tìm X biết
a) x-12= - 15
b) 123-5.(x+4)=38
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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
Giải:
a)-3/10-(-1/5)+x)=-3/2
-1/5+x =-3/10-(-3/2)
-1/5+x =6/5
x =6/5-(-1/5)
x =7/5
b)-(-x+3/4)-12/8.(-32/15)=-(-1/2)
x-3/4+16/5 =1/2
x-3/4 =1/2-16/5
x =-27/10
x =-27/10+3/4
x =-39/20
c)x-3/x+5=4/3
=>(x-3).3=4.(x+5)
3x-9 =4x+20
3x-4x =20+9
-1x =29
x =-29
Câu b cậu nên tính lại cho kĩ nhé, ấn máy tính dễ nhầm lắm đấy!
Mk phải ấn: -(39/20+3/4)-12/8.-32/15=1/2
Vì x là số âm mà đằng trước x là dấu ''-'' nên -(-39/20)=39/20 ; -(-1/2)=1/2
Chúc bạn học tốt!
b: \(\Leftrightarrow3x-36-5x-175+x-15=0\)
\(\Leftrightarrow-x-226=0\)
hay x=-226
\(a,\Rightarrow x=30-18=12\\ b,\Rightarrow x+6=45:5=9\\ \Rightarrow x=9-6=3\\ c,\Rightarrow38-3x=4^2=16\\ \Rightarrow3x=38-16=22\\ \Rightarrow x=\dfrac{22}{3}\)
a) 123 - 5 (x + 4) = 38
5 (x + 4) = 123 - 38
5 (x + 4) = 85
x + 4 = 85 : 5
x + 4 = 17
x = 17 - 4
x = 13
b) (3x - 24) . 73 = 2 . 74
3x - 24 = 2 . 74 : 73
3x - 24 = 2 . 7
3x - 16 = 14
3x = 14 + 16
3x = 30
x = 30 : 3
x = 10
c) (x : 3 - 4) . 5 = 15
x : 3 - 4 = 15 : 5
x : 3 - 4 = 3
x : 3 = 3 + 4
x : 3 = 7
x = 7 . 3
x = 21
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, \(\dfrac{3}{5}x-\dfrac{2}{3}=\dfrac{4}{5}:\dfrac{1}{5}=4\Leftrightarrow\dfrac{3}{5}x=4+\dfrac{2}{3}=\dfrac{14}{3}\Leftrightarrow x=\dfrac{14}{3}:\dfrac{3}{5}=\dfrac{70}{9}\)
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....
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: \(x=\dfrac{37}{70}\)
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 \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
\(a,\dfrac{12}{5}=\dfrac{x}{1,5}\Rightarrow x=\dfrac{12\cdot1,5}{5}=3,6\\ b,\dfrac{x}{5}=\dfrac{3}{20}\Rightarrow x=\dfrac{5\cdot3}{20}=\dfrac{3}{4}\\ c,\dfrac{4}{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{4\cdot9}{10}=\dfrac{18}{5}\\ d,\Rightarrow\dfrac{x}{15}=\dfrac{60}{x}\Rightarrow x^2=60\cdot15=900\Rightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\\ 2,\)
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 \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=k\Rightarrow x=3k;y=5k;z=6k\)
\(x^2-4y^2+2z^2=-475\\ \Rightarrow9k^2-100k^2+72z^2=-475\\ \Rightarrow-19k^2=-475\\ \Rightarrow k^2=25\Rightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=15;y=25;z=30\\x=-15;y=-25;z=-30\end{matrix}\right.\)
a) x-12=-15
x=-15+12
x=-3
b) 123-5(x+4)=38
5(x+4)=123-38
5(x+4)=85
x+4=85:5=17
x=17-4
x=13
\(a.x-12=-15\) \(b.123-5.\left(x+4\right)=38\)
\(x=\left(-15\right)+12\) \(5.\left(x+4\right)=123-38\)
\(x=-3\) \(5.\left(x+4\right)=85\)
\(x+4=85:5=17\)
\(x=17-4=13\)