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câu hỏi đâu??

5(x-3)+2(x-6)=1

5x-5.3+2x-2.6=1

5x-15+2x-12=1

5x+2x=(1+15+12)

(5+2)x=28

    7x=28

      x=28:7

       x=4

Vậy x=4

k đúng cho mk nha

\(5\left(x-3\right)+2\left(x-6\right)=1\)

\(5x-15+2x-12=1\)

\(7x-27=1\)

\(7x=1+27=28\)

\(x=4\)

5.x-5.3+2.x-2.6=1

5.x-15+2.x-12=1

(5.x+2.x)-(15-12)=1

7.x-3=1

7x=1+3

7x=4

x=4:7

x=4/7

Cái này bạn áp dụng cộng dãy số cách đều là ra à

Số số hạng là: \(\dfrac{58-1}{1}+1=58\left(số\right)\)

Tổng của dãy là; \(\dfrac{58\left(58+1\right)}{2}=29\cdot59=1711\)

1. She got the job as a secretary without being able to use a computer. 2. Our teachers don't force us to go to extra classes. 3. I used to be strong when I was young. 4. They didn't use to go out much before they bought a motorbike. 5. I am used to wearing contact lenses. 6. He spent two hours fixing his car. 7. This motorbike cost 500 dollars. 8. She is interested in walking in the rain. 9. It's no use talking without doing anything. 10. I am going to have my car fixed by a man. 11. What is the price of this dress? 12. She'd rather Vietnamese food than American one. 13. My dad'd rather 'do' than talk. 14. You'd better try new things in order to know your ability. 15. There's no point in persuading him to go with you. 16. Her English is too bad for her to work as an interpreter. 17. This car is too expensive for me to buy. 18. You don't learn hard enough. 19. It's foolish of him to say that. 20. You'd better spend more time learning English.

Ta có hệ: \(\left\{{}\begin{matrix}3sin^4x-cos^4x=\dfrac{1}{2}\\sin^2x+cos^2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3\left(1-cos^2x\right)^2-cos^4x=\dfrac{1}{2}\\sin^2x=1-cos^2x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4cos^4x-12cos^2x+5=0\left(1\right)\\sin^2x=1-cos^2x\left(2\right)\end{matrix}\right.\)

Từ (1) ta có: \(\Leftrightarrow\left[{}\begin{matrix}cos^2x=\dfrac{5}{2}\left(l\right)\\cos^2x=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow sin^2x=\dfrac{1}{2}\)

\(\Rightarrow sin^4x+3cos^4x=\left(\dfrac{1}{2}\right)^2+3\left(\dfrac{1}{2}\right)^2=1\)