\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)\(\Leftrightarrow\sqrt{x^2+x+\frac{1}{4}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)\(\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\Leftrightarrow2x+1=2x^3+x^2+2x+1\)\(\Leftrightarrow2x^3+x^2=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\)

\(\sqrt{x^2-\frac{1}{4}+\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\left(1\right)\)

\(\left(1\right)\Leftrightarrow\sqrt{x^2-\frac{1}{4}+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)

\(x^2+1\ge1\forall x\Rightarrow2x+1\ge0!2x+1!=2x+1\)

\(\left(1\right)\Leftrightarrow\sqrt{x^2+x+\frac{1}{4}}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)

\(\left(1\right)\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)

\(\left(1\right)\Leftrightarrow2x+1=\left(2x+1\right)\left(x^2+1\right)\Leftrightarrow\left(2x+1\right).\left(1-\left(x^2+1\right)\right)=0\)

\(\hept{\begin{cases}2x+1=0\\-x^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\x=0\end{cases}}}\)

Chúc bạn học tốt !!!

Tìm nhẩm nghiệm rồi nhân liên hợp

f: Ta có: \(\left(x+1\right)\left(x-2\right)-\left(2-x\right)\left(3-x\right)>0\)

\(\Leftrightarrow x^2-2x+x-2-\left(x-2\right)\left(x-3\right)>0\)

\(\Leftrightarrow x^2-x-2-x^2+5x-6>0\)

\(\Leftrightarrow4x>8\)

hay x>2

g: Ta có: \(\left(2x-1\right)^2\le2\left(x-1\right)^2\)

\(\Leftrightarrow4x^2-4x+1-2x^2+4x-2\le0\)

\(\Leftrightarrow2x^2\le1\)

\(\Leftrightarrow x^2\le\dfrac{1}{2}\)

\(\Leftrightarrow-\dfrac{\sqrt{2}}{2}\le x\le\dfrac{\sqrt{2}}{2}\)

1/ \(\Leftrightarrow\left\{{}\begin{matrix}2x^2-4xy+2x-4y+6=0\\y^2-x^2+2xy+2x-2=0\end{matrix}\right.\)

\(\Rightarrow x^2+y^2-2xy+4x-4y+4=0\)

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

\(\Leftrightarrow\left(x-y+2\right)^2=0\)

\(\Rightarrow y=x+2\)

Thay vào 1 trong 2 pt ban đầu là xong

2/ \(x^2-\left(y+2\right)x-6y^2+11y-3=0\)

\(\Delta=\left(y+2\right)^2-4\left(-6y^2+11y-3\right)\)

\(=25y^2-40y+16=\left(5y-4\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{y+2+5y-4}{2}\\x=\frac{y+2-5y+4}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3y-1\\x=-2y+3\end{matrix}\right.\)

Thay vào pt 2 là được

c/ \(S=\frac{2}{2\sqrt{1}}+\frac{2}{2\sqrt{2}}+\frac{2}{2\sqrt{3}}+...+\frac{2}{2\sqrt{100}}\)

\(S< 1+\frac{2}{1+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{99}+\sqrt{100}}\)

\(S< 1+2\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\right)\)

\(S< 1+2\left(\sqrt{100}-1\right)=19\)

\(S>\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{101}-\sqrt{100}}\)

\(S>2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{101}-\sqrt{100}\right)\)

\(S>2\left(\sqrt{101}-1\right)>2\left(\sqrt{100}-1\right)=18\)

\(\Rightarrow18< S< 19\Rightarrow S\) nằm giữa 2 số tự nhiên liên tiếp nên S không phải số tự nhiên

a) \(\sin x = \frac{{\sqrt 3 }}{2}\;\; \Leftrightarrow \sin x = \sin \frac{\pi }{3}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \pi  - \frac{\pi }{3} + k2\pi }\end{array}} \right.\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \frac{{2\pi }}{3} + k2\pi \;}\end{array}\;} \right.\left( {k \in \mathbb{Z}} \right)\)

b) \(2\cos x =  - \sqrt 2 \;\; \Leftrightarrow \cos x =  - \frac{{\sqrt 2 }}{2}\;\;\; \Leftrightarrow \cos x = \cos \frac{{3\pi }}{4}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{{3\pi }}{4} + k2\pi }\\{x =  - \frac{{3\pi }}{4} + k2\pi }\end{array}\;\;\left( {k \in \mathbb{Z}} \right)} \right.\)

c) \(\sqrt 3 \;\left( {\tan \frac{x}{2} + {{15}^0}} \right) = 1\;\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \frac{1}{{\sqrt 3 }}\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \tan \frac{\pi }{6}\)

\( \Leftrightarrow \frac{x}{2} + \frac{\pi }{{12}} = \frac{\pi }{6} + k\pi \;\;\;\; \Leftrightarrow \frac{x}{2} = \frac{\pi }{{12}} + k\pi \;\;\; \Leftrightarrow x = \frac{\pi }{6} + k\pi \;\left( {k \in \mathbb{Z}} \right)\)

d) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5}\;\;\;\; \Leftrightarrow 2x - 1 = \frac{\pi }{5} + k\pi \;\;\;\; \Leftrightarrow 2x = \frac{\pi }{5} + 1 + k\pi \;\; \Leftrightarrow x = \frac{\pi }{{10}} + \frac{1}{2} + \frac{{k\pi }}{2}\;\;\left( {k \in \mathbb{Z}} \right)\)

3.

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=cos3x\)

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

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

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

câu 2 mình sửa lại đề bài một chút là: sin(cosx)=1 ạ

1.

\(sin\left(sinx\right)=0\)

\(\Leftrightarrow sinx=k\pi\) (1)

Do \(-1\le sinx\le1\Rightarrow-1\le k\pi\le1\)

\(\Rightarrow-\dfrac{1}{\pi}\le k\le\dfrac{1}{\pi}\Rightarrow k=0\) do \(k\in Z\)

Thế vào (1)

\(\Rightarrow sinx=0\Rightarrow x=n\pi\)

2.

\(sin\left(cosx\right)=1\Leftrightarrow cosx=\dfrac{\pi}{2}+k2\pi\)

Do \(-1\le cosx\le1\Rightarrow-1\le\dfrac{\pi}{2}+k2\pi\le1\)

\(\Rightarrow-\dfrac{1}{2\pi}-\dfrac{1}{4}\le k\le\dfrac{1}{2\pi}-\dfrac{1}{4}\) 

\(\Rightarrow\) Không tồn tại k thỏa mãn

Pt vô nghiệm