tìm X:
a)4/5x8/3xX/7=96/105
b)7/4x3/Xx11/5=231/200
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29 tháng 5 2020
A. \(\frac{4}{5}\times\frac{8}{3}\times\frac{x}{7}=\frac{96}{105}\)
=\(\frac{4}{5}\times\left(\frac{8}{3}\times\frac{x}{7}\right)=\frac{96}{105}\)
= \(\frac{8}{3}\times\frac{x}{7}=\frac{96}{105}\div\frac{4}{5}\)
= \(\frac{8}{3}\times\frac{x}{7}=\frac{480}{420}=\frac{8}{7}\)
=\(\frac{x}{7}=\frac{8}{7}\div\frac{8}{3}\)
= \(X=\frac{24}{56}=\frac{3}{7}\)
\(X=3\)
@Moon
1 tháng 3 2018
a)\(\frac{4}{5}\cdot\frac{8}{3}\cdot\frac{x}{7}=\frac{96}{105}\)
\(\frac{4}{5}\cdot\frac{8}{3}\cdot\frac{x}{7}=\frac{32}{35}\)
\(\frac{4\cdot8\cdot x}{5\cdot3\cdot7}=\frac{32}{35}\)
\(\frac{32\cdot x}{105}=\frac{32}{35}\)
\(\Rightarrow32\cdot x\cdot35=32\cdot105\)( chỗ này là nhân tích chéo nha )
\(\Rightarrow x\cdot35=105\)
\(x=105:35\)
\(x=3\)Vậy x=3
b)\(\frac{7}{4}\cdot\frac{3}{x}\cdot\frac{11}{5}=\frac{231}{200}\)
\(\frac{7\cdot3\cdot11}{4\cdot x\cdot5}=\frac{231}{200}\)
\(\frac{231}{20\cdot x}=\frac{231}{200}\)
\(\Rightarrow20\cdot x=200\)
\(x=200:20\)
\(x=10\)Vậy x=10
1 tháng 3 2018
a) \(\frac{4}{5}\times\frac{8}{3}\times\frac{x}{7}=\frac{96}{105}\)
\(\frac{32}{15}\times\frac{x}{7}=\frac{32}{35}\)
\(\frac{x}{7}=\frac{32}{35}:\frac{32}{15}\)
\(\frac{x}{7}=\frac{3}{7}\)
\(\Rightarrow x=3\)
b) \(\frac{7}{4}\times\frac{3}{x}\times\frac{11}{5}=\frac{231}{200}\)
\(\frac{7}{4}\times\frac{11}{5}\times\frac{3}{x}=\frac{231}{200}\)
\(\frac{77}{20}\times\frac{3}{x}=\frac{231}{200}\)
\(\frac{3}{x}=\frac{231}{200}:\frac{77}{20}\)
\(\frac{3}{x}=\frac{3}{10}\)
\(\Rightarrow x=10\)
CHÚC BẠN HỌC TỐT!!!!!!!!!!
KK
8 tháng 9 2021
a. \(\dfrac{5}{7}+\dfrac{4}{3}:x=\dfrac{1}{7}\)
<=> \(\dfrac{5}{7}+\dfrac{4}{3}.\dfrac{1}{x}=\dfrac{1}{7}\)
<=> \(\dfrac{5}{7}+\dfrac{4}{3x}=\dfrac{1}{7}\) ĐKXĐ: x \(\ne\) 0
<=> \(\dfrac{15x}{21x}+\dfrac{28}{21x}=\dfrac{3x}{21x}\)
<=> 15x + 28 = 3x
<=> 15x - 3x = -28
<=> 12x = -28
<=> x = \(\dfrac{-28}{12}=-\dfrac{7}{3}\)
b. \(\dfrac{5}{3}x.\dfrac{-1}{4}=\dfrac{2}{6}\)
<=> \(\dfrac{-5x}{12}=\dfrac{2}{6}\)
<=> -5x . 6 = 12 . 2
<=> -30x = 24
<=> x = \(-\dfrac{4}{5}\)
7 tháng 3 2022
\(a,x-\dfrac{1}{4}=\dfrac{7}{2}.\dfrac{-3}{5}\\ \Rightarrow x-\dfrac{1}{4}=\dfrac{-21}{10}\\ \Rightarrow x=\dfrac{-21}{10}+\dfrac{1}{4}\\ \Rightarrow x=\dfrac{-37}{20}\\ b,\dfrac{x}{134}=\dfrac{9}{7}.\dfrac{5}{-11}\\ \Rightarrow\dfrac{x}{134}=\dfrac{-45}{77}\\ \Rightarrow x=\dfrac{-45}{77}.134\\ \Rightarrow x=\dfrac{-6030}{77}\)
O
7 tháng 3 2022
\(x-\dfrac{1}{4}=\dfrac{7}{2}\cdot\dfrac{-3}{5}\)
\(x-\dfrac{1}{4}=\dfrac{-21}{10}\)
\(x=\dfrac{-21}{10}+\dfrac{1}{4}\)
\(x=\dfrac{-37}{20}\)
b ) \(\dfrac{x}{134}=\dfrac{9}{7}\cdot\dfrac{5}{-11}\)
\(\dfrac{x}{134}=\dfrac{9}{7}\cdot\dfrac{-5}{11}\)
\(\dfrac{x}{134}=\dfrac{-45}{77}\)
\(x=\dfrac{-45}{77}\cdot134\)
\(x=-\dfrac{6034}{77}\)
4 tháng 9 2021
a) \(\left(3x+5\right)\left(7-2x\right)+6x\left(x+4\right)=0\)
\(\Leftrightarrow21x-6x^2+35-10x+6x^2+24x=0\)
\(\Leftrightarrow35x=-35\Leftrightarrow x=-1\)
b) \(x^3-25x=0\)
\(\Leftrightarrow x\left(x^2-25\right)=0\)
\(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
4 tháng 9 2021
a: Ta có: \(\left(3x+5\right)\left(7-2x\right)+6x\left(x+4\right)=0\)
\(\Leftrightarrow21x-6x^2+35-10x+6x^2+24x=0\)
\(\Leftrightarrow x=1\)
b: Ta có: \(x^3-25x=0\)
\(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
a)\(\frac{4}{5}\times\frac{8}{3}\times\frac{x}{7}=\frac{96}{105}\)
\(\frac{x}{7}=\frac{96}{105}:\frac{4}{5}:\frac{8}{3}=\frac{3}{7}\)
\(x=3\)
b) \(\frac{7}{4}\times\frac{3}{x}\times\frac{11}{5}=\frac{231}{200}\)
\(\frac{3}{x}=\frac{231}{200}:\frac{7}{4}:\frac{11}{5}=\frac{3}{10}\)
\(x=10\)