Đây là 1 lời giải sai em

Đơn giản vì phương trình gốc không thể giải được

Em cảm ơn ạ 

Đk:\(tanx\ne\pm1;tanx\ne0;sin\left(x+\dfrac{\pi}{4}\right)\ne0\)

Pt \(\Leftrightarrow\dfrac{\dfrac{sinx}{cosx}}{1-\dfrac{sin^2x}{cos^2x}}=\dfrac{1}{2}.cotx\left(x+\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow\dfrac{sinx.cosx}{cos^2x-sin^2x}=\dfrac{1}{2}.cotx\left(x+\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow\dfrac{\dfrac{1}{2}.sin2x}{cos2x}=\dfrac{1}{2}.tan\left(\dfrac{\pi}{4}-x\right)\)

\(\Leftrightarrow tan2x=tan\left(\dfrac{\pi}{4}-x\right)\)

\(\Leftrightarrow2x=\dfrac{\pi}{4}-x+k\pi\), k nguyên

\(\Leftrightarrow x=\dfrac{\pi}{12}+k.\dfrac{\pi}{3}\)

Ý D

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Đặt \(x-\dfrac{\pi}{4}=t\Rightarrow x=t+\dfrac{\pi}{4}\Rightarrow3x-\dfrac{\pi}{4}=3\left(t+\dfrac{\pi}{4}\right)-\dfrac{\pi}{4}=3t+\dfrac{\pi}{2}\)

\(\Rightarrow sin\left(3x-\dfrac{\pi}{4}\right)=sin\left(3t+\dfrac{\pi}{4}\right)=cos3t\)

Đồng thời: \(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\)

\(=1-\dfrac{1}{2}sin^22x=1-\dfrac{1}{2}sin^2\left(2t+\dfrac{\pi}{2}\right)=1-\dfrac{1}{2}cos^22t\)

Nên pt trở thành:

\(1-\dfrac{1}{2}cos^22t+cost.cos3t-\dfrac{3}{2}=0\)

\(\Leftrightarrow-1-cos^22t+cos4t+cos2t=0\)

\(\Leftrightarrow-1-cos^22t+2cos^22t-1+cos2t=0\)

\(\Leftrightarrow cos^22t+cos2t-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2t=1\\cos2t=-2\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow2t=k2\pi\)

\(\Leftrightarrow t=k\pi\)

\(\Leftrightarrow x-\dfrac{\pi}{4}=k\pi\)

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

\(\left\{{}\begin{matrix}6u_2+u_5=1\\3u_3+2u_4=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6u_1.q+u_1.q^4=1\\3u_1.q^2+2u_1.q^3=-1\end{matrix}\right.\)

\(\Rightarrow u_1\left(6q+q^4+3q^2+2q^3\right)=0\)

\(\Leftrightarrow q^3+2q^2+3q+6=0\)

\(\Leftrightarrow\left(q+2\right)\left(q^2+3\right)=0\)

\(\Leftrightarrow q=-\text{​​}2\)

\(\Rightarrow u_1=\dfrac{1}{4}\)

\(\Rightarrow u_n=u_1.q^{n-1}=\dfrac{1}{4}.\left(-2\right)^{n-1}=\left(-2\right)^{n-3}\)

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x-2}+1}{\sqrt[]{x+3}-2}=\lim\limits_{x\rightarrow1}\dfrac{\left(\sqrt[3]{x-2}+1\right)\left(\sqrt[3]{\left(x-2\right)^2}-\sqrt[3]{x-2}+1\right)\left(\sqrt[]{x+3}+2\right)}{\left(\sqrt[]{x+3}-2\right)\left(\sqrt[]{x+3}+2\right)\left(\sqrt[3]{\left(x-2\right)^2}-\sqrt[3]{x-2}+1\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(\sqrt[]{x+3}+2\right)}{\left(x-1\right)\left(\sqrt[3]{\left(x-2\right)^2}-\sqrt[3]{x-2}+1\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\sqrt[]{x+3}+2}{\sqrt[3]{\left(x-2\right)^2}-\sqrt[3]{x-2}+1}\)

\(=\dfrac{\sqrt[]{1+3}+2}{\sqrt[3]{\left(1-2\right)^2}-\sqrt[3]{1-2}+1}=\dfrac{4}{3}\)

em cảm ơn ạ

\(\lim\dfrac{3^n+2.6^n}{6^{n-1}+5.4^n}=\lim\dfrac{6^n\left[\left(\dfrac{3}{6}\right)^n+2\right]}{6^n\left[\dfrac{1}{6}+5\left(\dfrac{4}{6}\right)^n\right]}=\lim\dfrac{\left(\dfrac{3}{6}\right)^n+2}{\dfrac{1}{6}+5\left(\dfrac{4}{6}\right)^n}=\dfrac{0+2}{\dfrac{1}{6}+0}=12\)

\(\lim\left(\sqrt{n^2+9}-n\right)=\lim\dfrac{\left(\sqrt{n^2+9}-n\right)\left(\sqrt{n^2+9}+n\right)}{\sqrt{n^2+9}+n}=\lim\dfrac{9}{\sqrt{n^2+9}+n}\)

\(=\lim\dfrac{n\left(\dfrac{9}{n}\right)}{n\left(\sqrt{1+\dfrac{9}{n^2}}+1\right)}=\lim\dfrac{\dfrac{9}{n}}{\sqrt{1+\dfrac{9}{n^2}}+1}=\dfrac{0}{1+1}=0\)

\(\lim\dfrac{\sqrt{15+9n^2}-3}{5-n}=\lim\dfrac{n\sqrt{\dfrac{15}{n^2}+9}-3}{5-n}=\lim\dfrac{n\left(\sqrt{\dfrac{15}{n^2}+9}-\dfrac{3}{n}\right)}{n\left(\dfrac{5}{n}-1\right)}\)

\(=\lim\dfrac{\sqrt{\dfrac{15}{n^2}+9}-\dfrac{3}{n}}{\dfrac{5}{n}-1}=\dfrac{\sqrt{9}-0}{0-1}=-3\)

em cảm ơn ạ