\(cos\alpha=\dfrac{3}{5}=>cos^2\alpha=\dfrac{9}{25}\)

có \(sin^2\alpha+cos^2\alpha=1< =>sin^2\alpha+\dfrac{9}{25}=1=>sin\alpha=\dfrac{4}{5}\left(sin\alpha>0\right)\)

\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{4}{5}}{\dfrac{3}{5}}=\dfrac{4}{3}\)

\(cot\alpha\cdot tan\alpha=1=>cot\alpha\cdot\dfrac{4}{3}=1=>cot\alpha=\dfrac{3}{4}\)

`(3x-1)^2=25`

`=>3x-1=5` hoặc `3x-1=-5`

`=>x=2` hoặc `x=-4/3`

\(1+\left(x-1\right)^2+\left(x-1\right)^4+...+\left(x-1\right)^{2020}=\dfrac{17^{2022}-1}{\left(x-1\right)^2-1}\left(đk:x>2\right)\)

đặt 

\(A=1+\left(x-1\right)^2+\left(x-1\right)^4+...+\left(x-1\right)^{2020}\)

\(\left(x-1\right)^2A=\left(x-1\right)^2+\left(x-1\right)^4+\left(x-1\right)^6+...+\left(x-1\right)^{2022}\)

\(\left(x-1\right)^2A-A=\left[\left(x-1\right)^2+\left(x-1\right)^4+\left(x-1\right)^6+...+\left(x-1\right)^{2022}\right]-\left[1+\left(x-1\right)^2+\left(x-1\right)^4+...+\left(x-1\right)^{2020}\right]\)

\(\left[\left(x-1\right)^2-1\right]A=\left(x-1\right)^{2022}-1\)

\(A=\dfrac{\left(x-1\right)^{2022}-1}{\left(x-1\right)^2-1}\)

\(=>\dfrac{\left(x-1\right)^{2022}-1}{\left(x-1\right)^2-1}=\dfrac{17^{2022}-1}{\left(x-1\right)^2-1}\\ =>\left(x-1\right)^{2022}-1=17^{2022}-1\\ =>\left(x-1\right)^{2022}=17^{2022}\\ =>x-1=17\\ =>x=18\left(tm\right)\)

a)

`x^2+3x+2=0`

`<=>x^2+2x+x+2=0`

`<=>x(x+2)+(x+2)=0`

`<=>(x+2)(x+1)=0`

`<=>x+2=0` hoặc `x+1=0`

`<=>x=-2` hoặc `x=-1`

b)

\(\left\{{}\begin{matrix}x-3y=5\\3x+2y=4\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}3x-9y=15\\3x+2y=4\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}-11y=11\\x-3y=5\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}y=-1\\x-3\cdot\left(-1\right)=5\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}y=-1\\x=2\end{matrix}\right.\)

a)

`2/7` của số đó là `145` '

`=>` số đó là `145:2/7=507,5`

b)

số đó là

`-36:3/8=-96`

=)))

ừ cháu=))

già phải ôn thi nên ko chăm như mấy kì trước được=))