cái câu 1 kia lạ thật, phần phía trc có ngoặc thì phải nhân vs hạng tử nào đó chứ nhỉ? Và mk tính ra kq là \(-\cos^22\alpha\)

\(VT=\cos^4\alpha+\sin^4\alpha-2\cos^6\alpha-2\sin^6\alpha\)

\(=\sin^4\alpha\left(1-2\sin^2\alpha\right)-\cos^4\alpha\left(2\cos^2\alpha-1\right)\)

\(=\sin^4\alpha.\cos2\alpha-\cos^4\alpha.\cos2\alpha\)

\(=\cos2\alpha\left(\sin^2\alpha.\sin^2\alpha-\cos^4\alpha\right)\)

\(=\cos2\alpha.\left[\left(1-\cos^2\alpha\right)^2-\cos^4\alpha\right]\)

\(=\cos2\alpha.\left(1-2\cos^2\alpha\right)\)

\(=-\cos^22\alpha\)

2/ \(VT=\frac{1-\cos^2\alpha+\cos^2\alpha}{1+\sin2\alpha}=\frac{1}{1+\sin2\alpha}\)

\(VP=\frac{\frac{\sin\alpha}{\cos\alpha}-1}{\frac{\sin\alpha}{\cos\alpha}+1}=\frac{\frac{\sin\alpha-\cos\alpha}{\cos\alpha}}{\frac{\sin\alpha+\cos\alpha}{\cos\alpha}}=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}\)

hmm, câu 2 có vẻ vô lí, bn thử nhân chéo lên mà xem, nó ko ra KQ = nhau đâu

1)

\((\cos^4a+\sin ^4a)-2(\cos^6a+\sin ^6a)=(\cos ^4a+\sin ^4a)-2(\cos ^2a+\sin ^2a)(\cos ^4a-\cos ^2a\sin ^2a+\sin ^4a)\)

\(=(\cos ^4a+\sin ^4a)-2(\cos ^4a-\cos ^2a\sin ^2a+\sin ^4a)\)

\(=-(\cos ^4a-2\sin ^2a\cos ^2a+\sin ^4a)=-(\cos ^2a-\sin ^2a)^2=-\cos ^22a\)

(bạn xem lại đề. Nếu thay $(\cos ^4a+\sin ^4a)$ thành $3(\cos ^4a+\sin ^4a)$ thì kết quả thu được là $(\cos ^2a+\sin ^2a)^2=1$ như yêu cầu)

2) Sửa đề:

\(\frac{\sin ^2a-\cos ^2a}{1+2\sin a\cos a}=\frac{(\sin a-\cos a)(\sin a+\cos a)}{\sin ^2a+\cos ^2a+2\sin a\cos a}=\frac{(\sin a-\cos a)(\sin a+\cos a)}{(\sin a+\cos a)^2}\)

\(=\frac{\sin a-\cos a}{\sin a+\cos a}=\frac{\frac{\sin a}{\cos a}-1}{\frac{\sin a}{\cos a}+1}=\frac{\tan a-1}{\tan a+1}\)

Bạn lưu ý viết đề bài chuẩn hơn.

3)

\(\sin ^4a+\cos ^4a-\sin ^6a-\cos ^6a=\sin ^4a+\cos ^4a-[(\sin ^2a)^3+(\cos ^2a)^3]\)

\(=\sin ^4a+\cos ^4a-(\sin ^2a+\cos ^2a)(\sin ^4a-\sin ^2a\cos ^2a+\cos ^4a)\)

\(=\sin ^4a+\cos ^4a-(\sin ^4a-\sin ^2a\cos ^2a+\cos ^4a)\)

\(=\sin ^2a\cos ^2a\) (đpcm)

4)

\(\frac{\cos a}{1+\sin a}+\tan a=\frac{\cos a}{1+\sin a}+\frac{\sin a}{\cos a}=\frac{\cos ^2a+\sin^2a+\sin a}{\cos a(1+\sin a)}=\frac{1+\sin a}{\cos a(1+\sin a)}=\frac{1}{\cos a}\)

5)

\(\frac{\tan a}{1-\tan ^2a}.\frac{\cot ^2a-1}{\cot a}=\frac{\tan a}{(tan a\cot a)^2-\tan ^2a}.\frac{\cot ^2a-1}{\cot a}\)

\(=\frac{\tan a}{\tan ^2a(\cot ^2a-1)}.\frac{\cot ^2a-1}{\cot a}=\frac{1}{\tan a\cot a}=\frac{1}{1}=1\)

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Mấu chốt của các bài này là bạn sử dụng 2 công thức sau:

1. \(\sin ^2x+\cos^2x=1\)

2. \(\tan x.\cot x=1\)

=5 đó pn ko chắc 100 phần trăm nha

ta có: \(\frac{1+2y}{18}=\frac{1+4y}{24}\)

\(\Rightarrow\left(1+2y\right).24=18.\left(1+4y\right)\)

\(24+48y=18+72y\)

\(48y-72y=18-24\)

\(-24y=-6\)

\(y=\frac{1}{4}\)

thay vào \(\frac{1+4.\frac{1}{4}}{24}=\frac{1+6.\frac{1}{4}}{6x}\)

\(\frac{1+1}{24}=\frac{1+\frac{3}{2}}{6x}\)

\(\frac{1}{12}=\frac{5}{2}:6x\)

\(6x=\frac{5}{2}:\frac{1}{12}\)

\(6x=30\)

\(x=30:6\)

\(x=5\)

KL: x =5; y = 1/4

\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}=\frac{1+2y+1+6y}{18+6x}=\frac{2+8y}{6\left(3+x\right)}=\frac{1+4y}{3\left(3+x\right)}\)

\(\Rightarrow3\left(3+x\right)=24\)\(\Rightarrow3+x=8\)\(\Rightarrow x=5\)

Vậy \(x=5\)

Ta có: \(\frac{1+2y}{18}=\frac{1+4y}{24}\)

\(\Leftrightarrow24\left(1+2y\right)=18\left(1+4y\right)\)

\(\Leftrightarrow24+48y=18+72y\)

\(\Leftrightarrow24y-6=0\Leftrightarrow y=\frac{1}{4}\)

\(\Rightarrow\frac{1+2y}{18}=\frac{1+6y}{6x}\Leftrightarrow\frac{1+\frac{1}{2}}{18}=\frac{1+\frac{3}{2}}{6x}\)

\(\Leftrightarrow x=5\)

Vậy x = 5 và \(y=\frac{1}{4}\)

1 + 2y/18=1 + 6y/6x=1 + 2y + 1 + 6y/18 + 6x=2 + 8y/18 + 6x=2.(1 + 4y)/2.(9 + 3x)=1 + 4y/9 + 3x

Suy ra:1 + 4y/9 + 3x=1 + 4y/24=>9 + 3x=24

3x=15

x=5

\(\Rightarrow\)\(\frac{1+2y+1+4y+1+6y}{18+24+6x}\)=\(\frac{\left(1+1+1\right)+2y+4y+6y}{6\left(3+4+x\right)}=\frac{y\left(2+4+6\right)+3}{6\left(3+4+x\right)}=\frac{3+y.12}{6\left(7+x\right)}\)

=\(\frac{3\left(1+4y\right)}{3.2\left(7+x\right)}=\frac{1+4y}{14+2x}\)

\(\Rightarrow\)\(\frac{1}{14}=\frac{2y}{x}\Rightarrow x=14.2y=28y\)

\(\frac{x}{y}=28\)

Ta có: \(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}=\frac{1+2y+1+6y}{18+6x}=\frac{2\left(1+4y\right)}{6\left(x+3\right)}=\frac{1+4y}{3x+9}\)

\(=>\frac{1+4y}{24}=\frac{1+4y}{3x+9}\)\(=>3x+9=24\)

<=>3x=15

<=>x=5

Vậy x có giá trị bằng 5

Chúc bạn học tốt!

Ta có: 1 + 2y/18 = 2.(1+2y)/2.18 = 2+4y/36

Sử dụng tc dãy tỉ số bằng nhau ta có:

2+4y/36 = 1+4y/24 = 2+4y-1-4y/36-24 = 1/12

Do 1+2y/18 = 1/12=> y = 1/4

1+6y/6x = 1/12=> x = 5

Vậy x = 5; y = 1/4

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}=\frac{1+2y+1+6y}{18+6x}\)

                                                       \(=\frac{2\left(1+4y\right)}{2\left(3x+9\right)}=\frac{1+4y}{3x+9}\)

\(\Rightarrow\frac{1+4y}{24}=\frac{1+4y}{3x+9}\)

\(\Rightarrow3x+9=24\)

\(\Rightarrow x+3=24:3=8\)

\(\Rightarrow x=5\)

\(\frac{1+2y}{18}=\frac{2+4y}{36}=\frac{1+4y}{24}=\frac{2+4y-\left(1+4y\right)}{36-24}=\frac{1}{12}\)

\(\Rightarrow\frac{1+4y}{24}=\frac{1}{12}\Rightarrow4y+1=2\Rightarrow y=\frac{1}{4}\)

\(\Rightarrow\frac{1+6y}{6x}=\frac{1}{12}\Rightarrow x=2\cdot\left(1+6y\right)=2+12y=2+12\cdot\frac{1}{4}=5\)

Vậy x = 5.