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Vd5:
a: \(cos42^0=\sin\left(90^0-42^0\right)=\sin48^0\)
\(cos67^0=\sin\left(90^0-67^0\right)=\sin23^0\)
Ta có: \(23^0<30^0<38^0<48^0<75^0\)
=>Sin23<sin30<sin38<sin48<sin 75
=>cos67<sin30<sin38<cos42<sin75
b: \(\cot49^0=\tan\left(90^0-49^0\right)=\tan41^0\)
\(\cot50^0=\tan\left(90^0-50^0\right)=\tan40^0\)
Ta có: 25<27<40<41<80
=>tan25<tan27<tan40<tan41<tan80
=>tan25<tan27<cot49<cot50<tan80
Vd6:
a: \(\sin^2a+cos^2a=1\)
=>\(cos^2a=1-0,8^2=1-0,64=0,36=0,6^2\)
=>cosa=0,6
tan a=sin a/cosa
=0,8/0,6
=4/3
cot a=cosa/sin a
=0,6/0,8=3/4
b: \(\sin^2a+cos^2a=1\)
=>\(\sin^2a=1-0,6^2=1-0,36=0,64\)
=>sin a=0,8
tan a=sin a/cosa
=0,8/0,6
=4/3
cot a=cosa/sin a
=0,6/0,8=3/4
c: \(\tan a\cdot\cot a=1\)
=>\(\cot a=\frac13\)
\(1+\tan^2a=\frac{1}{cos^2a}\)
=>\(\frac{1}{cos^2a}=1+3^2=1+9=10\)
=>\(cos^2a=\frac{1}{10}\)
=>\(cosa=\frac{1}{\sqrt{10}}=\frac{\sqrt{10}}{10}\)
Ta có: \(\sin^2a+cos^2a=1\)
=>\(\sin^2a=1-\frac{1}{10}=\frac{9}{10}\)
=>\(\sin a=\frac{3}{\sqrt{10}}\)
d: \(\tan a\cdot\cot a=1\)
=>\(\tan a=\frac12\)
Ta có: \(1+\tan^2a=\frac{1}{cos^2a}\)
=>\(\frac{1}{cos^2a}=1+\left(\frac12\right)^2=1+\frac14=\frac54\)
=>\(cos^2a=\frac45\)
=>\(cosa=\frac{2}{\sqrt5}\)
\(\sin^2a+cos^2a=1\)
=>\(\sin^2a=1-\frac45=\frac15\)
=>\(\sin a=\frac{1}{\sqrt5}\)
Vd5:
a: \(cos42^0=\sin\left(90^0-42^0\right)=\sin48^0\)
\(cos67^0=\sin\left(90^0-67^0\right)=\sin23^0\)
Ta có: 23<30<38<48<75
=>sin23<sin30<sin38<sin48<sin75
=>cos67<sin30<sin38<cos42<sin75
b: \(\cot49^0=\tan\left(90^0-49^0\right)=tan41^0\)
\(\cot50^0=\tan\left(90^0-50^0\right)=\tan40^0\)
Ta có: 25<27<40<41<80
=>tan 25<tan 27<tan 40<tan41<tan80
=>tan 25<tan27<cot49<cot50<tan80
Vd6:
a: \(\sin^2a+cos^2a=1\)
=>\(cos^2a=1-0,8^2=1-0,64=0,36=0,6^2\)
=>cosa=0,6
\(\tan a=\frac{\sin a}{cosa}=\frac{0.8}{0.6}=\frac43\)
\(\cot a=\frac{1}{\tan a}=1:\frac43=\frac34\)
b: \(\sin^2a+cos^2a=1\)
=>\(\sin^2a=1-0,6^2=1-0,36=0,64=0,8^2\)
=>sin a=0,8
\(\tan a=\frac{\sin a}{cosa}=\frac{0.8}{0.6}=\frac43\)
\(\cot a=\frac{1}{\tan a}=1:\frac43=\frac34\)
c: \(\tan a\cdot\cot a=1\)
=>\(\cot a=\frac13\)
Ta có: \(1+\tan^2a=\frac{1}{cos^2a}\)
=>\(\frac{1}{cos^2a}=1+3^2=10\)
=>\(cos^2a=\frac{1}{10}\)
=>\(cosa=\frac{1}{\sqrt{10}}\)
Ta có: \(\sin^2a+cos^2a=1\)
=>\(\sin^2a=1-\frac{1}{10}=\frac{9}{10}\)
=>\(\sin a=\frac{3}{\sqrt{10}}\)
d: \(\tan a\cdot\cot a=1\)
=>\(\tan a=\frac{1}{\cot a}=\frac12\)
Ta có: \(1+\tan^2a=\frac{1}{cos^2a}\)
=>\(\frac{1}{cos^2a}=1+\left(\frac12\right)^2=\frac54\)
=>\(cos^2a=\frac45\)
=>\(cosa=\frac{2}{\sqrt5}\)
Ta có: \(\sin^2a+cos^2a=1\)
=>\(sin^2a=1-\frac45=\frac15\)
=>sin a=\(\frac{1}{\sqrt5}\)
Vd5:
a: \(cos42^0=\sin\left(90^0-42^0\right)=\sin48^0\)
\(cos67^0=\sin\left(90^0-67^0\right)=\sin23^0\)
Ta có: 23<30<38<48<75
=>sin23<sin30<sin38<sin48<sin75
=>cos67<sin30<sin38<cos42<sin75
b: \(\cot49^0=\tan\left(90^0-49^0\right)=tan41^0\)
\(\cot50^0=\tan\left(90^0-50^0\right)=\tan40^0\)
Ta có: 25<27<40<41<80
=>tan 25<tan 27<tan 40<tan41<tan80
=>tan 25<tan27<cot49<cot50<tan80
Vd6:
a: \(\sin^2a+cos^2a=1\)
=>\(cos^2a=1-0,8^2=1-0,64=0,36=0,6^2\)
=>cosa=0,6
\(\tan a=\frac{\sin a}{cosa}=\frac{0.8}{0.6}=\frac43\)
\(\cot a=\frac{1}{\tan a}=1:\frac43=\frac34\)
b: \(\sin^2a+cos^2a=1\)
=>\(\sin^2a=1-0,6^2=1-0,36=0,64=0,8^2\)
=>sin a=0,8
\(\tan a=\frac{\sin a}{cosa}=\frac{0.8}{0.6}=\frac43\)
\(\cot a=\frac{1}{\tan a}=1:\frac43=\frac34\)
c: \(\tan a\cdot\cot a=1\)
=>\(\cot a=\frac13\)
Ta có: \(1+\tan^2a=\frac{1}{cos^2a}\)
=>\(\frac{1}{cos^2a}=1+3^2=10\)
=>\(cos^2a=\frac{1}{10}\)
=>\(cosa=\frac{1}{\sqrt{10}}\)
Ta có: \(\sin^2a+cos^2a=1\)
=>\(\sin^2a=1-\frac{1}{10}=\frac{9}{10}\)
=>\(\sin a=\frac{3}{\sqrt{10}}\)
d: \(\tan a\cdot\cot a=1\)
=>\(\tan a=\frac{1}{\cot a}=\frac12\)
Ta có: \(1+\tan^2a=\frac{1}{cos^2a}\)
=>\(\frac{1}{cos^2a}=1+\left(\frac12\right)^2=\frac54\)
=>\(cos^2a=\frac45\)
=>\(cosa=\frac{2}{\sqrt5}\)
Ta có: \(\sin^2a+cos^2a=1\)
=>\(sin^2a=1-\frac45=\frac15\)
=>sin a=\(\frac{1}{\sqrt5}\)
22.
ĐKXĐ: \(y\ne1\)
\(\left\{{}\begin{matrix}x^2-\dfrac{1}{y-1}=2\\2x^2+\dfrac{3}{1-y}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2+\dfrac{2}{1-y}=4\\2x^2+\dfrac{3}{1-y}=2\end{matrix}\right.\)
Trừ pt dưới cho trên:
\(\Rightarrow\dfrac{1}{1-y}=-2\)
\(\Rightarrow1-y=-\dfrac{1}{2}\Rightarrow y=\dfrac{3}{2}\)
Thế vào \(x^2-\dfrac{1}{y-1}=2\)
\(\Rightarrow x^2=4\Rightarrow x=\pm2\)
Vậy nghiệm của hệ là \(\left(x;y\right)=\left(2;\dfrac{3}{2}\right);\left(-2;\dfrac{3}{2}\right)\)
b.
ĐKXĐ: \(x\ne-\dfrac{1}{2}\)
\(Hệ\Leftrightarrow\left\{{}\begin{matrix}2y^2-\dfrac{10}{2x+1}=8\\2y^2-\dfrac{11}{2x+1}=7\end{matrix}\right.\)
Trừ pt trên cho dưới:
\(\Rightarrow\dfrac{1}{2x+1}=1\)
\(\Rightarrow2x+1=1\)
\(\Rightarrow x=0\)
Thế vào \(y^2-\dfrac{5}{2x+1}=4\)
\(\Rightarrow y^2=9\Rightarrow y=\pm3\)
Vậy nghiệm của hệ là \(\left(x;y\right)=\left(0;3\right);\left(0;-3\right)\)
1) Vì x=25 thỏa mãn ĐKXĐ nên Thay x=25 vào biểu thức \(A=\dfrac{\sqrt{x}-2}{x+1}\), ta được:
\(A=\dfrac{\sqrt{25}-2}{25+1}=\dfrac{5-2}{25+1}=\dfrac{3}{26}\)
Vậy: Khi x=25 thì \(A=\dfrac{3}{26}\)
2) Ta có: \(B=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}+\dfrac{2x+8\sqrt{x}-6}{x-\sqrt{x}-2}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}+\dfrac{2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-5\sqrt{x}+6+2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3x+3\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}-2}\)
Bài 3:
1: ĐKXĐ: \(x\ge1\)
2: ĐKXĐ: \(x\in R\)
3: ĐKXĐ: \(x\le1\)
4: ĐKXĐ: \(x>\dfrac{3}{2}\)
a: \(x=\dfrac{6^2}{3}=12\left(cm\right)\)
\(y=\sqrt{6^2+12^2}=6\sqrt{5}\)
b: \(x=\sqrt{4\cdot9}=6\)
c: \(x=5\cdot\tan40^0\simeq4,2\left(cm\right)\)
ghi đầy đủ đc ko ạ