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Bài giải
a, \(x=-15\text{ }\Rightarrow\text{ }\left|x\right|=\left|-15\right|=15\)
b, \(x=-\frac{3}{5}\text{ }\Rightarrow\text{ }\left|x\right|=\left|-\frac{3}{5}\right|=\frac{3}{5}\)
c, \(x=-0,345\text{ }\Rightarrow\text{ }\left|x\right|=\left|-0,345\right|=0,345\)
d, \(x=2\frac{3}{5}=\frac{13}{5}\text{ }\Rightarrow\text{ }\left|x\right|=\left|\frac{13}{5}\right|=\frac{13}{5}\)
e, \(x=-\frac{5}{-7}=\frac{5}{7}\text{ }\Rightarrow\text{ }\left|x\right|=\left|\frac{5}{7}\right|=\frac{5}{7}\)
Bài làm :
\(a\text{ )}x=-15\text{ }\Rightarrow\text{ }\left|x\right|=\left|-15\right|=15\)
\(b\text{ )}x=-\frac{3}{5}\text{ }\Rightarrow\text{ }\left|x\right|=\left|-\frac{3}{5}\right|=\frac{3}{5}\)
\(c\text{ )}x=-0,345\text{ }\Rightarrow\text{ }\left|x\right|=\left|-0,345\right|=0,345\)
\(d\text{ )}x=2\frac{3}{5}=\frac{13}{5}\text{ }\Rightarrow\text{ }\left|x\right|=\left|\frac{13}{5}\right|=\frac{13}{5}\)
\(e\text{ )}x=-\frac{5}{-7}=\frac{5}{7}\text{ }\Rightarrow\text{ }\left|x\right|=\left|\frac{5}{7}\right|=\frac{5}{7}\)
ÁP DỤNG TÍNH CHẤT DÃY TỈ SỐ BẰNG NHAU, TA ĐƯỢC :
`(x)/(3)=(y)/(4)=(x+y)/(3+4)=(90)/(7)`
`->` $\begin{cases}x=\dfrac{90}{7}.3=\dfrac{30}{7} \\ y=\dfrac{90}{7}.4=\dfrac{360}{7} \end{cases}$
1)\(\dfrac{x}{5}=\dfrac{y}{3}\) áp dụng...ta đc:
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x-y}{5-3}=\dfrac{20}{2}=10\)
x=50
y=30
\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{32}\left(1+2+3+...+32\right)\)
\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3\left(3+1\right)}{2}+....+\frac{1}{32}.\frac{32.\left(32+1\right)}{2}\)
\(=1+\frac{2+1}{2}+\frac{3+1}{2}+....+\frac{32+1}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+....+\frac{33}{2}\)
\(\frac{2+3+4+....+33}{2}\)
\(=\frac{\frac{33\left(33+1\right)}{2}-1}{2}=280\)
Ta có 12 + 22 + 32 + …102 = 385
Suy ra ( 12 +22 + 32 +…+102 ) .32 = 385.32
Do đó ta tính được A = 32 + 62 + 92 + …+302 = 3465
a) \(A=2+2^2+2^3+...+2^{2017}\)
\(2A=2^2+2^3+2^4+...+2^{2018}\)
\(2A-A=\left(2^2+2^3+2^4+...+2^{2018}\right)-\left(2+2^2+2^3+...+2^{2017}\right)\)
\(A=2^{2018}-2\)
b) \(C=1+3^2+3^4+...+3^{2018}\)
\(3^2\cdot C=3^2+3^4+3^6+...+3^{2020}\)
\(9C-C=\left(3^2+3^4+3^6+...+3^{2020}\right)-\left(1+3^2+3^4+...+3^{2018}\right)\)
\(8C=3^{2020}-1\)
\(\Rightarrow C=\dfrac{3^{2020}-1}{8}\)
\(Toru\)