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

a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

a/ ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\sqrt{1-x}=a\ge0\\\sqrt{1+x}=b\ge0\end{matrix}\right.\) được hệ:

\(\left\{{}\begin{matrix}\sqrt{1+ab}\left(a^3-b^3\right)=2+ab\\a^2+b^2=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{1+ab}\left(a-b\right)\left(a^2+ab+b^2\right)=a^2+b^2+ab\\a^2+b^2=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{1+ab}\left(a-b\right)=1\\a^2+b^2=2\end{matrix}\right.\) \(\left(a\ge b\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(1+ab\right)\left(a-b\right)^2=1\\a^2+b^2=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(1+ab\right)\left(2-2ab\right)=1\\a^2+b^2=2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}1-a^2b^2=\frac{1}{2}\\a^2+b^2=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2b^2=\frac{1}{2}\\a^2+b^2=2\end{matrix}\right.\)

Theo Viet đảo, \(a^2;b^2\) là nghiệm của:

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

\(\Rightarrow\left[{}\begin{matrix}1-x=\frac{2+\sqrt{2}}{2}\\1-x=\frac{2-\sqrt{2}}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\frac{\sqrt{2}}{2}\\x=\frac{\sqrt{2}}{2}\end{matrix}\right.\)

2 phần còn lại ko biết giải theo kiểu lớp 10, chỉ biết lượng giác hóa, bạn tham khảo thôi :(

b/ Đặt \(x=cos2t\) pt trở thành:

\(\sqrt{1-cos2t}-2cos2t.\sqrt{1-cos^22t}-\left(2cos^22t-1\right)=0\)

\(\Leftrightarrow\sqrt{2}sint-2sin2t.cos2t-cos4t=0\)

\(\Leftrightarrow\sqrt{2}sint-sin4t-cos4t=0\)

\(\Leftrightarrow\sqrt{2}sint=sin4t+cos4t=\sqrt{2}sin\left(4t+\frac{\pi}{4}\right)\)

\(\Leftrightarrow sin\left(4t+\frac{\pi}{4}\right)=sint\)

\(\Leftrightarrow\left[{}\begin{matrix}4t+\frac{\pi}{4}=t+k2\pi\\4t+\frac{\pi}{4}=\pi-t+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-\frac{\pi}{12}+\frac{k2\pi}{3}\\t=-\frac{\pi}{20}+\frac{k2\pi}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=cos\left(-\frac{\pi}{6}+\frac{k4\pi}{3}\right)\\x=cos\left(-\frac{\pi}{10}+\frac{k4\pi}{5}\right)\end{matrix}\right.\) với \(k\in Z\)

c/ Đặt \(x=cost\)

\(64cos^6t-112cos^4t+56cos^2t-7=2\sqrt{1-cos^2t}\)

\(\Leftrightarrow64cos^6t-112cos^4t+56cos^2t-7=2sint\)

Nhận thấy \(cost=0\) không phải nghiệm, pt tương đương:

\(64cos^7t-112cos^5t+56cos^3t-7cost=2sint.cost\)

\(\Leftrightarrow cos7t=sin2t=cos\left(\frac{\pi}{2}-2t\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}7t=\frac{\pi}{2}-2t+k2\pi\\7t=2t-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\frac{\pi}{18}+\frac{k2\pi}{9}\\t=-\frac{\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=cos\left(\frac{\pi}{18}+\frac{k2\pi}{9}\right)\\x=\left(-\frac{\pi}{10}+\frac{k2\pi}{5}\right)\end{matrix}\right.\)

Ý tưởng của người ra đề khá kì quặc, công thức \(cos7a\) kia thực sự là chứng minh rất mất thời gian

a) - 3 =0

<=> (√x-3)(√x+3)-3√x-3=0

<=> (√x-3)(√x+3-3)=0

<=> (√x-3)√x=0

<=> √x-3=0

<=>x=9

b)=x - 3

<=> √(2x -3)² =x-3

<=> 2x-3=x-3

<=>2x-x=-3+3

<=>x=0

c)=3x-1

<=> √(x+3)² =3x-1

<=> x+3=3x-1

<=> -2x=-4

<=>  x=2

Nhớ cho mình 1 tim nha bạn

Lời giải:

a. ĐKXĐ: $x\geq 3$

PT $\Leftrightarrow \sqrt{(x-3)(x+3)}-3\sqrt{x-3}=0$

$\Leftrightarrow \sqrt{x-3}(\sqrt{x+3}-3)=0$

$\Leftrightarrow \sqrt{x-3}=0$ hoặc $\sqrt{x+3}-3=0$

$\Leftrightarrow \sqrt{x-3}=0$ hoặc $\sqrt{x+3}=3$

$\Leftrightarrow x=3$ hoặc $x=6$ (tm)

b.

PT \(\Rightarrow \left\{\begin{matrix} x-3\geq 0\\ 4x^2-12x+9=(x-3)^2=x^2-6x+9\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 3\\ 3x^2-6x=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 3\\ 3x(x-2)=0\end{matrix}\right.\)

$\Rightarrow$ không có giá trị $x$ nào thỏa mãn 

Vậy pt vô nghiệm.

c.

PT \(\Rightarrow \left\{\begin{matrix} 3x-1\geq 0\\ x^2+6x+9=(3x-1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{3}\\ x^2+6x+9=9x^2-6x+1\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{3}\\ 8x^2-12x-8=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{3}\\ 4(x-2)(2x+1)=0\end{matrix}\right.\Leftrightarrow x=2\)

a: \(=6+2\sqrt{11}-4+\sqrt{11}=2+3\sqrt{11}\)

b: \(=\dfrac{3x+9\sqrt{x}-2x+4\sqrt{x}}{\left(\sqrt{x}+3\right)\left(x-2\sqrt{x}\right)}\cdot\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}=\dfrac{\sqrt{x}+3}{x-2\sqrt{x}}\)

a) Áp dụng bđt AM-GM có:

\(\sqrt[3]{\left(9-x\right).8.8}\le\dfrac{9-x+8+8}{3}=\dfrac{25-x}{3}\)\(\Leftrightarrow\sqrt[3]{9-x}\le\dfrac{25-x}{12}\)

\(\sqrt[3]{\left(7+x\right).8.8}\le\dfrac{7+x+8+8}{3}=\dfrac{23+x}{3}\)\(\Leftrightarrow\sqrt[3]{7+x}\le\dfrac{23+x}{12}\)

Cộng vế với vế \(\Rightarrow\sqrt[3]{9-x}+\sqrt[3]{7+x}\le4\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}9-x=8\\7+x=8\end{matrix}\right.\)\(\Rightarrow x=1\)

Vậy...

b)Đk:\(x\ge2\)

Pt \(\Leftrightarrow\left(x-1\right)^2.\left(x^2-4\right)=\left(x-2\right)^2.\left(x^2-1\right)\)

\(\Leftrightarrow\left(x-1\right)^2\left(x-2\right)\left(x+2\right)=\left(x-2\right)^2\left(x+1\right)\left(x-1\right)\)

Do \(x\ge2\Rightarrow x-1>0\)

Chia cả hai vế của pt cho x-1 ta được:

\(\left(x-1\right)\left(x-2\right)\left(x+2\right)=\left(x-2\right)^2\left(x+1\right)\)

\(\Leftrightarrow\left(x-2\right)\left[\left(x-1\right)\left(x+2\right)-\left(x-2\right)\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left[x^2+x-2-x^2+3x-2\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)

Vậy S={2}

c)Đk:\(\left\{{}\begin{matrix}9-x^2\ge0\\x^2-1\ge0\\x-3\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-3\le x\le3\\\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\\x\ge3\end{matrix}\right.\)\(\Rightarrow x=3\)

Thay x=3 vào pt thấy thỏa mãn

Vậy S={3}

a) \(\sqrt{x+3}-\sqrt{x-1}=\sqrt{2x+2}\)

Điều kiện: \(\hept{\begin{cases}x+3\ge0\\x-1\ge0\\2x+2\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-3\\x\ge1\\x\ge-1\end{cases}\Leftrightarrow x\ge1}\)

    \(\Leftrightarrow\left(\sqrt{x+3}-\sqrt{x-1}\right)^2=\left(\sqrt{2x+2}\right)^2\)

     \(\Leftrightarrow x+3-2\sqrt{\left(x+3\right)\left(x-1\right)}+x-1=2x+2\)

     \(\Leftrightarrow2x+2-2\sqrt{\left(x+3\right)\left(x-1\right)}=2x+2\)

     \(\Leftrightarrow-2\sqrt{\left(x+3\right)\left(x-1\right)}=0\)

     \(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)

      \(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\left(l\right)\\x=1\left(n\right)\end{cases}}\)

Vậy \(S=\left\{1\right\}\)