C) Pt \(\Rightarrow m\cdot\dfrac{1-cos2x}{2}-\left(m-1\right)sin2x+\left(2m+1\right)\cdot\dfrac{1+cos2x}{2}=0\)

\(\Rightarrow\left(m+1\right)cos2x-\left(2m-2\right)sin2x=-1-3m\)

Pt có nghiệm:  \(\Leftrightarrow\) \(\left(m+1\right)^2+\left[-\left(2m-2\right)\right]^2\ge\left(1+3m\right)^2\)

                        \(\Rightarrow\dfrac{-3-\sqrt{13}}{2}\le m\le\dfrac{-3+\sqrt{13}}{2}\)

Pt vô nghiệm: \(\Rightarrow\left\{{}\begin{matrix}m>\dfrac{-3+\sqrt{13}}{2}\\m< \dfrac{-3-\sqrt{13}}{2}\end{matrix}\right.\)

a) Dễ thấy cosx = 0 không thỏa mãn phương trình đã cho nên chiaw phương trình cho cos²x ta được phương trình tương đương 2tan²x + tanx - 3 = 0.

Đặt t = tanx thì phương trình này trở thành

2t² + t - 3 = 0 ⇔ t ∈ {1 ; }.

Vậy

b) Thay 2 = 2(sin²x + cos²x), phương trình đã cho trở thành

3sin²x - 4sinxcosx + 5cos²x = 2sin²x + 2cos²x

⇔ sin²x - 4sinxcosx + 3cos²x = 0

⇔ tan²x - 4tanx + 3 = 0

⇔

⇔ x = + kπ ; x = arctan3 + kπ, k ∈ Z.

c) Thay sin2x = 2sinxcosx ; = (sin²x + cos²x) vào phương trình đã cho và rút gọn ta được phương trình tương đương

sin²x + 2sinxcosx - cos²x = 0 ⇔ tan²x + 4tanx - 5 = 0 ⇔

⇔ x = + kπ ; x = arctan(-5) + kπ, k ∈ Z.

d) 2cos²x - 3√3sin2x - 4sin²x = -4

⇔ 2cos²x - 3√3sin2x + 4 - 4sin²x = 0

⇔ 6cos²x - 6√3sinxcosx = 0 ⇔ cosx(cosx - √3sinx) = 0

⇔

b.

\(\Leftrightarrow\dfrac{3}{2}\left(1-cos2x\right)-sin2x+m=0\)

\(\Leftrightarrow sin2x+\dfrac{3}{2}cos2x-\dfrac{3}{2}=m\)

\(\Leftrightarrow\dfrac{\sqrt{13}}{2}\left(\dfrac{2}{\sqrt{13}}sin2x+\dfrac{3}{\sqrt{13}}cos2x\right)-\dfrac{3}{2}=m\)

Đặt \(\dfrac{2}{\sqrt{13}}=cosa\) với \(a\in\left(0;\dfrac{\pi}{2}\right)\)

\(\Rightarrow\dfrac{\sqrt{13}}{2}sin\left(2x+a\right)-\dfrac{3}{2}=m\)

Phương trình có nghiệm khi và chỉ khi:

\(\dfrac{-\sqrt{13}-3}{2}\le m\le\dfrac{\sqrt{13}-3}{2}\)

Lý thuyết đồ thị:

Phương trình \(f\left(x\right)=m\) có nghiệm khi và chỉ khi \(f\left(x\right)_{min}\le m\le f\left(x\right)_{max}\)

Hoặc sử dụng điều kiện có nghiệm của pt lương giác bậc nhất (tùy bạn)

a.

\(\dfrac{\sqrt{3}}{2}\left(1-cos2x\right)+\dfrac{1}{2}sin2x=m\)

\(\Leftrightarrow\dfrac{1}{2}sin2x-\dfrac{\sqrt{3}}{2}cos2x+\dfrac{\sqrt{3}}{2}=m\)

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{3}\right)+\dfrac{\sqrt{3}}{2}=m\)

\(\Rightarrow\) Pt có nghiệm khi và chỉ khi:

\(-1+\dfrac{\sqrt{3}}{2}\le m\le1+\dfrac{\sqrt{3}}{2}\)

c.

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos2x+\left(m-1\right)sin2x-\left(m+1\right)\left(\dfrac{1}{2}+\dfrac{1}{2}cos2x\right)=m\)

\(\Leftrightarrow\left(2m-2\right)sin2x-\left(m+2\right)cos2x=3m\)

Theo điều kiện có nghiệm của pt lượng giác bậc nhất, pt có nghiệm khi:

\(\left(2m-2\right)^2+\left(m+2\right)^2\ge9m^2\)

\(\Leftrightarrow m^2+m-2\le0\)

\(\Leftrightarrow-2\le m\le\)

a) \(\sqrt{3}\left(\dfrac{1+cos2x}{2}\right)+\dfrac{1}{2}sin2x=m\) ↔ \(\dfrac{\sqrt{3}}{2}cos2x+\dfrac{1}{2}sin2x=m-\dfrac{\sqrt{3}}{2}\) 

→\(\sqrt{3}cos2x+sin2x=2m-\sqrt{3}\) ↔ \(2cos\left(\dfrac{\pi}{6}-2x\right)=2m-\sqrt{3}\)

→\(cos\left(\dfrac{\pi}{6}-2x\right)=m-\dfrac{\sqrt{3}}{2}\) 

Pt có nghiệm khi và chỉ khi \(-1\le m-\dfrac{\sqrt{3}}{2}\le1\) 

b)  \(\left(3+m\right)sin^2x-2sinx.cosx+mcos^2x=0\)

 cosx=0→ sinx=0=> vô lý 

→ sinx#0 chia cả 2 vế của pt cho cos²x ta đc:

\(\left(3+m\right)tan^2x-2tanx+m=0\)

pt có nghiệm ⇔ △^' ≥0

Tự giải phần sau 

c) \(\left(1-m\right)sin^2x+2\left(m-1\right)sinx.cosx-\left(2m+1\right)cos^2x=0\) 

⇔cosx=0→sinx=0→ vô lý

⇒ cosx#0 chia cả 2 vế pt cho cos²x

\(\left(1-m\right)tan^2x+2\left(m-1\right)tanx-\left(2m+1\right)=0\)

pt có nghiệm khi và chỉ khi △^' ≥ 0

Tự giải

a, \(cos^2x-cosx=0\)

\(\Leftrightarrow cosx\left(cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=0\end{matrix}\right.\)

b, \(2sin2x+\sqrt{2}sin4x=0\)

\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)

\(\Leftrightarrow sin2x\left(1+\sqrt{2}cos2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\1+\sqrt{2}cos2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\cos2x=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\dfrac{3\pi}{8}+k\pi\\x=\dfrac{\pi}{8}+k\pi\end{matrix}\right.\)

a, \(cos^2x-cosx=0\)

\(\Leftrightarrow cosx\left(cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=k2\pi\end{matrix}\right.\) (k ∈ Z)

Vậy...

b, \(2sin2x+\sqrt{2}sin4x=0\)

\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)

\(\Leftrightarrow2sin2x\left(1+\sqrt{2}cos2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=\dfrac{-\sqrt{2}}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\2x=\pm\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\pm\dfrac{3\pi}{8}+k\pi\end{matrix}\right.\)

Vậy...

c, \(8cos^2x+2sinx-7=0\)

\(\Leftrightarrow8\left(1-sin^2x\right)+2sinx-7=0\)

\(\Leftrightarrow8sin^2x-2sinx-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-\dfrac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.\)

Vậy...

d, \(4cos^4x+cos^2x-3=0\)

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

\(\Leftrightarrow\dfrac{cos2x+1}{2}=\dfrac{3}{4}\)

\(\Leftrightarrow cos2x=\dfrac{1}{2}\)

\(\Leftrightarrow2x=\pm\dfrac{\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+k\pi\)

Vậy...

e, \(\sqrt{3}tanx-6cotx+\left(2\sqrt{3}-3\right)=0\) (ĐK: \(x\ne\dfrac{k\pi}{2}\))

\(\Leftrightarrow\sqrt{3}tanx-\dfrac{6}{tanx}+\left(2\sqrt{3}-3\right)=0\)

\(\Leftrightarrow\sqrt{3}tan^2x+\left(2\sqrt{3}-3\right)tanx-6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=\sqrt{3}\\tanx=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\left(tm\right)\\x=arctan\left(-2\right)+k\pi\end{matrix}\right.\)

Vậy...

c, \(8cos^2x+2sinx-7=0\)

\(\Leftrightarrow-8sin^2x+2sinx+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-\dfrac{1}{4}\end{matrix}\right.\)

Với \(sinx=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

Với \(sinx=-\dfrac{1}{4}\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.\)

d, \(4cos^4x+cos^2x-3=0\)

\(\Leftrightarrow\left(4cos^2x-3\right)\left(cos^2x+1\right)=0\)

\(\Leftrightarrow4cos^2x-3=0\left(\text{Vì }cos^2x+1>0\right)\)

\(\Leftrightarrow cos^2x=\dfrac{3}{4}\)

\(\Leftrightarrow cosx=\pm\dfrac{\sqrt{3}}{2}\)

Với \(cosx=\dfrac{\sqrt{3}}{2}\Leftrightarrow x=\pm\dfrac{\pi}{3}+k2\pi\)

Với \(cosx=-\dfrac{\sqrt{3}}{2}\Leftrightarrow x=\pm\dfrac{5\pi}{6}+k2\pi\)

1.

Đề là \(x\in\left(0;\frac{\pi}{4}\right)\) hay \(x\in\left[0;\frac{\pi}{4}\right]\) ?

2.

\(sin3x-4sinx.cos2x=0\)

\(\Leftrightarrow sin3x-\left(2sin3x-2sinx\right)=0\)

\(\Leftrightarrow2sinx-sin3x=0\)

\(\Leftrightarrow2sinx-3sinx+4sin^3x=0\)

\(\Leftrightarrow sinx\left(4sin^2x-1\right)=0\)

\(\Leftrightarrow sinx\left(1-2cos2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos2x=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\frac{\pi}{6}+k\pi\end{matrix}\right.\)

3.

\(sin^2x.cosx=0\)

\(\Leftrightarrow sin2x=0\)

\(\Leftrightarrow x=\frac{k\pi}{2}\)

4.

\(\sqrt{3}sin2x+1-cos2x=3\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x=1\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)=1\)

\(\Leftrightarrow2x-\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\frac{\pi}{3}+k\pi\)

5.

Ko có 4 đáp án thì làm sao biết, có vô số pt tương đương với pt này :)

6.

\(sinx+cosx-2sinx.cosx+1=0\)

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=t^2-1\end{matrix}\right.\)

Pt trở thành:

\(t+1-t^2+1=0\)

\(\Leftrightarrow-t^2+t+2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-1\\t=2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow2sinx.cosx=t^2-1=0\)

\(\Leftrightarrow sin2x=0\)

\(\Leftrightarrow x=\frac{k\pi}{2}\)

1, \(sin\left(x+\dfrac{\pi}{6}\right)+cos\left(x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{6}}{2}\)

⇔  \(\dfrac{\sqrt{2}}{2}sin\left(x+\dfrac{\pi}{6}\right)+\dfrac{\sqrt{2}}{2}cos\left(x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}\)

⇔ \(sin\left(x+\dfrac{\pi}{6}+\dfrac{\pi}{4}\right)=sin\dfrac{\pi}{4}\)

2, \(\left(\sqrt{3}-1\right)sinx+\left(\sqrt{3}+1\right)cosx=1-\sqrt{3}\)

⇔ \(\dfrac{\left(\sqrt{3}-1\right)}{2\sqrt{2}}sinx+\dfrac{\left(\sqrt{3}+1\right)}{2\sqrt{2}}cosx=\dfrac{1-\sqrt{3}}{2\sqrt{2}}\)

⇔ sinx . si

\(\text{a) }cos^2x+sin2x-1=0\\ \Leftrightarrow2sinx\cdot cosx-sin^2x=0\\ \Leftrightarrow sinx\left(2cosx-sinx\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=2cosx\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=0\\tanx=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=a\pi\\x=arctan\left(2\right)+b\pi\end{matrix}\right.\)

\(\text{b) }\sqrt{3}sin2x+cos^4x-sin^4x=\sqrt{2}\\ \Leftrightarrow\sqrt{3}sin2x+\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)=\sqrt{2}\\ \Leftrightarrow\frac{\sqrt{3}}{2}\cdot sin2x+\frac{1}{2}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sin2x+sin\frac{\pi}{6}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sin2x+sin\frac{\pi}{6}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=sin\frac{\pi}{4}\\ \\ \Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=\frac{\pi}{4}+a2\pi\\2x+\frac{\pi}{6}=\frac{3\pi}{4}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{24}+a\pi\\x=\frac{7\pi}{24}+b\pi\end{matrix}\right.\)

\(c\text{) }cos^2x-sin^2x=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\\ \Leftrightarrow cos^2x-sin^2x=\sqrt{2}\left(sinx\cdot\frac{\sqrt{2}}{2}+cosx\cdot\frac{\sqrt{2}}{2}\right)\\ \Leftrightarrow\left(cosx-sinx\right)\left(sinx+cosx\right)=sinx+cosx\\ \Leftrightarrow\left[{}\begin{matrix}cosx-sinx=1\\sinx=-cosx\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}cos^2x+\left(cosx-1\right)^2=1\\tanx=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\\tanx=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+a\pi\\x=b2\pi\\x=\frac{3\pi}{4}=c\pi\end{matrix}\right.\)

\(d\text{) }4\left(sin^4x+cos^4x\right)+\sqrt{3}sin4x=2\\ \Leftrightarrow4\left(1-2sin^2x\cdot cos^2x\right)+\sqrt{3}sin4x=2\\ \Leftrightarrow-8sin^2x\cdot cos^2x+\sqrt{3}sin4x=-2\\ \Leftrightarrow-2sin^22x+\sqrt{3}sin4x=-2\\ \Leftrightarrow cos4x-1+\sqrt{3}sin4x=-2\\ \Leftrightarrow\frac{1}{2}cos4x+\frac{\sqrt{3}}{2}sin4x=-\frac{1}{2}\\ \Leftrightarrow sin\frac{\pi}{6}\cdot cos4x+cos\frac{\pi}{6}\cdot sin4x=-\frac{1}{2}\\ \Leftrightarrow sin\left(4x+\frac{\pi}{6}\right)=sin\frac{-\pi}{6}\\ \Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{6}=\frac{-\pi}{6}+a2\pi\\4x+\frac{\pi}{6}=\frac{7\pi}{6}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-\pi}{12}+\frac{a\pi}{2}\\x=\frac{\pi}{4}+\frac{b\pi}{2}\end{matrix}\right.\)

\(e\text{) }4sinx\cdot cosx\cdot cos2x+cos4x=\sqrt{2}\\ \Leftrightarrow sin4x+cos4x=\sqrt{2}\\ \Leftrightarrow sin4x\cdot\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}cos4x=1\\ \Leftrightarrow sin4x\cdot cos\frac{\pi}{4}+cos4x\cdot sin\frac{\pi}{4}=1\\ \Leftrightarrow sin\left(4x+\frac{\pi}{4}\right)=1=sin\frac{\pi}{2}\\ \Leftrightarrow4x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\\ \Leftrightarrow x=\frac{\pi}{16}+\frac{k\pi}{2}\)

1.Pt \(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=sin\left(x+\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\left(\dfrac{\pi}{6}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\\2x-\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)

\(\Rightarrow x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\)\(\left(k\in Z\right)\)

2.\(sin^22x+cos^23x=1\)

\(\Leftrightarrow\dfrac{1-cos4x}{2}+\dfrac{1+cos6x}{2}=1\)

\(\Leftrightarrow cos6x=cos4x\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{k\pi}{5}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow x=\dfrac{k\pi}{5}\)\(\left(k\in Z\right)\) (Gộp nghiệm)

Vậy...

3. \(Pt\Leftrightarrow\left(sinx+sin3x\right)+\left(sin2x+sin4x\right)=0\)

\(\Leftrightarrow2.sin2x.cosx+2.sin3x.cosx=0\)

\(\Leftrightarrow2cosx\left(sin2x+sin3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin3x=-sin2x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\sin3x=sin\left(\pi+2x\right)\end{matrix}\right.\)(\(k\in Z\))

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\)(\(k\in Z\))\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\) (\(k\in Z\))

Vậy...

4. Pt\(\Leftrightarrow\dfrac{1-cos2x}{2}+\dfrac{1-cos4x}{2}=\dfrac{1-cos6x}{2}\)

\(\Leftrightarrow cos2x+cos4x=1+cos6x\)

\(\Leftrightarrow2cos3x.cosx=2cos^23x\)

\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\cosx=cos3x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=-k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)

Vậy...