Tìm x
\(\dfrac{x-291}{1700}\)+\(\dfrac{x-293}{1698}\)+\(\dfrac{x-295}{1696}\)+\(\dfrac{x-297}{1694}\)=4
(X-291):1700+(X-293):1698+(x-295):1696+(x-297):1694=4
Giải phương trình:
a) \(\dfrac{x-3}{113}+\dfrac{x-5}{115}=\dfrac{x-7}{117}+\dfrac{x-9}{119}\)
b) \(\dfrac{x-291}{1700}+\dfrac{x-293}{1698}+\dfrac{x-295}{1696}+\dfrac{x-297}{1694}=4\)
Giúp mình với!!! Mình cần gấp!!!
https://giaibaitapvenha.blogspot.com/2017/12/toan-lop-8-ai-so-giai-phuong-trinh.html
Nếu lời giải giúp ích được cho bạn, truy cập
https://giaibaitapvenha.blogspot.com/2017/12/en-voi-do-homework-for-you-e-trai.html
a,\(\dfrac{x-3}{113}+\dfrac{x-5}{115}=\dfrac{x-7}{117}+\dfrac{x-9}{119}\)
=>\(\dfrac{x-3}{113}+1+\dfrac{x-5}{115}+1=\dfrac{x-7}{117}+1+\dfrac{x-9}{119}+1\)
=>\(\dfrac{x+110}{113}+\dfrac{x+110}{115}=\dfrac{x+110}{117}+\dfrac{x+110}{119}\)
=>(x+110)(\(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\))=0
vì 1/113+1/115-1/117-1/119 khác 0
=>x+110=0=>x=-110
\(\dfrac{x-291}{1700}+\dfrac{x-293}{1698}+\dfrac{x-295}{1696}+\dfrac{x-297}{1694}=4\)
=>\(\dfrac{x-291}{1700}-1+\dfrac{x-293}{1698}-1+\dfrac{x-295}{1696}-1+\dfrac{x-297}{1694}-1+4=4\)=>\(\dfrac{x-1991}{1700}+\dfrac{x-1991}{1698}+\dfrac{x-1991}{1696}+\dfrac{x-1991}{1694}=0\)
=>(x-1991)(1/1700+1/1698+1/1696+1/1694)=0
vì 1/1700+1/1698+1/1696+1/1694 khác 0
=>x-1991=0=>x=1991
a) (x-1)* (x-3) *(x+5)*(x +7) - 297 = 0
b) x4 - 2x2 + 144x - 1295 = 0
c) \(\dfrac{1927-x}{91}+\dfrac{1925-x}{93}+\dfrac{1923-x}{95}+\dfrac{1921-x}{97}+4=0\)
d) \(\dfrac{x-85}{15}+\dfrac{x-74}{13}+\dfrac{x-67}{11}+\dfrac{x-64}{9}=10\)
a) Đặt x -3 = a
<=> a(a+2)(a+8)(a+10) - 297=0
<=> \(\left[a\left(a+10\right)\right]\left[\left(a+2\right)\left(a+8\right)\right]\)-297=0
<=> \(\left(a^2+10a\right)\left(a^2+10a+16\right)-297=0\)
Đặt \(a^2+10a=b\)
\(b^2+16b-297=0\)
\(\Rightarrow\left[{}\begin{matrix}b=11\\b=-27\end{matrix}\right.\)\(b=11\Rightarrow\left[{}\begin{matrix}a=1\\a=-11\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)
b= -27 \(\Rightarrow a=\varnothing\Rightarrow x=\varnothing\)
b) bấm máy ra nhân tử chung :D
c)
\(\Leftrightarrow\left(\frac{1927-X}{91}+1\right)+\left(\frac{1925-x}{93}+1\right)+...=0\)
\(\Leftrightarrow\frac{2018-x}{91}+\frac{2018-x}{93}+\frac{2018-x}{95}+\frac{2018-x}{97}=0\)
\(\Leftrightarrow\left(2018-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)
<=> x = 2018
d) \(\Leftrightarrow\left(\frac{x-85}{15}-1\right)+\left(\frac{x-74}{13}-2\right)+\left(\frac{x-67}{11}-3\right)+\left(\frac{x-64}{9}-3\right)=0\)
giống câu c
Bài 4. Tìm x biết:
a. \(\dfrac{x}{5}=\dfrac{2}{5},\dfrac{3}{8}=\dfrac{6}{x},\dfrac{1}{9}=\dfrac{x}{27}\)
b. \(\dfrac{4}{x}=\dfrac{8}{6},\dfrac{3}{x-5}=\dfrac{-4}{x+2},\dfrac{x}{-2}=\dfrac{-8}{x}\)
a,2/5 = 2/5 ; 3/8=6/16 ; 1/9=3/27
b, 4/3=8/6 ; -1=-1 ; -4/-2=-8/4
tick cho mik nhé
a) x= 2, x= 8.(6 : 3) = 16, x= 1. (27 : 9)= 3
b) x= 6 : (8 : 4) = 3, x= -1, x= -2 . -8 = x.x => 16 = x2 => 42 = x2 => x=4
Tick cho mình đi
tìm x e Q
a) \(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)
b) \(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)
c) \(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)
\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)
\(\left(x+1\right)\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)
\(\)vì \(\dfrac{1}{3}>\dfrac{1}{6};\dfrac{1}{4}>\dfrac{1}{6};\dfrac{1}{5}>\dfrac{1}{6}=>\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}>0\)
\(=>x+1=0\)
\(=>x=-1\)
b,
\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)
\(\left(\dfrac{x+1}{2020}+1\right)+\left(\dfrac{x+2}{2019}+1\right)=\left(\dfrac{x+3}{2018}+1\right)+\left(\dfrac{x+4}{2017}+1\right)\)
\(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}=\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}\)
\(=>\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}-\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}=0\)
\(=>\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}\right)=0\)
Vì \(\dfrac{1}{2020}< \dfrac{1}{2018};\dfrac{1}{2019}< \dfrac{1}{2017}=>\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}< 0\)
\(=>x+2021=0\)
\(=>x=-2021\)
c,
\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
\(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)
\(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)
\(=>\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)
Vì \(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}>0\)
\(=>x+329=0\)
\(=>x=-329\)
\(\dfrac{x}{3}-\dfrac{1}{4}=\dfrac{-5}{6}\)
\(\dfrac{2x}{3}=\dfrac{6}{x}\)
\(\dfrac{x-1}{14}=\dfrac{4}{x}\)
\(\dfrac{-x}{8}=\dfrac{-50}{x}\)
Tìm x
Lời giải:
$\frac{x}{3}-\frac{1}{4}=\frac{-5}{6}$
$\frac{x}{3}=\frac{1}{4}-\frac{5}{6}=\frac{-7}{12}$
$x=\frac{-7}{12}.3=\frac{-7}{4}$
-------------------
$\frac{2x}{3}=\frac{6}{x}$ ($x\neq 0$)
$\Rightarrow 2x^2=18$
$x^2=18:2=9=(-3)^2=3^2$
$\Rightarrow x=\pm 3$
---------------------
$\frac{x-1}{14}=\frac{4}{x}$ ($x\neq 0$)
$\Rightarrow x(x-1)=14.4=56$
$x^2-x-56=0$
$(x+7)(x-8)=0$
$\Rightarrow x+7=0$ hoặc $x-8=0$
$\Leftrightarrow x=-7$ hoặc $x=8$
---------------------------
$\frac{-x}{8}=\frac{-50}{x}$ ($x\neq 0$)
$\Rightarrow -x^2=8(-50)$
$x^2=400=20^2=(-20)^2$
$\Rightarrow x=\pm 20$
Tìm x, biết
a)\(\dfrac{1}{2}\)x\(x\)-\(\dfrac{7}{3}\)=\(\dfrac{-5}{6}\)+\(\dfrac{3}{4}\)x\(x\)
b)\(\dfrac{4}{5}\)x\(x\)-\(\dfrac{6}{5}\)=\(\dfrac{1}{2}\)+\(\dfrac{3}{2}\)x\(x\)
c)\(\dfrac{2}{5}\)x(3x\(x\)+\(\dfrac{3}{4}\))=\(1\dfrac{1}{5}\)-\(\dfrac{1}{3}\)x\(x\)
d)2x(3x\(x \)+\(\dfrac{3}{4}\))+\(\dfrac{4}{5}\)=\(\dfrac{1}{2}\)-2x\(x\)
giúp mình giải bài toán trên với. Mình cảm ơn rất nhiều
a: =>1/2x-3/4x=-5/6+7/3
=>-1/4x=14/6-5/6=3/2
=>x=-3/2*4=-6
b: =>4/5x-3/2x=1/2+6/5
=>-7/10x=17/10
=>x=-17/7
c: =>6/5x+6/20=6/5-1/3x
=>6/5x+1/3x=6/5-3/10=12/10-3/10=9/10
=>x=27/46
d: =>6x+3/2+4/5=1/2-2x
=>8x=1/2-3/2-4/5=-1-4/5=-9/5
=>x=-9/40
tìm x
\(x-\dfrac{1}{2}=\dfrac{3}{4}\)
\(x+\dfrac{7}{8}=\dfrac{3}{4}\)
\(\dfrac{1}{2}.x-\dfrac{1}{4}=\dfrac{-1}{2}\)
\(\dfrac{1}{2}-\dfrac{5}{6}:x=\dfrac{2}{3}\)
\(x-\dfrac{1}{2}=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}+\dfrac{1}{2}\)
\(x=\dfrac{5}{4}\)
\(x+\dfrac{7}{8}=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}-\dfrac{7}{8}\)
\(x=\dfrac{-1}{8}\)
\(\dfrac{1}{2}\cdot x-\dfrac{1}{4}=\dfrac{-1}{2}\)
\(\dfrac{1}{2}\cdot x=\dfrac{-1}{2}+\dfrac{1}{4}\)
\(\dfrac{1}{2}\cdot x=\dfrac{-1}{4}\)
\(x=\dfrac{-1}{4}\div\dfrac{1}{2}\)
\(x=\dfrac{-1}{2}\)
Câu D ko bt
Tìm x, biết:
a) \(\dfrac{2}{5}\) + \(\dfrac{3}{4}\): x = \(\dfrac{-1}{2}\)
b) \(\dfrac{5}{7}\) - \(\dfrac{2}{3}\) . x = \(\dfrac{4}{5}\)
c) \(\dfrac{1}{2}\) x + \(\dfrac{2}{3}\) x = \(\dfrac{-2}{3}\)
d) \(\dfrac{4}{7}\)x - x= \(\dfrac{-9}{14}\)
a, 2/5 + 3/4 : x = -1/2
3/4 : x = -1/2 - 2/5
3/4 : x = -9/10
x = 3/4 : -9/10
x = -5/6
b, 5/7 - 2/3 . x = 4/5
2/3 . x = 4/5 + 5/7
2/3 . x = 53/35
x = 53/35 : 2/3
x = 159/70
c và d mình làm dược nhưng ko ghi được cái suy ra