HOC24
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Môn học
Chủ đề / Chương
Bài học
`8^7:8^4:8` (đề như thế này phải không ạ?)
`=8^(7-4-1)`
`=8^2`
`=64`
`2^x-26=6`
`2^x=6+26`
`2^x=32`
`2^x=2^5`
`=>x=5`
\(cos\alpha=\dfrac{3}{5}=>cos^2\alpha=\dfrac{9}{25}\)
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)
\(\left\{{}\begin{matrix}x-3y=5\\3x+2y=4\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}3x-9y=15\\3x+2y=4\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}-11y=11\\x-3y=5\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}y=-1\\x-3\cdot\left(-1\right)=5\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}y=-1\\x=2\end{matrix}\right.\)
`2/7` của số đó là `145` '
`=>` số đó là `145:2/7=507,5`
số đó là
`-36:3/8=-96`
=)))