Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(\frac{15}{7}\right)^2+\left(\frac{20}{7}\right)^2=\frac{225}{49}+\frac{400}{49}=\frac{625}{49}\)
\(A=sin^210^o+cos^220^o+sin^280^o+cos^270^o\)
\(A=\left(sin^210^o+sin^280^o\right)+\left(cos^220^o+cos^270^o\right)\)
\(A=0+0\)
\(A=0\)
a: \(A=sin^210^0+sin^280^0+cos^220^0+sin^270^0\)
\(=sin^210^0+cos^210^0+sin^270^0+sin^270^0\)
\(=2\cdot sin^270^0+1\)
b: \(=sin^215^0+sin^275^0+sin^235^0+sin^255^0\)
\(=sin^215^0+cos^215^0+sin^235^0+cos^235^0\)
=1+1
=2
\(A=sin^210^0+sin^280^0+cos^220^0+sin^270^0\)
\(=sin^210^0+cos^210^0+sin^270^0+sin^270^0\)
\(=2sin^270^0+1\)
\(B=sin^215^0+sin^275^0+sin^235^0+sin^255^0\)
\(=sin^215^0+cos^215^0+sin^235^0+cos^235^0\)
=1+1
=2
Đặt \(x^{10}=a\ge0\)
Khi đó:
\(a^{10}-10a+2029\)
\(=\left(a^{10}+1+1+1+1\right)-10a+2025\)
\(\ge5\sqrt[5]{a^{10}}-10a+2025\)
\(=5a^2-10a+2025\)
\(=5\left(a^2-2a+1\right)+2020\)
\(=5\left(a-1\right)^2+2020\ge2020\)
Đẳng thức xảy ra tại x=1 hoặc x=-1
Câu 2:
a: \(=\sqrt{\left(37-35\right)\left(37+35\right)}=\sqrt{72\cdot2}=12\)
b: \(=\sqrt{\left(65-63\right)\left(65+63\right)}=\sqrt{128\cdot2}=16\)
c: \(=\sqrt{\left(221-220\right)\left(221+220\right)}=\sqrt{441}=21\)
d: \(=\sqrt{\left(117-108\right)\left(117+108\right)}=\sqrt{225\cdot9}=3\cdot15=45\)
\(\sqrt{33-20\sqrt{10}}+\dfrac{4}{\sqrt{2}}\)
\(\approx\sqrt{-30,25}+2\sqrt{2}\)
Vì \(\sqrt{...}\ge0\)
Mà \(\sqrt{-30,25}< 0\) (vô lí)
\(\Rightarrow\sqrt{33-20\sqrt{10}}+\dfrac{4}{\sqrt{2}}=\varnothing\)
\(\sqrt{42-10\sqrt{17}}+\sqrt{33-8\sqrt{17}}\\ =\sqrt{\sqrt{25}^2-2.\sqrt{25}.\sqrt{17}+\sqrt{17}^2}+\sqrt{\sqrt{17}^2-2.\sqrt{17}.\sqrt{16}+\sqrt{16}^2}\\ =\sqrt{\left(5-\sqrt{17}\right)^2}+\sqrt{\left(\sqrt{17}-\sqrt{16}\right)^2}\\ =\left|5-\sqrt{17}\right|+\left|\sqrt{17}-\sqrt{16}\right|\\ =5-\sqrt{17}+\sqrt{17}-\sqrt{16}\\ =5-4\\ =1\)
1033 = 100..00 (33 chữ số 0)