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\(a,4\frac{5}{9}:\frac{\left(-5\right)}{7}+\frac{4}{9}:\frac{-5}{7}\)
\(=\frac{41}{9}.\frac{-7}{5}+\frac{4}{9}.\frac{-7}{5}\)
\(=\frac{-7}{5}.\left(\frac{41}{9}+\frac{4}{9}\right)\)
\(=-\frac{7}{9}.5\)
\(=-7\)
a)Bn Kaito Kid làm rùi!
B)Không viết lại đề
\(=\frac{11}{7}\cdot\left(-\frac{3}{5}+\frac{4}{9}-\frac{2}{5}+\frac{5}{9}\right)=\frac{11}{7}\cdot0=0\)
c)Không viết lại đề
\(A=\left(2+4+...+100\right)\left(\frac{3}{5}\cdot\frac{10}{7}-\frac{6}{7}\right):\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(2+4+6+...+100\right)\cdot0\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)=0\)
\(=\frac{7}{6}\cdot\left(\frac{3}{26}-\frac{3}{13}+\frac{1}{10}-\frac{8}{5}\right)=\frac{7}{6}\left(\frac{-3}{26}+\frac{-17}{10}\right)=\frac{7}{6}\cdot\frac{236}{130}=\frac{413}{195}\)
D)
a) \(\frac{3}{5}.x-\frac{1}{5}=\frac{4}{5}\)
\(\Leftrightarrow\frac{3}{5}.x=\frac{4}{5}+\frac{1}{5}\)
\(\Leftrightarrow\frac{3}{5}.x=1\)
\(\Leftrightarrow x=1:\frac{3}{5}\)
\(\Leftrightarrow x=\frac{5}{3}\)
Vậy : \(x=\frac{5}{3}\)
b) \(\frac{4}{7}+\frac{5}{7}:x=1\)
\(\Leftrightarrow\frac{5}{7}:x=1-\frac{4}{7}\)
\(\Leftrightarrow\frac{5}{7}:x=\frac{3}{7}\)
\(\Leftrightarrow x=\frac{5}{7}:\frac{3}{7}\)
\(\Leftrightarrow x=\frac{5}{3}\)
Vậy : \(x=\frac{5}{3}\)
c) \(-\frac{12}{7}.\left(\frac{3}{4}-x\right).\frac{1}{4}=-1\)
\(\Leftrightarrow\frac{-12.1}{7.4}.\left(\frac{3}{4}-x\right)=-1\)
\(\Leftrightarrow-\frac{3}{7}.\left(\frac{3}{4}-x\right)=-1\)
\(\Leftrightarrow\frac{3}{4}-x=-1:\left(-\frac{3}{7}\right)\)
\(\Leftrightarrow\frac{3}{4}-x=\frac{7}{3}\)
\(\Leftrightarrow x=\frac{3}{4}-\frac{7}{3}=-\frac{19}{12}\)
Vậy : \(x=-\frac{19}{12}\)
d) \(x:\frac{17}{8}=-\frac{2}{5}.-\frac{9}{17}+3\)
\(\Leftrightarrow x:\frac{17}{8}=\frac{273}{85}\)
\(\Leftrightarrow x=\frac{273}{85}.\frac{17}{8}\)
\(\Leftrightarrow x=\frac{273}{40}\)
Vậy : \(x=\frac{273}{40}\)
\(\)
a)Ta có : B = (1-\(\frac{z}{x}\))(1-\(\frac{x}{y}\))(1+\(\frac{y}{z}\))
=> B=\(\frac{x-z}{x}\).\(\frac{y-x}{y}\).\(\frac{z+y}{z}\)
Từ : x-y-z = 0
=>x – z = y; y – x = – z và y + z = x
b) Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}=\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}\)
\(=\frac{12x-8y+6z-12x+8y-6z}{16+9+4}=\frac{0}{16+9+4}=0\)
\(\left\{\begin{matrix}\frac{12x-8y}{16}=0\\\frac{6z-12x}{9}=0\\\frac{8y-6z}{4}=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}12x-8y=0\\6z-12x=0\\8y-6z=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}12x=8y\\6z=12x\\8y=6z\end{matrix}\right.\Rightarrow12x=8y=6z\)
\(\Rightarrow\frac{12x}{24}=\frac{8y}{24}=\frac{6z}{24}\)
\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\left(đpcm\right)\)
d: \(D=-8\cdot\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-8\cdot\dfrac{1}{2}:\dfrac{27-14}{12}\)
\(=-4:\dfrac{13}{12}\)
\(=-4\cdot\dfrac{12}{13}=-\dfrac{48}{13}\)
e: \(E=5\cdot4-4\cdot3+5-0.3\cdot20\)
=20-12+5-6
=8+5-6
=13-6=7
f: \(F=\dfrac{9}{4}+\dfrac{5}{6}-\dfrac{3}{2}:6\)
\(=\dfrac{9}{4}+\dfrac{5}{6}-\dfrac{3}{12}\)
\(=\dfrac{27}{12}+\dfrac{10}{12}-\dfrac{3}{12}=\dfrac{34}{12}=\dfrac{17}{6}\)
a) 4. ( 1.1/4)2 + [(3/4)2 : (5/4)3] : (3/2)3
= 4.1/16 + [9/16 : 125/64] : 27/8
= \(\frac{1}{4}+\frac{9}{16}:\frac{125}{64}:\frac{27}{8}=\frac{1}{4}+\frac{36}{125}:\frac{27}{8}\)
= \(\frac{1}{4}+\frac{36}{125}.\frac{8}{27}\)
=\(\frac{1}{4}+\frac{32}{375}=\frac{375}{1500}+\frac{128}{1500}=\frac{503}{1500}\)
b] = 2^3 + 3 x 1 - 1 + ( 2^2 x 2 ) x 2^3
= 2^3 + 3 - 1 + 2^3 x 2^3
= 2^3 + 2 + 2^6 = 74
a] = 4 x ( 1/4 )2 + ( 32/42 : 53/43 ) : 27/8
= 4 x 1/16 + ( 32 x 4/53 ) x 8/27
= 1/4 + 36/53 x 8/27 = 1/4 + 4/125 x 8/3 = 503/1500 sấp sỉ 0,335333
Bấm máy tính:
E = \(\frac{4}{3}+\frac{1}{4}+\frac{3}{5}:\frac{4}{5}\)
E = \(\frac{4}{3}+\frac{1}{4}+\frac{3}{4}\)
E = \(\frac{7}{3}\)
Vậy E = \(\frac{7}{3}\)
\(=\frac{1}{3}\)
3 1/4:8/5=0,25:x
13/4:8/5=1/4:x
1/4:x =65/12
x =1/4:65/12
x =3/65