Câu `1`
\(a,A=3\sqrt{7}-\dfrac{5}{2}\sqrt{28}-\dfrac{2}{3}\sqrt{63}-\dfrac{14}{\sqrt{7}}\\ =3\sqrt{7}-\dfrac{5}{2}\sqrt{4\cdot7}-\dfrac{2}{3}\sqrt{9\cdot7}-2\sqrt{7}\\ =3\sqrt{7}-5\sqrt{2}-2\sqrt{7}-2\sqrt{7}\\ =\left(3-5-2-2\right)\sqrt{7}\\ =-6\sqrt{7}\)
\(b,B=\sqrt{6-2\sqrt{5}}-\sqrt{\left(2-\sqrt{5}\right)^2}\\ =\sqrt{5-2\sqrt{5}+1}-\left|2-\sqrt{5}\right|\\ =\sqrt{\left(\sqrt{5}\right)^2-2\cdot\sqrt{5}\cdot2+1^2}-\left(\sqrt{5}-2\right)\\ =\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{5}+2\\ =\sqrt{5}+1-\sqrt{5}+2\\ =3\)
3:
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >9\end{matrix}\right.\)
a: Khi x=16 thì \(A=\dfrac{4}{4-3}=\dfrac{4}{1}=4\)
b: \(B=\dfrac{7}{\sqrt{x}+1}+\dfrac{12}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{7\sqrt{x}-21+12}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{7\sqrt{x}-9}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(M=A-B=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{7\sqrt{x}-9}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\left(x+\sqrt{x}-7\sqrt{x}+9\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
c: Để M là số nguyên thì \(\sqrt{x}-3⋮\sqrt{x}+1\)
=>\(\sqrt{x}+1-4⋮\sqrt{x}+1\)
=>\(\sqrt{x}+1\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(\sqrt{x}+1\in\left\{1;2;4\right\}\)
=>\(\sqrt{x}\in\left\{0;1;3\right\}\)
=>\(x\in\left\{0;1;9\right\}\)
Kết hợp ĐKXĐ, ta được: x=0 hoặc x=1
Khi x=0 thì \(M=\dfrac{0-3}{0+1}=-3\)
Khi x=1 thì \(M=\dfrac{1-3}{1+1}=\dfrac{-2}{2}=-1\)
Vậy: M là số nguyên nhỏ nhất khi x=0
Câu 5:
\(GT\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
Ta có:
\(K=\dfrac{a^2}{c\left(c^2+a^2\right)}+\dfrac{b^2}{a\left(a^2+b^2\right)}+\dfrac{c^2}{b\left(b^2+c^2\right)}\)
\(=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\left(\dfrac{c}{a^2+c^2}+\dfrac{a}{a^2+b^2}+\dfrac{b}{b^2+c^2}\right)\)
\(\ge\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-\left(\dfrac{c}{2ac}+\dfrac{a}{2ab}+\dfrac{c}{2bc}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{3}{2}\)
Vậy \(MinK=\dfrac{3}{2}\), đạt được khi và chỉ khi \(a=b=c=1\)
Câu 1:c)
\(C=cos^4\alpha-sin^4\alpha-cos^2\alpha+sin^2\alpha\)
\(=\left(cos^2\alpha-sin^2\alpha\right)\left(cos^2\alpha+sin^2\alpha\right)-\left(cos^2\alpha-sin^2\alpha\right)\)
\(=cos^2\alpha-sin^2\alpha-\left(cos^2\alpha-sin^2\alpha\right)=0\)
Câu 4
Ta có:
∠ABC = 90⁰ - 41⁰ = 49⁰
∆ABC vuông tại A
⇒ sin B = AC/BC
⇒ AC = BC . sin B
= BC . sin 49⁰
≈ 3 . 0,75
= 2,25 (m)
Khoảng cách từ thuyền đến bờ:
2,25 - 1,2 = 1,05 (m)
Câu 1
c) cos⁴a - sin⁴a - cos²a + sin²a
= (cos⁴a - sin⁴a) + sin²a - cos²a
= (cos²a - sin²a)(cos²a + sin²a) + sin²a - cos²a
= cos²a - sin²a + sin²a - cos²a
= (cos²a - cos²a) + (-sin²a + sin²a)
= 0
