Tìm x, biết:
a) 4 , 5 − 2. x .1 4 7 = 11 14
b) x 4 − x 5 = − 1 1 2
K
Khách
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8 tháng 9 2021
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
8 tháng 9 2021
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
AH
Akai Haruma
Giáo viên
9 tháng 5 2021
Lời giải:
a)
$\frac{4}{7}x=\frac{2}{3}+\frac{1}{5}=\frac{13}{15}$
$x=\frac{13}{15}:\frac{4}{7}=\frac{91}{60}$
b)
$\frac{5}{7}:x=\frac{1}{6}-\frac{4}{5}$
$\frac{5}{7}:x=\frac{-19}{30}$
$x=\frac{5}{7}:\frac{-19}{30}=\frac{-150}{133}$
9 tháng 5 2021
a) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{1}{5}\)
\(\dfrac{4}{7}.x=\dfrac{1}{5}+\dfrac{2}{3}\)
\(\dfrac{4}{7}.x=\dfrac{13}{15}\)
\(x=\dfrac{13}{15}:\dfrac{4}{7}\)
\(x=\dfrac{91}{60}\)
b) \(\dfrac{4}{5}+\dfrac{5}{7}:x=\dfrac{1}{6}\)
\(\dfrac{5}{7}:x=\dfrac{1}{6}-\dfrac{4}{5}\)
\(\dfrac{5}{7}:x=\dfrac{-19}{30}\)
\(x=\dfrac{5}{7}:\dfrac{-19}{30}\)
\(x=\dfrac{-150}{133}\)
16 tháng 7 2021
`a)(4,5-2x)*1 4/7=11/14`
`=>(4,5-2x)*11/7=11/14`
`=>4,5-2x=1/2`
`=>2x=4,5-0,5=4`
`=>x=2`
Vậy `x=2`
`b)(2,8x-32):2/3=-90`
`=>2,8x-32=-90*2/3=-60`
`=>2,8x=-28`
`=>x=-10`
Vậy `x=-10`
HN
3
TG
18 tháng 6 2021
a) \(x+1^3=2^5-\left(-1^3\right)\)
\(\Rightarrow x+1=33\)
=> x = 32
b) \(3^7-x=1^4-\left(-3^5\right)\)
\(\Rightarrow2187-x=1+243=244\)
=> x = 1943
2 tháng 8 2020
a ) \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\left|\frac{4}{35}-\frac{\left(-11\right)}{70}\right|\)
=> \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\left|\frac{4}{35}+\frac{11}{70}\right|\)
=> \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\left|\frac{19}{70}\right|\)
=> \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\frac{19}{70}=\frac{3}{35}\)
=> \(\frac{2}{5}+x+\frac{3}{2}=\frac{3}{7}-\frac{3}{35}=\frac{12}{35}\)
=> \(\frac{2}{5}+x=\frac{12}{35}-\frac{3}{2}=-\frac{81}{70}\)
=> \(x=-\frac{81}{70}-\frac{2}{5}=-\frac{109}{70}\)
b) \(\frac{3}{4}\left(x-8\right)=\frac{5}{7}\left(4-\frac{1}{2}\right)\)
=> \(\frac{3}{4}x-6=\frac{5}{2}\)
=> \(\frac{3}{4}x=\frac{17}{2}\)
=> \(x=\frac{17}{2}:\frac{3}{4}=\frac{34}{3}\)
Câu c,d tự làm nhé
KN
2 tháng 8 2020
a. \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\left|\frac{4}{35}-\frac{-11}{70}\right|\)
\(\Rightarrow\frac{3}{7}-\left(\frac{19}{10}+x\right)=\frac{5}{14}-\left|\frac{4}{35}+\frac{11}{70}\right|\)
\(\Rightarrow\frac{3}{7}-\frac{19}{10}-x=\frac{5}{14}-\left|\frac{19}{70}\right|=\frac{5}{14}-\frac{19}{70}\)
\(\Rightarrow-\frac{103}{70}-x=\frac{3}{35}\)
\(\Rightarrow x=-\frac{103}{70}-\frac{3}{35}\)
\(\Rightarrow x=-\frac{109}{70}\)
b. \(\frac{3}{4}\left(x-8\right)=\frac{5}{7}\left(4-\frac{1}{2}\right)\)
\(\Rightarrow\frac{3}{4}\left(x-8\right)=\frac{5}{7}.\frac{7}{2}=\frac{5}{2}\)
\(\Rightarrow x-8=\frac{10}{3}\)
\(\Rightarrow x=\frac{34}{3}\)
c. \(\frac{3}{2}-4\left(\frac{1}{4}-x\right)=\frac{2}{3}-7x\)
\(\Rightarrow\frac{3}{2}-1+4x=\frac{2}{3}-7x\)
\(\Rightarrow\frac{1}{2}=\frac{2}{3}-7x-4x=\frac{2}{3}-11x\)
\(\Rightarrow11x=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{66}\)
d. \(4\left(\frac{1}{2}-x\right)-5\left(x-\frac{3}{10}\right)=\frac{7}{4}\)
\(\Rightarrow2-4x-5x+\frac{3}{2}=\frac{7}{4}\)
\(\Rightarrow2-9x=\frac{1}{4}\)
\(\Rightarrow9x=\frac{7}{4}\)
\(\Rightarrow x=\frac{7}{36}\)
GH
3
HQ
Hà Quang Minh
Giáo viên
16 tháng 9 2023
a)
\(\begin{array}{l}x + \left( { - \frac{1}{5}} \right) = \frac{{ - 4}}{{15}}\\x = \frac{{ - 4}}{{15}} + \frac{1}{5}\\x = \frac{{ - 4}}{{15}} + \frac{3}{{15}}\\x = \frac{{ - 1}}{{15}}\end{array}\)
Vậy \(x = \frac{{ - 1}}{{15}}\).
b)
\(\begin{array}{l}3,7 - x = \frac{7}{{10}}\\x = 3,7 - \frac{7}{{10}}\\x = \frac{{37}}{{10}} - \frac{7}{{10}}\\x=\frac{30}{10}\\x = 3\end{array}\)
Vậy \(x = 3\).
c)
\(\begin{array}{l}x.\frac{3}{2} = 2,4\\x.\frac{3}{2} = \frac{{12}}{5}\\x = \frac{{12}}{5}:\frac{3}{2}\\x = \frac{{12}}{5}.\frac{2}{3}\\x = \frac{8}{5}\end{array}\)
Vậy \(x = \frac{8}{5}\)
d)
\(\begin{array}{l}3,2:x = - \frac{6}{{11}}\\\frac{{16}}{5}:x = - \frac{6}{{11}}\\x = \frac{{16}}{5}:\left( { - \frac{6}{{11}}} \right)\\x = \frac{{16}}{5}.\frac{{ - 11}}{6}\\x = \frac{{ - 88}}{{15}}\end{array}\)
Vậy \(x = \frac{{ - 88}}{{15}}\).
13 tháng 11 2021
\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a) 4 , 5 − 2. x .1 4 7 = 11 14 ⇒ 4 , 5 − 2. x = 11 14 : 11 7 ⇒ 9 2 − 2. x = 1 2 ⇒ 2. x = 9 2 − 1 2 = 4 ⇒ x = 2.
b) x 4 − x 5 = − 1 1 2 ⇒ x . 1 4 − 1 5 = − 3 2 ⇒ x . 1 20 = − 3 2 ⇒ x = − 3 2 . 1 20 = − 3 40 .