a ) 1 4 + 3 4 + 2 8 + − 5 13 + 8 13 = 2 8

b ) − 7 31 + 7 31 + 24 19 + − 5 19 + 1 15 = 16 15 c ) − 46 43 + 3 43 + 5 17 + 11 17 + 1 17 = 0 d ) 21 31 + 10 31 + 44 53 + 9 53 + − 16 7 = − 2 7 e ) 15 16 + − 15 16 + 13 33 + 7 33 + 10 33 + 3 33 = 1

\(A=-\frac{7}{813}+496.-\frac{7}{813}+-\frac{7}{813}.316\)

\(A=-\frac{7}{813}.\left(1+496+316\right)\)

\(A=-\frac{7}{813}.813\)

\(A=-7\)

C=-7/813+496 x (-7/813)+(-7/813) x 316

  =-7/813 x (1+496+316)

=-7/813 x 813

=-7

\(C=\frac{-7}{813}+496.\frac{-7}{813}+\frac{-7}{813}.316\)

\(C=\frac{-7}{813}.1+496.\frac{-7}{813}+\frac{-7}{813}.316\)

\(C=\frac{-7}{813}.\left(1+496+316\right)\)

\(C=\frac{-7}{813}.813\)

\(C=\frac{-5691}{813}=-7\)

\(C=\frac{-7}{813}+496\cdot\frac{-7}{813}+\frac{-7}{813}\cdot316\)

\(C=\frac{-7}{813}\cdot1+496\cdot\frac{-7}{813}+\frac{-7}{813}\cdot316\)

\(C=\frac{-7}{813}\cdot\left[1+496+316\right]\)

\(C=\frac{-7}{813}\cdot813=-7\)

= \(\frac{-7}{813}.\left(1+496+316\right)=\frac{-7}{813}.813=-7\)

1) \(x:y:z=2:3:4\) ⇒ \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\)

⇒ x=4;y=6;z=8

\(1,\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

Áp dụng t/c dtsbn

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot2=4\\y=2\cdot3=6\\z=2\cdot4=8\end{matrix}\right.\)

\(2,\) Áp dụng t/c dtsbn

\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{3y}{-9}=\dfrac{2z}{8}=\dfrac{4x-3y-2z}{8-\left(-9\right)-8}=\dfrac{81}{9}=9\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot\left(-3\right)=-6\\z=2\cdot4=8\end{matrix}\right.\)

\(3,4y=3z\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{6}=\dfrac{z}{8};\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{9}=\dfrac{y}{6}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}\)

Áp dụng t/c dtsbn

\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}=\dfrac{x+y+z}{9+6+8}=\dfrac{46}{23}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot6=12\\z=2\cdot8=16\end{matrix}\right.\)

\(4,5x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\Rightarrow\dfrac{x}{9}=\dfrac{y}{15};\dfrac{y}{z}=\dfrac{3}{2}\Rightarrow\dfrac{y}{3}=\dfrac{z}{2}\Rightarrow\dfrac{y}{15}=\dfrac{z}{10}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{2x}{18}=\dfrac{3y}{45}=\dfrac{4z}{40}=\dfrac{2x+3y-4z}{18+45-40}=\dfrac{34}{23}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{34}{23}\cdot9=\dfrac{306}{23}\\y=\dfrac{34}{23}\cdot15=\dfrac{510}{23}\\z=\dfrac{34}{23}\cdot10=\dfrac{340}{23}\end{matrix}\right.\)

5\(^{x+1}\) - 5\(^x\) = 2.2⁸ + 8

5\(^x\).(5 - 1) = 520

5\(^x\).4        = 520

5\(^x\)           = 520 : 4

5\(^x\)           = 130

Với \(x\) = 0 ⇒ 5\(^x\) = 5⁰ = 1 < 130 (loại)

Với \(x\) > 0 ⇒ 5\(^x\) = \(\overline{...5}\) \(\ne\) 130 (loại)

Vậy \(x\) \(\in\) \(\varnothing\) 

\(5^{x+1}-5^x=2.2^8+8\\ 5^x\left(5-1\right)=512+8\\ 5^x.4=520\\ 5^x=\dfrac{520}{4}=130\)

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