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b)\(\left(x-8\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x-8=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=2\end{cases}}\)
c) \(\left(x+1\right)+\left(x+2\right)+...+\left(x+10\right)=9x+200\)
\(\Leftrightarrow\left(x+x+...+x\right)+\left(1+2+...+10\right)=9x+200\) (10 số hạng x)
\(\Leftrightarrow10x+55=9x+200\Leftrightarrow x+55=200\)
\(\Leftrightarrow x=145\)
A. x = 2
B. \(\dfrac{3}{8}=\dfrac{6}{x}\)\(\Leftrightarrow x=\dfrac{6.8}{3}=16\)
C. x = 3
D. \(x=\dfrac{4.6}{8}=3\)
E. \(x=\dfrac{7}{3}\)
G.\(\dfrac{14}{13}=\dfrac{28}{10-x}\)
<=>\(14\left(10-x\right)=364\)
<=> 10 - x = 26
<=> x = -16
H. \(3\left(x+2\right)=4\left(x-5\right)\)
<=> 3x + 6 = 4x - 20
<=> -x = -26
<=> x = 26
K. \(\dfrac{x}{2}=\dfrac{8}{x}\)
<=> \(x^2=16\)
<=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
M. \(\left(x-2\right)^2=100\)
<=> \(\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)
a=2
b=16
c=3
d=3
mik chỉ biết thế này thôi(ko chắc đúng=3)
a)\(\frac{7}{13}.\frac{7}{15}-\frac{5}{12}.\frac{21}{39}+\frac{49}{21}.\frac{8}{15}\)
=\(\frac{7}{13}.\frac{7}{15}-\frac{5}{12}.\frac{7}{13}+\frac{7}{13}.\frac{8}{15}\)
=\(\frac{7}{13}.\left(\frac{7}{15}-\frac{5}{12}-\frac{8}{15}\right)\)
=\(\frac{7}{13}.\frac{7}{12}\)
=\(\frac{49}{156}\)
b)\(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
=\(a.\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
=a . 0
=0
Bài 2
a)Có
\(3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)
\(2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)
Vì 8<9 =>\(8^{100}< 9^{100}\) =>\(3^{200}>2^{300}\)
\(a,\dfrac{2}{3}.x=\dfrac{2}{7}\\ x=\dfrac{2}{7}:\dfrac{2}{3}=\dfrac{3}{7}\\ ---\\ b,x.\dfrac{3}{5}=\dfrac{2}{5}\\ x=\dfrac{2}{5}:\dfrac{3}{5}=\dfrac{2}{3}\\ ---\\ c,x:\dfrac{8}{13}=\dfrac{13}{7}\\x=\dfrac{13}{7}.\dfrac{8}{13}=\dfrac{8}{7}\\ ----\\ d,\dfrac{3}{2}:x=\dfrac{7}{4}\\ x=\dfrac{3}{2}:\dfrac{7}{4}=\dfrac{3}{2}.\dfrac{4}{7}=\dfrac{6}{7}\)