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ấm áp

Ấm chén

mik chỉ nghĩ ra ấm áp thui

Đáp án: with

Giải thích: be happy with: vui vẻ với điều gì đó

with

be happy with: vui vẻ, hạnh phúc

with

Chọn B

a.

\(0< x< \dfrac{\pi}{2}\Rightarrow cosx>0\Rightarrow cosx=\sqrt{1-sin^2x}=\dfrac{\sqrt{6}}{3}\)

\(cos\left(x+\dfrac{\pi}{3}\right)=cosx.cos\left(\dfrac{\pi}{3}\right)-sinx.sin\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{6}-3}{6}\)

b.

\(\pi< x< \dfrac{3\pi}{2}\Rightarrow sinx< 0\)

\(\Rightarrow sinx=-\sqrt{1-cos^2x}=-\dfrac{5}{13}\)

\(B=sin\left(\dfrac{\pi}{3}-x\right)=sin\left(\dfrac{\pi}{3}\right).cosx-cos\left(\dfrac{\pi}{3}\right).sinx=...\) (bạn tự thay số bấm máy)

c.

\(A=cos^2x+cos^2y+2cosx.cosy+sin^2x+sin^2y+2sinx.siny\)

\(=\left(cos^2x+sin^2x\right)+\left(cos^2y+sin^2y\right)+2\left(cosx.cosy+sinx.siny\right)\)

\(=1+1+2cos\left(x-y\right)\)

\(=2+2cos\left(\dfrac{\pi}{3}\right)=...\)

d.

\(B=cos^2x+sin^2y+2cosx.siny+cos^2y+sin^2x-2sinx.cosy\)

\(=\left(cos^2x+sin^2x\right)+\left(cos^2y+sin^2y\right)-2\left(sinx.cosy-cosx.siny\right)\)

\(=2-2sin\left(x-y\right)=2-2sin\left(\dfrac{\pi}{3}\right)=...\)

jup j 3 ????

`sin(2x-π/3)+1=0`
`<=>sin(2x-π/3)=-1`
`<=>2x-π/3=-π/2=k2π`
`<=>x=(5π)/12+kπ (k \in ZZ)`
Có: `-2020π < (5π)/12+kπ < 2020π`
`<=> -2020 < 5/12+k<2020`
`<=>-2020-5/12 <k<2020+5/12`
`=> k \in {-2020;.....;2020}`
`=>` Có `4041` giá trị của `k` thỏa mãn.

a.

Đặt \(sinx+cosx=t\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(\Rightarrow1+2sinx.cosx=t^2\Rightarrow2sinx.cosx=t^2-1\)

Phương trình trở thành:

\(3t=2\left(t^2-1\right)\)

\(\Leftrightarrow2t^2-3t-2=0\)

\(\Rightarrow\left[{}\begin{matrix}t=2>\sqrt{2}\left(loại\right)\\t=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow sinx+cosx=-\dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{8}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=arcsin\left(-\dfrac{\sqrt{2}}{8}\right)+k2\pi\\x+\dfrac{\pi}{4}=\pi-arcsin\left(-\dfrac{\sqrt{2}}{8}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+arcsin\left(-\dfrac{\sqrt{2}}{8}\right)+k2\pi\\x=\dfrac{3\pi}{4}-arcsin\left(-\dfrac{\sqrt{2}}{8}\right)+k2\pi\end{matrix}\right.\)

b.

ĐKXĐ: \(x\ne\dfrac{\pi}{2}+k\pi\)

\(1+\dfrac{sinx}{cosx}=2\sqrt{2}sinx\)

\(\Rightarrow sinx+cosx=2\sqrt{2}sinx.cosx\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=\sqrt{2}sin2x\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=sin2x\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=x+\dfrac{\pi}{4}+k2\pi\\2x=\dfrac{3\pi}{4}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{\pi}{4}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k2\pi}{3}\)

c.

\(\Leftrightarrow1+sinx+cosx+sinx.cosx=2\)

\(\Leftrightarrow sinx+cosx+sinx.cosx=1\)

Đặt \(sinx+cosx=t\in\left[-\sqrt[]{2};\sqrt{2}\right]\)

\(\Rightarrow sinx.cosx=\dfrac{t^2-1}{2}\)

Phương trình trở thành:

\(t+\dfrac{t^2-1}{2}=1\)

\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow sinx+cosx=1\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow...\)