help bài 7 vs huhu
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21 tháng 1
s: \(\dfrac{-21}{46}\cdot\left(-13\right)+\dfrac{3^2}{-9}-\dfrac{-1}{2}\cdot\left(-10\right)\)
\(=\dfrac{21}{46}\cdot13-1-\dfrac{1}{2}\cdot10\)
\(=\dfrac{273}{46}-1-5=\dfrac{273}{46}-5=\dfrac{43}{46}\)
t: \(T=\left(-\dfrac{1}{7}\right)+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2024}\)
=>\(\left(-\dfrac{1}{7}\right)\cdot T=\left(-\dfrac{1}{7}\right)^2+\left(-\dfrac{1}{7}\right)^3+...+\left(-\dfrac{1}{7}\right)^{2025}\)
=>\(\left(-\dfrac{1}{7}\right)\cdot T-T=\left(-\dfrac{1}{7}\right)^2+\left(-\dfrac{1}{7}\right)^3+...+\left(-\dfrac{1}{7}\right)^{2025}-\left(-\dfrac{1}{7}\right)-\left(-\dfrac{1}{7}\right)^2-...-\left(-\dfrac{1}{7}\right)^{2024}\)
=>\(-\dfrac{8}{7}T=\left(-\dfrac{1}{7}\right)^{2025}+\dfrac{1}{7}\)
=>\(-\dfrac{8}{7}\cdot T=-\dfrac{1}{7^{2025}}+\dfrac{1}{7}\)
=>\(-\dfrac{8}{7}\cdot T=\dfrac{-1+7^{2024}}{7^{2025}}\)
=>\(T\cdot\dfrac{8}{7}=\dfrac{-7^{2024}+1}{7^{2025}}\)
=>\(T=\dfrac{-7^{2024}+1}{7^{2025}}:\dfrac{8}{7}=\dfrac{-7^{2024}+1}{7^{2024}}\cdot8\)
u: \(U=\dfrac{1}{5}-\dfrac{1}{5^2}+\dfrac{1}{5^3}-...-\dfrac{1}{5^{2024}}\)
=>\(5\cdot U=1-\dfrac{1}{5}+\dfrac{1}{5^2}-...-\dfrac{1}{5^{2023}}\)
=>\(5U+U=1-\dfrac{1}{5}+\dfrac{1}{5^2}-...-\dfrac{1}{5^{2023}}+\dfrac{1}{5}-\dfrac{1}{5^2}+...-\dfrac{1}{5^{2024}}\)
=>\(6U=1-\dfrac{1}{5^{2024}}=\dfrac{5^{2024}-1}{5^{2024}}\)
=>\(U=\dfrac{5^{2024}-1}{5^{2024}\cdot6}\)
29 tháng 10 2017
= 2^10 . ( 13 + 65 ) / 2^8 . 104
= 2^10 . 78 / 2^8 . 104
= 2 . 2 . 2^8 . 78 / 2^8 . 104
= 2^8 . 312 / 104
= 3
LN
2
N
3 tháng 1 2019
( -12 + 8 - 9 ) - (5 - 12 - 7) -4
= (-12) + 8 - 9 - 5 + 12 + 7 - 4
= (-4) -9 - 5 + 12 + 7 - 4
= (-13) - 5 + 12 + 7 - 4
= (-18) +12 + 7 -4
= (-6) + 7 - 4
= 1 - 4
= -3
3 tháng 1 2019
\(\left(-12+8-9\right)\) -\(\left(5-12-7\right)-4\)
=\(\left(-4-9\right)-\left(-7-7\right)-4\)
\(=-13-\left(-14\right)-4\)
=\(-13+14-4\)
\(=1-4\)
\(=-3\)
15 tháng 12 2023
k: \(\left(4x-16\right)\left(-72+9x\right)=0\)
=>\(4\cdot\left(x-4\right)\cdot9\left(x-8\right)=0\)
=>\(36\left(x-4\right)\left(x-8\right)=0\)
=>\(\left(x-4\right)\left(x-8\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=8\end{matrix}\right.\)
m: \(\left(20+5x\right)\left(4x-8\right)=0\)
=>\(5\cdot\left(x+4\right)\cdot4\left(x-2\right)=0\)
=>\(\left(x+4\right)\left(x-2\right)=0\)
=>\(\left[{}\begin{matrix}x+4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)
n: \(\left(-4x+48\right)\left(2x-24\right)=0\)
=>\(-4\left(x-12\right)\cdot2\left(x-12\right)=0\)
=>\(\left(x-12\right)^2=0\)
=>x-12=0
=>x=12
o: \(\left(4x+16\right)\left(-2x+20\right)\left(-40+x\right)=0\)
=>\(4\cdot\left(x+4\right)\cdot\left(-2\right)\left(x-10\right)\left(x-40\right)=0\)
=>\(\left(x+4\right)\left(x-10\right)\left(x-40\right)=0\)
=>\(\left[{}\begin{matrix}x+4=0\\x-10=0\\x-40=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=10\\x=40\end{matrix}\right.\)
p: \(\left(-5x+40\right)\left(-x+2023\right)\left(2x-2\right)=0\)
=>\(-5\left(x-8\right)\cdot\left(-1\right)\cdot\left(x-2023\right)\cdot2\left(x-1\right)=0\)
=>\(\left(x-8\right)\left(x-2023\right)\left(x-1\right)=0\)
=>\(\left[{}\begin{matrix}x-8=0\\x-1=0\\x-2023=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=1\\x=2023\end{matrix}\right.\)
q: \(2024x\left(4x-8\right)\left(5+5x\right)=0\)
=>\(x\cdot4\left(x-2\right)\cdot5\left(x+1\right)=0\)
=>\(x\left(x-2\right)\left(x+1\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-1\end{matrix}\right.\)
r: \(-4x\left(3x+9\right)\left(2x-16\right)=0\)
=>\(-4x\cdot3\left(x+3\right)\cdot2\left(x-8\right)=0\)
=>\(x\left(x+3\right)\left(x-8\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x+3=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=8\end{matrix}\right.\)
s: \(\left(-100+5x\right)\left(2x-10\right)\left(6x+6\right)=0\)
=>\(5\cdot\left(x-20\right)\cdot2\left(x-5\right)\cdot6\left(x+1\right)=0\)
=>\(\left(x-20\right)\left(x-5\right)\left(x+1\right)=0\)
=>\(\left[{}\begin{matrix}x-20=0\\x-5=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=20\\x=5\\x=-1\end{matrix}\right.\)
t: \(\left(-2x+4\right)\left(2x+16\right)\cdot\left(7-x\right)=0\)
=>\(-2\left(x-2\right)\cdot2\left(x+8\right)\cdot\left(-1\right)\cdot\left(x-7\right)=0\)
=>\(\left(x-2\right)\left(x+8\right)\left(x-7\right)=0\)
=>\(\left[{}\begin{matrix}x-2=0\\x-7=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-8\\x=7\end{matrix}\right.\)
12 tháng 1 2017
Vì : \(\left(x^2+7\right)\left(x^2-27\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2+7=0\\x^2-27=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-7\Rightarrow x\in\varphi\\x^2=27\Rightarrow x\in\varphi\end{cases}}}\)
Vậy ...
12 tháng 1 2017
\(\left(x^2+7\right)\left(x^2-27\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2+7=0\\x^2-27=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x^2=-7\left(vl\right)\\x^2=27\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\sqrt{27}\\x=-\sqrt{27}\end{cases}}\)
vậy \(x=\sqrt{27}\) hay \(x=-\sqrt{27}\)
sau 3 giờ vòi chảy dc : \(\dfrac{1}{5}+\dfrac{2}{7}+\dfrac{11}{35}=\dfrac{7}{35}+\dfrac{10}{35}+\dfrac{11}{35}=\dfrac{7+10+11}{35}=\dfrac{28}{35}\left(b\text{ể}\right)\)
Vậy sau 3 giờ , vòi vẫn chưa đầy bể