Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(\text{Δ}=\left(-5\right)^2-4\cdot3\cdot8=25-96< 0\)
Do đó: Phươbg trình vô nghiệm
b: \(\text{Δ}=\left(-3\right)^2-4\cdot15\cdot5=9-300< 0\)
Do đó: Phương trình vô nghiệm
c: \(\Leftrightarrow x^2-4x+4-3=0\)
\(\Leftrightarrow\left(x-2\right)^2=3\)
hay \(x\in\left\{2+\sqrt{3};2-\sqrt{3}\right\}\)
d: \(\Leftrightarrow3x^2+6x+x+2=0\)
=>(x+2)(3x+1)=0
=>x=-2 hoặc x=-1/3
ta có : \(M=2cot37.cot53+sin^228\dfrac{3tan54}{cot36}+sin^262\)
\(=2.cot37.cot\left(90-37\right)+sin^228\dfrac{3tan54}{cot\left(90-54\right)}+sin^262\)
\(=2.cot37.tan37+sin^228\dfrac{3tan54}{tan54}+sin^262\)\(=2+3sin^228+sin^262=2+2sin^228+sin^228+sin^2\left(90-28\right)\)
\(=2+2sin^228+sin^228+cos^228=3+2sin^228\)
1: \(sin^6x+cos^6x+3sin^2x\cdot cos^2x\)
\(=\left(sin^2x+cos^2x\right)^2-3\cdot sin^2x\cdot cos^2x\cdot\left(sin^2x+cos^2x\right)+3\cdot sin^2x\cdot cos^2x\)
=1
2: \(sin^4x-cos^4x\)
\(=\left(sin^2x+cos^2x\right)\left(sin^2x-cos^2x\right)\)
\(=1-2\cdot cos^2x\)
ta có : \(A=cot\alpha+\dfrac{sin\alpha}{1+cos\alpha}=\dfrac{cos\alpha}{sin\alpha}+\dfrac{sin\alpha}{1+cos\alpha}\)
\(=\dfrac{cos\alpha\left(1+cos\alpha\right)+sin^2\alpha}{sin\alpha\left(1+cos\alpha\right)}=\dfrac{cos\alpha+cos^2\alpha+sin^2\alpha}{sin\alpha\left(1+cos\alpha\right)}\)
\(=\dfrac{1+cos\alpha}{sin\alpha\left(1+cos\alpha\right)}=\dfrac{1}{sin\alpha}\)
Đặt \(\sqrt{2+\sqrt{3}}=a\left(a>0\right)\)
Ta có x=\(\sqrt{2+a}-\sqrt{3\left(2-a\right)}\Rightarrow x^2=2+a+3\left(2-a\right)-2\sqrt{3\left(2+a\right)\left(2-a\right)}\)\(=8-2a-2\sqrt{3\left(4-a^2\right)}=8-2a-2\sqrt{3\left(4-2-\sqrt{3}\right)}=8-2a-\sqrt{6}\sqrt{4-2\sqrt{3}}\)
\(=8-2\sqrt{2+\sqrt{3}}-\sqrt{6}\left(\sqrt{3}-1\right)=8-\sqrt{2}\sqrt{4+2\sqrt{3}}-3\sqrt{2}+\sqrt{6}\)
\(=8-\sqrt{2}\left(\sqrt{3}+1\right)-3\sqrt{2}+\sqrt{6}=8-\sqrt{6}-\sqrt{2}-3\sqrt{2}+\sqrt{6}=8-4\sqrt{2}\)
\(\Rightarrow x^2-8=-4\sqrt{2}\Rightarrow\left(x^2-8\right)^2=32\Rightarrow x^4-16x^2+64=32\Rightarrow x^4-16x^2+32=0\left(ĐPCM\right)\)
a) 1 + tan22 a =1 +(\(\dfrac{sina}{cosa}\))2 =\(\dfrac{sina+cosa}{cos^2a}\)=\(\dfrac{1}{cos^2a}\)
b) 1 + cot2 a= 1 +(\(\dfrac{cosa}{sina}\))2 = \(\dfrac{cosa+sina}{sin^2a}\)=\(\dfrac{1}{sin^2a}\)
c) tan2 a (2 sin2a + 3 cos2 a - 2)
=tan2 a[cos2 a +2 (\(sina^2+cos^2a\))-2 ]
=\(\dfrac{sin^2a}{cos^2a}\)×\(cos^2a=sin^2a\)
b: \(1+cot^2a=1+\left(\dfrac{cosa}{sina}\right)^2=\dfrac{1}{sin^2a}\)
c: \(=tan^2a\left[2\left(1-cos^2a\right)+3cos^2a-2\right]\)
\(=tan^2a\left[cos^2a\right]\)
\(=\dfrac{sin^2a}{cos^2a}\cdot cos^2a=sin^2a\)
A