\right|=\sqrt{10}-1\)
b) Ta có: \(\sqrt{9-2\sqrt{14}}\)
\(=\sqrt{7-2\cdot\sqrt{7}\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{7}-\sqrt{2}\right|\)
\(=\sqrt{7}-\sqrt{2}\)
c) Ta có: \(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3+2\cdot\sqrt{3}\cdot1+1}+\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|\)
\(=\sqrt{3}+1+\sqrt{3}-1\)
\(=2\sqrt{3}\)
d) Ta có: \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5+2\cdot\sqrt{5}\cdot2+4}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\left|\sqrt{5}-2\right|-\left|\sqrt{5}+2\right|\)
\(=\sqrt{5}-2-\left(\sqrt{5}+2\right)\)
\(=\sqrt{5}-2-\sqrt{5}-2\)
\(=-4\)
e) Ta có: \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
\(=\frac{\sqrt{2}\cdot\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{2}\cdot\left(\sqrt{4-\sqrt{7}}\right)-\sqrt{2}\cdot\left(\sqrt{4+\sqrt{7}}\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)
\(=\frac{\sqrt{7-2\cdot\sqrt{7}\cdot1+1}-\sqrt{7+2\cdot\sqrt{7}\cdot1+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|}{\sqrt{2}}\)
\(=\frac{\sqrt{7}-1-\left(\sqrt{7}+1\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}\)
\(=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)
g) Ta có: \(\sqrt{3}+\sqrt{11+6\sqrt{2}}+\sqrt{5+2\sqrt{6}}\)
\(=\sqrt{3}+\sqrt{9+2\cdot3\cdot\sqrt{2}+2}+\sqrt{2+2\cdot\sqrt{2}\cdot\sqrt{3}+3}\)
\(=\sqrt{3}+\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}\)
\(=\sqrt{3}+\left|3+\sqrt{2}\right|+\left|\sqrt{2}+\sqrt{3}\right|\)
\(=\sqrt{3}+3+\sqrt{2}+\sqrt{2}+\sqrt{3}\)
\(=3+2\sqrt{3}+2\sqrt{2}\)
h) Ta có: \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\cdot\sqrt{3+2\cdot\sqrt{3}\cdot2+4}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\cdot\sqrt{\left(\sqrt{3}+2\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{48-10\cdot\left(\sqrt{3}+2\right)}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{48-10\sqrt{3}-20}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{25-2\cdot5\cdot\sqrt{3}+3}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\left(5-\sqrt{3}\right)}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{25}=5\)
k) Ta có: \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
\(=\sqrt{49-2\cdot7\cdot\sqrt{45}+45}-\sqrt{49+2\cdot7\cdot\sqrt{45}+45}\)
\(=\sqrt{\left(7-\sqrt{45}\right)^2}-\sqrt{\left(7+\sqrt{45}\right)^2}\)
\(=\left|7-\sqrt{45}\right|-\left|7+\sqrt{45}\right|\)
\(=7-\sqrt{45}-\left(7+\sqrt{45}\right)\)
\(=7-\sqrt{45}-7-\sqrt{45}\)
\(=-2\sqrt{45}=-6\sqrt{5}\)
i) Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Leftrightarrow A^2=\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)^2\)
\(=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\cdot\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(=8+2\cdot\sqrt{16-\left(10+2\sqrt{5}\right)}\)
\(=8+2\cdot\sqrt{6-2\sqrt{5}}\)
\(=8+2\cdot\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=8+2\cdot\left(\sqrt{5}-1\right)\)
\(=8+2\sqrt{5}-2\)
\(=6+2\sqrt{5}\)
\(=\left(\sqrt{5}+1\right)^2\)
\(\Leftrightarrow A=\sqrt{5}+1\)
Theo bài ra ta có: k + 4 ⋮ 11
⇒ k - (-4) ⋮ 11
⇒ k \(\equiv\) - 4 (mod 11)
⇒ k2 \(\equiv\) (-4)2 (mod 11)
3k \(\equiv\) 3.(-4)(mod 11)
5 \(\equiv\) 5 (mod 11)
Cộng vế với vế ta có: k2 + 3k + 5 \(\equiv\) 16 - 12 + 5 (mod 11)
⇒ k2 + 3k + 5 \(\equiv\) 9 (mod 11)
Giả sử điều cần chứng minh là đúng thì
k2 + 3k + 5 ⋮ 11 ⇔ 9 ⋮ 11 ( vô lý)
Nên điều giả sử là sai
Vậy với k \(\in\) Z chứng minh rằng k2 + 3k + 5 ⋮ 11 ⇔ k + 4 ⋮ 11 là điều không thể xảy ra.
Bạn xem lại đề có đúng không theo tôi k-4⋮11