\(\frac{2sina+3cosa}{4sina-5cosa}=\frac{\frac{2sina}{cosa}+\frac{3cosa}{cosa}}{\frac{4sina}{cosa}-\frac{5cosa}{cosa}}=\frac{2tana+3}{4tana-5}=\frac{6+3}{12-5}=\frac{9}{7}\)

\(\frac{3sina-2cosa}{5sina+4cos^3a}=\frac{\frac{3sina}{cosa}-\frac{2cosa}{cosa}}{\frac{5sina}{cosa}+\frac{4cos^3a}{cosa}}=\frac{3tana-2}{5tana+4cos^2a}=\frac{3tana-2}{5tana+\frac{4}{1+tan^2a}}=\frac{9-2}{15+\frac{4}{10}}=\frac{5}{11}\)

\(A=\frac{3sina-2cosa}{12sin^3a+4cos^3a}=\frac{\frac{3sina}{sin^3a}-\frac{2cosa}{sin^3a}}{12+\frac{4cos^3a}{sin^3a}}=\frac{3.\frac{1}{sin^2a}-2cota.\frac{1}{sin^2a}}{12+4cot^3a}\)

\(=\frac{3\left(1+cot^2a\right)-2cota\left(1+cot^2a\right)}{12+4cot^3a}=\frac{3\left(1+3^2\right)-2.3.\left(1+3^2\right)}{12+4.3^3}=...\)

4.

Gọi H là chân đường cao kẻ từ C xuống đường thẳng d.

Ta có: \(CH=d\left(C;d\right)=\dfrac{\left|-3.2-4.5+4\right|}{\sqrt{3^2+4^2}}=\dfrac{22}{5}\)

Khi đó: \(S_{ABC}=\dfrac{1}{2}CH.AB=\dfrac{1}{2}.\dfrac{22}{5}.AB=15\Rightarrow AB=\dfrac{75}{11}\)

\(\Rightarrow IA=IB=\dfrac{75}{22}\)

Gọi \(A=\left(4m;3m+1\right)\) là điểm cần tìm.

Ta có: \(IA=\dfrac{75}{22}\Leftrightarrow\sqrt{\left(4m-2\right)^2+\left(3m-\dfrac{3}{2}\right)^2}=\dfrac{75}{22}\)

\(\Leftrightarrow\sqrt{25m^2-25m+\dfrac{25}{4}}=\dfrac{75}{22}\)

\(\Leftrightarrow\left|m-\dfrac{1}{2}\right|=\dfrac{15}{22}\)

\(\Leftrightarrow\left[{}\begin{matrix}m-\dfrac{1}{2}=\dfrac{15}{22}\\m-\dfrac{1}{2}=-\dfrac{15}{22}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{13}{11}\\m=-\dfrac{2}{11}\end{matrix}\right.\)

\(m=\dfrac{13}{11}\Rightarrow A=\left(\dfrac{52}{11};\dfrac{50}{11}\right)\Rightarrow B=\left(-\dfrac{8}{11};\dfrac{5}{11}\right)\)

Vậy \(A=\left(\dfrac{52}{11};\dfrac{50}{11}\right);B=\left(-\dfrac{8}{11};\dfrac{5}{11}\right)\)

1.

\(P=\left(m;m+1\right)\) là điểm cần tìm 

\(\Rightarrow NP=\sqrt{\left(m-3\right)^2+m^2}=\sqrt{2m^2-6m+9}\)

Ta có: \(NM=NP\)

\(\Leftrightarrow\sqrt{\left(-1-3\right)^2+\left(2-1\right)^2}=\sqrt{2m^2-6m+9}\)

\(\Leftrightarrow m^2-3m-4=0\)

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

\(\Rightarrow\left[{}\begin{matrix}P=\left(4;5\right)\\P=\left(-1;0\right)\end{matrix}\right.\)

Vậy \(P=\left(4;5\right)\) hoặc \(P=\left(-1;0\right)\)

2.

\(tanx=-2\Leftrightarrow\dfrac{sinx}{cosx}=-2\Leftrightarrow sinx=-2cosx\)

Khi đó:

\(A=\dfrac{sin^2x+3sinx.cosx-cos^2x+1}{3sin^2x+4sinx.cosx+5cos^2x-2}\)

\(=\dfrac{2sin^2x+3sinx.cosx}{2\left(sin^2x-1\right)+sin^2x+5cos^2x+4sinx.cosx}\)

\(=\dfrac{2sin^2x+3sinx.cosx}{sin^2x+3cos^2x+4sinx.cosx}\)

\(=\dfrac{8cos^2x-3.2cos^2x}{4cos^2x+3cos^2x-4.2cos^2x}\)

\(=\dfrac{2cos^2x}{-cos^2x}=-2\)

d.

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^4x\)

\(tan^4x-3tan^2x-4tanx-3=0\)

\(\Leftrightarrow\left(tan^2x+tanx+1\right)\left(tan^2x-tanx-3\right)=0\)

\(\Leftrightarrow tan^2x-tanx-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1-\sqrt{13}}{2}\\tanx=\frac{1+\sqrt{13}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=arctan\left(\frac{1-\sqrt{13}}{2}\right)+k\pi\\x=arctan\left(\frac{1+\sqrt{13}}{2}\right)+k\pi\end{matrix}\right.\)

mọi người giúp hộ mình nhanh với

a.

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(2tan^3x+4=3tanx\left(1+tan^2x\right)\)

\(\Leftrightarrow2tan^3x+4=3tanx+3tan^3x\)

\(\Leftrightarrow tan^3x+3tanx-4=0\)

\(\Leftrightarrow\left(tanx-1\right)\left(tan^2x+tanx+4\right)=0\)

\(\Leftrightarrow tanx=1\Rightarrow x=\frac{\pi}{4}+k\pi\)

\(cos^4a+sin^4a-6sin^2a.cos^2a\)

\(=cos^4a+sin^4a-2sin^2a.cos^2a-4sin^2a.cos^2a\)

\(=\left(cos^2a-sin^2a\right)^2-\left(2sina.cosa\right)^2\)

\(=cos^22a-sin^22a\)

\(=cos4a\)

Lời giải:

$\cos ^2a=1-\sin ^2a=1-(\frac{1}{2})^2=\frac{3}{4}$

$\Rightarrow \cos a=\pm \frac{\sqrt{3}}{2}$

Nếu $\cos a=\frac{\sqrt{3}}{2}$ thì:

$A=3\sin a+4\cos a=3.\frac{1}{2}+4.\frac{\sqrt{3}}{2}=\frac{3+4\sqrt{3}}{2}$

Nếu $\cos a=\frac{-\sqrt{3}}{2}$ thì:

$A=3\sin a+4\cos a=3.\frac{1}{2}+4.\frac{-\sqrt{3}}{2}=\frac{3-4\sqrt{3}}{2}$

:v bn ns v là bn bik hết là dạng gì rr mà lm ko đc á :))

\(3sin^4x-\left(1-sin^2x\right)^2=\frac{1}{2}\Leftrightarrow3sin^4x-\left(sin^4x-2sin^2x+1\right)=\frac{1}{2}\)

\(\Leftrightarrow2sin^4x+2sin^2x-\frac{3}{2}=0\) \(\Rightarrow\left[{}\begin{matrix}sin^2x=\frac{1}{2}\\sin^2x=-\frac{3}{2}< 0\left(l\right)\end{matrix}\right.\)

\(\Rightarrow cos^2x=1-\frac{1}{2}=\frac{1}{2}\)

\(\Rightarrow B=\left(\frac{1}{2}\right)^2+3\left(\frac{1}{2}\right)^2=1\)

\(4sin^4x+3\left(1-sin^2x\right)^2=\frac{7}{4}\Leftrightarrow4sin^4x+3\left(sin^4x-2sin^2x+1\right)=\frac{7}{4}\)

\(\Leftrightarrow7sin^4x-6sin^2x+\frac{5}{4}=0\Rightarrow\left[{}\begin{matrix}sin^2x=\frac{1}{2}\Rightarrow cos^2x=\frac{1}{2}\\sin^2x=\frac{5}{14}\Rightarrow cos^2x=\frac{9}{14}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}C=3\left(\frac{1}{2}\right)^2+4\left(\frac{1}{2}\right)^2=\frac{7}{4}\\C=3\left(\frac{5}{14}\right)^2+4\left(\frac{9}{14}\right)^2=\frac{57}{28}\end{matrix}\right.\)