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Lên cốc cốc tìm cốc cốc toán thay 2 vào mà tìm vậy cũng phải đăng
\(A=\left(\sin\alpha+\cos\alpha+\sin\alpha-\cos\alpha\right)^2-2\left(\sin\alpha+\cos\alpha\right)\left(\sin\alpha-\cos\alpha\right)\)
\(=4\sin^2\alpha-2\sin^2\alpha+2\cos^2\alpha=2\left(\sin^2\alpha+\cos^2\alpha\right)=2\)
\(B=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\)
\(=\left(\sin^2\alpha+\cos^2\alpha\right)^2-1=0\)
\(C=3\left(\sin^4\alpha+\cos^4\alpha\right)-2\sin^2\alpha.\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)=3\left(\sin^4\alpha+\cos^4\alpha\right)-2\sin^2\alpha.\cos^2\alpha\)
\(=3\left(\sin^2\alpha+\cos^2\alpha-\frac{1}{9}\right)^2-\frac{1}{9}=\frac{61}{27}\)
tạm thời chưa nghĩ ra cách dùng \(a^3+b^3\ge a^2b+ab^2=ab\left(a+b\right)\) :'<
Có: \(\sqrt[3]{4\left(a^3+b^3\right)}=\sqrt[3]{2\left(a+b\right)\left(2a^2-2ab+2b^2\right)}\)
\(=\sqrt[3]{2\left(a+b\right)\left[\frac{1}{2}\left(a+b\right)^2+\frac{3}{2}\left(a-b\right)^2\right]}=\sqrt[3]{2\left(a+b\right)\frac{1}{2}\left(a+b\right)^2}=a+b\)
Tương tự cộng lại ta có đpcm
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c\)
ư ư.. ra r :))))))))) cộng thêm Cauchy-Schwarz nữa nhé
Có: \(a^3+b^3\ge a^2b+ab^2\)\(\Leftrightarrow\)\(2\left(a^3+b^3\right)\ge a^3+b^3+a^2b+ab^2=\left(a+b\right)\left(a^2+b^2\right)\)
\(\Rightarrow\)\(\sqrt[3]{4\left(a^3+b^3\right)}\ge\sqrt[3]{2\left(a+b\right)\left(a^2+b^2\right)}\ge\sqrt[3]{2\left(a+b\right).\frac{\left(a+b\right)^2}{2}}=a+b\)
Tương tự cộng lại ra đpcm
Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)có :
\(C\ge\frac{4}{1+\left(a+b\right)^2}\ge\frac{4}{1+1}=2\)
Dấu = khi a=b=1/2
a) \(\dfrac{2sina+3cosa}{3sina-4cosa}=\dfrac{9}{5}\)
b) \(\dfrac{sina.cosa}{sin^2a-sina.cosa+cos^2a}=0\)
\(a.\dfrac{2\sin\alpha+3\cos\alpha}{3\sin\alpha-4\cos\alpha}=\dfrac{2\left(3cos\alpha\right)+3cos\alpha}{3\left(3cos\alpha\right)-4cos\alpha}=\dfrac{9cos\alpha}{5cos\alpha}=\dfrac{9}{5}\)
\(b.\dfrac{sin\alpha cos\alpha}{sin^2\alpha-sin\alpha cos\alpha+cos^2\alpha}=\dfrac{3cos^2\alpha}{9cos^2\alpha-3cos^2\alpha+cos^2\alpha}=\dfrac{3cos^2\alpha}{7cos^2\alpha}=\dfrac{3}{7}\)
uk
\(A=\sin ^{2}\alpha (2-3\cot \alpha +7\cot ^{2}\alpha )\)Sử dụng công thức \(\sin^2\alpha = \frac{1}{1 + \cot^2\alpha}\):\(A=\frac{2-3\cot \alpha +7\cot ^{2}\alpha }{1+\cot ^{2}\alpha }\)Thay \(\cot \alpha = 3\) vào:\(A=\frac{2-3(3)+7(3^{2})}{1+3^{2}}=\frac{2-9+63}{10}=\frac{56}{10}\)
Kết quả: \(A = 5,6\)Bài 2: Cho \(\cot \alpha = 10\). Tính \(B = \frac{5\sin\alpha - \cos\alpha}{4\sin\alpha + 5\cos\alpha}\)Để tính \(B\), ta chia cả tử và mẫu cho \(\sin\alpha\):\(B=\frac{\frac{5\sin \alpha }{\sin \alpha }-\frac{\cos \alpha }{\sin \alpha }}{\frac{4\sin \alpha }{\sin \alpha }+\frac{5\cos \alpha }{\sin \alpha }}\)
\(B=\frac{5-\cot \alpha }{4+5\cot \alpha }\)Thay \(\cot \alpha = 10\) vào:\(B=\frac{5-10}{4+5(10)}=\frac{-5}{4+50}=\frac{-5}{54}\)
Kết quả: \(B = -\frac{5}{54}\)
a, Ta có: $\cot \alpha = 3$
$A = \dfrac{2\sin^2 \alpha - 3\sin \alpha \cos \alpha + 7\cos^2 \alpha}{\sin^2 \alpha + \cos^2 \alpha}$
$ = \dfrac{2 - 3\cot \alpha + 7\cot^2 \alpha}{1 + \cot^2 \alpha}$
$ = \dfrac{2 - 3 \cdot 3 + 7 \cdot 3^2}{1 + 3^2}$
$ = \dfrac{2 - 9 + 63}{10}$
$= \dfrac{56}{10} = 5,6$
a: Ta có: \(1+\cot^2a=\frac{1}{\sin^2a}\)
=>\(\frac{1}{\sin^2a}=1+3^2=10\)
=>\(\sin^2a=\frac{1}{10}\)
=>\(cos^2a=1-\frac{1}{10}=\frac{9}{10}\)
\(A=2\cdot\sin^2a-3\cdot\sin a\cdot cosa+7\cdot cos^2a\)
\(=2\cdot\frac{1}{10}+7\cdot\frac{9}{10}-3\cdot\sqrt{\frac{1}{10}\cdot\frac{9}{10}}\)
\(=\frac{2}{10}+\frac{63}{10}-3\cdot\frac{3}{10}=\frac{65}{10}-\frac{9}{10}=\frac{56}{10}=5,6\)
b: \(B=\frac{5\cdot\sin a-cosa}{4\cdot\sin a+5\cdot cosa}\)
\(=\frac{5\cdot\frac{\sin a}{\sin a}-\frac{cosa}{\sin a}}{4\cdot\frac{\sin a}{\sin a}+5\cdot\frac{cosa}{\sin a}}=\frac{5-\cot a}{4+5\cdot\cot a}=\frac{5-10}{4+5\cdot10}=\frac{-5}{54}\)