Tìm x biết
a, 5 ( x + 35 ) = 515
b, 12x - 33 = 32 . 33
c, 6.x - 5 = 9
d, 4.( x - 12 ) + 9 = 17
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13 tháng 8 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
`5(x + 35) = 515`
`=> x + 35 = 515 \div 5`
`=> x+ 35 = 103`
`=> x = 103 - 35`
`=> x = 68`
Vậy, `x = 68`
`b)`
`12x - 33 = 3^2 * 3^3`
`=> 12x - 33 = 3^5`
`=> 12x = 3^5 + 33`
`=> 12x = 276`
`=> x = 276 \div 12`
`=> x = 23`
Vậy, `x = 23`
`c)`
`6x - 5 = 19`
`=> 6x = 19 + 5`
`=> 6x = 24`
`=> x = 24 \div 6`
`=> x = 4`
Vậy, `x = 4`
`d)`
`4(x - 12) + 9 = 17`
`=> 4(x - 12) = 17 - 9`
`=> 4(x - 12) = 8`
`=> x - 12 = 8 \div 4`
`=> x - 12 = 2`
`=> x = 14`
Vậy, `x = 14`
`e)`
123 - 5(x + 4) = 38`
`=> 5(x + 4)= 123 - 38`
`=> 5(x + 4) =85`
`=> x + 4 = 85 \div 5`
`=> x + 4 = 17`
`=> x = 13`
Vậy, `x = 13`
`f)`
`(3x - 2^4) * 7^3 = 2 * 7^4`
`=> 3x - 2^4 = 2 * 7^4 \div 7^3`
`=> 3x - 2^4 = 2* 7`
`=> 3x - 2^4 = 14`
`=> 3x = 14 + 2^4`
`=> 3x = 30`
`=> x = 30 \div 3`
`=> x = 10`
Vậy, `x = 10.`
21 tháng 1 2022
b: =>x(8-7)=-33
=>x=-33
c: =>-12x+60+21-7x=5
=>-19x=-76
hay x=4
d: =>-2x-2-x+5+2x=0
=>3-x=0
hay x=3
12 tháng 7 2021
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
16 tháng 10 2021
\(a,\dfrac{12}{5}=\dfrac{x}{1,5}\Rightarrow x=\dfrac{12\cdot1,5}{5}=3,6\\ b,\dfrac{x}{5}=\dfrac{3}{20}\Rightarrow x=\dfrac{5\cdot3}{20}=\dfrac{3}{4}\\ c,\dfrac{4}{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{4\cdot9}{10}=\dfrac{18}{5}\\ d,\Rightarrow\dfrac{x}{15}=\dfrac{60}{x}\Rightarrow x^2=60\cdot15=900\Rightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\\ 2,\)
a, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x+y-z}{3+5-6}=\dfrac{8}{2}=4\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=20\\z=24\end{matrix}\right.\)
b, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x-y+z}{3-5+6}=\dfrac{-4}{4}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-6\end{matrix}\right.\)
c, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{2y}{10}=\dfrac{3z}{18}=\dfrac{x-2y+3z}{3-10+18}=\dfrac{-33}{11}=-3\\ \Rightarrow\left\{{}\begin{matrix}x=-9\\y=-15\\z=-18\end{matrix}\right.\)
d, Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=k\Rightarrow x=3k;y=5k;z=6k\)
\(x^2-4y^2+2z^2=-475\\ \Rightarrow9k^2-100k^2+72z^2=-475\\ \Rightarrow-19k^2=-475\\ \Rightarrow k^2=25\Rightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=15;y=25;z=30\\x=-15;y=-25;z=-30\end{matrix}\right.\)
TM
6 tháng 3 2021
a/ \(x-\dfrac{3}{7}=\dfrac{2}{5}\cdot\dfrac{1}{4}\)
\(x-\dfrac{3}{7}=\dfrac{1}{10}\)
\(x=\dfrac{1}{10}+\dfrac{3}{7}=\dfrac{37}{70}\)
Vậy....
b/ \(x+\dfrac{4}{5}=-\dfrac{5}{12}\cdot\dfrac{3}{25}\)
\(x+\dfrac{4}{5}=-\dfrac{1}{20}\)
\(x=-\dfrac{1}{20}-\dfrac{4}{5}=-\dfrac{17}{20}\)
Vậy....
c/ \(\dfrac{x}{182}=-\dfrac{6}{12}\cdot\dfrac{35}{91}\)
\(\dfrac{x}{182}=-\dfrac{5}{26}\)
\(=>x\cdot26=-5\cdot182\)
\(26x=-910\)
\(x=-910:26=-35\)
Vậy....
6 tháng 3 2021
a) Ta có: \(x-\dfrac{3}{7}=\dfrac{2}{5}\cdot\dfrac{1}{4}\)
\(\Leftrightarrow x-\dfrac{3}{7}=\dfrac{1}{10}\)
\(\Leftrightarrow x=\dfrac{1}{10}+\dfrac{3}{7}=\dfrac{7}{70}+\dfrac{30}{70}\)
hay \(x=\dfrac{37}{70}\)
Vậy: \(x=\dfrac{37}{70}\)
NT
1
6 tháng 8 2016
3/2+5/4+9/8/+17/16+33/32-6+x-1/x+1=31/32-2/2015
=(1+1/2)+(1+1/4)+(1+1/8)+(1+1/16)+(1+1/32-6+x-1/x+1=31/32-2/2015
=(1/2+1/4+1/8+1/16+1/32)+(1+1+1+1+1)-6+x-1/x+1=31/32-2/2015
=31/32+5-6+x-1/x+1=31/32-2/2015
=5-6+x-1/x+1=31/32-2/2015-31/32
=-1+x-1/x+1=-2/2015
=x-1/x+1=-2/2015- -1
=x-1/x+1=2013/2015
=>x=2014
21 tháng 1
a: Sửa đề: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{2}{-z}=\dfrac{-t}{-9}\)
=>\(\dfrac{x}{5}=\dfrac{y}{-3}=\dfrac{-2}{z}=\dfrac{t}{9}=-2\)
=>\(x=-2\cdot5=-10;y=-2\cdot\left(-3\right)=6;z=\dfrac{-2}{-2}=1;t=9\cdot\left(-2\right)=-18\)
b: \(\dfrac{-24}{-6}=\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{z^3}{-2}\)
=>\(\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{z^3}{-2}=4\)
=>\(\left\{{}\begin{matrix}x=4\cdot3=12\\y^2=\dfrac{4}{4}=1\\z^3=-2\cdot4=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=12\\y\in\left\{1;-1\right\}\\z=-2\end{matrix}\right.\)
23 tháng 3 2022
a, \(x=-\dfrac{5}{6}-\dfrac{7}{12}=\dfrac{-10-7}{12}=-\dfrac{17}{12}\)
b, \(\dfrac{2}{9}-x=-\dfrac{4}{3}.\dfrac{5}{6}=-\dfrac{24}{18}=-\dfrac{4}{3}\Leftrightarrow x=\dfrac{2}{9}+\dfrac{4}{3}=\dfrac{14}{9}\)
c, \(-3=x-1\Leftrightarrow x=-2\)
d, \(\dfrac{3}{5}x-\dfrac{2}{3}=\dfrac{4}{5}:\dfrac{1}{5}=4\Leftrightarrow\dfrac{3}{5}x=4+\dfrac{2}{3}=\dfrac{14}{3}\Leftrightarrow x=\dfrac{14}{3}:\dfrac{3}{5}=\dfrac{70}{9}\)
12 tháng 10 2021
Câu 2:
\(\dfrac{7}{9}\cdot\dfrac{3}{35}=\dfrac{1}{5}\cdot\dfrac{1}{3}=\dfrac{1}{15}\)
\(\dfrac{9}{22}\cdot55=\dfrac{9\cdot55}{22}=\dfrac{9\cdot5}{2}=\dfrac{45}{2}\)
đây là cách giải của mk:
a, 5(x + 35) = 515
=> x + 35 = 515 : 5
=> x + 35 = 103
=> x = 103 - 35
=> x = 68
b, 12x - 33 = 32 . 33
=> 12x - 33 = 32 + 3
=> 12x - 33 = 35
=> 12x - 33 = 243
=> 12x = 243 + 33
=> 12x = 276
=> x = 276 : 12
=> x = 23
c, 6 . x - 5 = 9
=> 6 . x = 9 + 5
=> 6 . x = 14
=> x = 14 : 6
=> x = \(\frac{2}{3}\)
d, 4(x - 12) + 9 = 17
=> 4(x - 12) = 17 - 9
=> 4(x - 12) = 8
=> 4 . x - 12 = 8
=> 4 . x = 8 + 12
=> 4 . x = 20
=> x = 20 : 4
=> x = 5
a, 5 ( x + 35 ) = 515
x + 35 = 515 : 5
x + 35 = 103
x = 103 - 35
x = 68
b, 12x - 33 = 32 . 33
12x - 33 = 35
12x - 33 = 243
12x = 243 + 33
12x = 276
x = 276 : 12
x = 23
c, 6.x - 5 = 9
6x = 9 + 5
6x = 14
x = 14 : 6
x = \(\frac{7}{3}\)
d, 4.( x - 12 ) + 9 = 17
4 . ( x - 12 ) = 17 - 9
4 . ( x - 12 ) = 8
x - 12 = 8 : 4
x - 12 = 2
x = 2 + 12
x = 14